W12-W13. Limits and Continuity

1. Summary

This chapter extends the concept of limits from sequences to functions and provides a precise definition of continuous functions. We explore various types of limits, theorems about limits, and applications to understanding function behavior.

1.1 Limit of a Function

When studying a function \(y = f(x)\), we are often interested in its behavior near a particular point \(c\), even if the function is not defined at \(c\) itself. This situation commonly arises when evaluating a function at \(c\) leads to division by zero.

Example: Consider the function \(f(x) = \frac{x^2 - 1}{x - 1}\). This function is undefined at \(x = 1\) because the denominator becomes zero. However, for any \(x \neq 1\), we can simplify:

\[f(x) = \frac{x^2 - 1}{x - 1} = \frac{(x-1)(x+1)}{x-1} = x + 1, \quad \forall x \neq 1\]

The graph of \(f\) is the line \(y = x + 1\) with the point \((1, 2)\) removed. Even though \(f(1)\) is not defined, we can make \(f(x)\) as close as we want to 2 by choosing \(x\) close enough to 1.

1.1.1 Epsilon-Delta Definition of Limit (Cauchy’s Definition)

The epsilon-delta definition provides a rigorous mathematical formalization of the limit concept.

Let \(f(x)\) be a function defined on an open interval about \(c\), except possibly at \(c\) itself. We say that the limit of \(f(x)\) is \(L \in \mathbb{R}\) as \(x\) approaches \(c\), written

\[\lim_{x \to c} f(x) = L\]

if for every \(\varepsilon > 0\), there exists \(\delta = \delta(\varepsilon) > 0\) (depending on \(\varepsilon\)) such that

\[|f(x) - L| < \varepsilon \quad \text{whenever} \quad 0 < |x - c| < \delta\]

This means that when \(x\) is close enough to \(c\) (within distance \(\delta\)), the function value \(f(x)\) is close enough to \(L\) (within distance \(\varepsilon\)).

Important Note: The point \(c\) in the definition must be a limit point of the domain of \(f\). A point \(c\) is a limit point of a set \(A\) if each neighborhood (open interval) of \(c\) contains at least one point of \(A\) different from \(c\).

Example of limit points: Let \(A = (1, 2) \cup \{3\}\). The set of limit points of \(A\) is \([1, 2]\). Note that 3 is not a limit point because we can find a neighborhood of 3 that contains no other points of \(A\).

1.1.2 How to Find Delta for a Given Epsilon

Finding the appropriate \(\delta > 0\) for a given \(\varepsilon > 0\) can be accomplished in two steps:

1. Solve the inequality: Solve \(|f(x) - L| < \varepsilon\) to find an open interval \((a, b)\) containing \(c\) on which the inequality holds for all \(x \neq c\).
2. Find delta: Find a value of \(\delta > 0\) that places the open \(\delta\)-interval \((c - \delta, c + \delta)\) centered at \(c\) inside the interval \((a, b)\).

Example 1: Show that \(\lim_{x \to 1} (5x - 3) = 2\).

Set \(f(x) = 5x - 3\), \(c = 1\), and \(L = 2\). For any given \(\varepsilon > 0\), we need to find \(\delta > 0\) such that if \(x \neq 1\) and \(0 < |x - 1| < \delta\), then \(|f(x) - 2| < \varepsilon\).

We have: \[|f(x) - 2| < \varepsilon \implies |(5x - 3) - 2| = |5x - 5| < \varepsilon \implies 5|x - 1| < \varepsilon \implies |x - 1| < \frac{\varepsilon}{5}\]

Thus, we can take \(\delta = \frac{\varepsilon}{5}\). Then if \(0 < |x - 1| < \delta = \frac{\varepsilon}{5}\): \[|(5x - 3) - 2| = |5x - 5| = 5|x - 1| < 5 \cdot \frac{\varepsilon}{5} = \varepsilon\]

This proves that \(\lim_{x \to 1} (5x - 3) = 2\). Note that \(\delta = \frac{\varepsilon}{5}\) is not the only value that works—any smaller positive \(\delta\) will also work.

Example 2: Prove basic limits:

1. \(\lim_{x \to c} x = c\): Let \(\varepsilon > 0\) be given. We need \(|x - c| < \varepsilon\) whenever \(0 < |x - c| < \delta\). If we take \(\delta = \varepsilon\), then \(|x - c| < \varepsilon\) immediately holds.
2. \(\lim_{x \to c} k = k\) (where \(k\) is a constant): Let \(\varepsilon > 0\) be given. We need \(|k - k| < \varepsilon\) whenever \(0 < |x - c| < \delta\). Since \(k - k = 0\), we always have \(|k - k| < \varepsilon\), so any positive \(\delta\) works.

Example 3: Prove that \(\lim_{x \to 5} \sqrt{x - 1} = 2\).

For any given \(\varepsilon > 0\), we need to find \(\delta > 0\) such that: \[|\sqrt{x - 1} - 2| < \varepsilon \quad \text{whenever} \quad 0 < |x - 5| < \delta\]

For all \(x \neq 5\): \[|\sqrt{x - 1} - 2| < \varepsilon \implies -\varepsilon < \sqrt{x - 1} - 2 < \varepsilon \implies 2 - \varepsilon < \sqrt{x - 1} < 2 + \varepsilon\] \[\implies (2 - \varepsilon)^2 < x - 1 < (2 + \varepsilon)^2 \implies 1 + (2 - \varepsilon)^2 < x < 1 + (2 + \varepsilon)^2\]

We want to place the interval \((5 - \delta, 5 + \delta)\) inside \((1 + (2 - \varepsilon)^2, 1 + (2 + \varepsilon)^2)\).

For \(\varepsilon < 4\), take \(\delta = \min\{4 - (2 - \varepsilon)^2, -4 + (2 + \varepsilon)^2\} = \min\{4\varepsilon - \varepsilon^2, 4\varepsilon + \varepsilon^2\}\). Then \(0 < |x - 5| < \delta\) will place \(x\) in the required interval.

For \(\varepsilon \geq 4\), take \(\delta = \min\{5, -4 + (2 + \varepsilon)^2\}\).

Example 4: Let \(f(x) = \begin{cases} x^2, & x \neq 2 \\ 1, & x = 2 \end{cases}\). Prove that \(\lim_{x \to 2} f(x) = 4\).

Note that even though \(f(2) = 1 \neq 4\), the limit can still be 4 because the limit only cares about values near \(c = 2\), not at \(c = 2\) itself.

For \(x \neq 2\), we need to solve \(|x^2 - 4| < \varepsilon\): \[|x^2 - 4| < \varepsilon \implies 4 - \varepsilon < x^2 < 4 + \varepsilon\]

For \(\varepsilon < 4\): \(\sqrt{4 - \varepsilon} < x < \sqrt{4 + \varepsilon}\).

Take \(\delta = \min\{2 - \sqrt{4 - \varepsilon}, \sqrt{4 + \varepsilon} - 2\}\). Then \(0 < |x - 2| < \delta\) will place \(x\) in the required interval.

For \(\varepsilon \geq 4\): Take \(\delta = \min\{2, \sqrt{4 + \varepsilon} - 2\}\).

1.1.3 Heine’s Definition of Limit

An alternative definition of limit uses sequences.

Heine’s definition: Let \(f: A \to \mathbb{R}\), and let \(c\) be a limit point of \(A\). We say that \(f\) has a limit \(L\) at \(c\) if, for any sequence \(\{x_n\}\) with \(x_n \to c\) as \(n \to \infty\) and \(x_n \neq c\) for all \(n \in \mathbb{N}\), we have \(f(x_n) \to L\) as \(n \to \infty\).

Theorem (Equivalence of Definitions): The Cauchy definition and the Heine definition are equivalent. That is: \[\lim_{x \to c} f(x) = L \iff \lim_{n \to \infty} f(x_n) = L\] for every sequence \(\{x_n\} \subset A\) such that \(x_n \neq c\) for all \(n \in \mathbb{N}\) and \(x_n \to c\) as \(n \to \infty\).

Theorem (Uniqueness of the Limit): If \(f\) has a limit at \(c\), then this limit is unique.

Proof: Suppose \(f\) has two different limits \(L_1 \neq L_2\). Let \(\{x_n\} \subset A \setminus \{c\}\) be a sequence that converges to \(c\). Then the sequence \(\{f(x_n)\}\) converges to two different limits \(L_1\) and \(L_2\). This contradicts the uniqueness of the limit of a sequence. Therefore, the limit must be unique.

1.2 Limit Theorems

1.2.1 Limit Laws

If \(L\), \(M\), \(c\), and \(k\) are real numbers with \(\lim_{x \to c} f(x) = L\) and \(\lim_{x \to c} g(x) = M\), then:

1. Sum Rule: \(\lim_{x \to c} (f(x) + g(x)) = L + M\)
2. Difference Rule: \(\lim_{x \to c} (f(x) - g(x)) = L - M\)
3. Constant Multiple Rule: \(\lim_{x \to c} (kf(x)) = kL\)
4. Product Rule: \(\lim_{x \to c} (f(x) \cdot g(x)) = L \cdot M\)
5. Quotient Rule: \(\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M}\), provided \(M \neq 0\)
6. Power Rule: \(\lim_{x \to c} [f(x)]^n = L^n\), \(n\) a positive integer
7. Root Rule: \(\lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{L}\), \(n\) a positive integer (if \(n\) is even, assume \(f(x) \geq 0\))

Proof of Sum Rule (using Heine’s definition): Let \(\{x_n\}\) be a sequence in \(A\) that converges to \(c\) with \(x_n \neq c\) for every \(n\). Then \(\lim_{n \to \infty} f(x_n) = L\) and \(\lim_{n \to \infty} g(x_n) = M\). By the limit laws for sequences: \[\lim_{n \to \infty} (f(x_n) + g(x_n)) = \lim_{n \to \infty} (f + g)(x_n) = L + M\] This means \(\lim_{x \to c} (f(x) + g(x)) = L + M\).

Proof of Sum Rule (using epsilon-delta definition): Let \(\varepsilon > 0\) be given. Since \(\lim_{x \to c} f(x) = L\), there exists \(\delta_1 > 0\) such that: \[|f(x) - L| < \frac{\varepsilon}{2} \quad \text{whenever} \quad 0 < |x - c| < \delta_1\]

Similarly, since \(\lim_{x \to c} g(x) = M\), there exists \(\delta_2 > 0\) such that: \[|g(x) - M| < \frac{\varepsilon}{2} \quad \text{whenever} \quad 0 < |x - c| < \delta_2\]

Let \(\delta = \min\{\delta_1, \delta_2\}\). If \(0 < |x - c| < \delta\), then both inequalities hold, and by the triangle inequality: \[|f(x) + g(x) - (L + M)| \leq |f(x) - L| + |g(x) - M| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon\]

This shows that \(\lim_{x \to c} (f(x) + g(x)) = L + M\). The other limit laws can be proved similarly.

1.2.2 Limits of Polynomials and Rational Functions

Theorem:

1. Limits of Polynomials: If \(P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\), then: \[\lim_{x \to c} P(x) = P(c) = a_n c^n + a_{n-1} c^{n-1} + \cdots + a_1 c + a_0\]
2. Limits of Rational Functions: If \(P(x)\) and \(Q(x)\) are polynomials and \(Q(c) \neq 0\), then: \[\lim_{x \to c} \frac{P(x)}{Q(x)} = \frac{P(c)}{Q(c)}\]

1.2.3 Indeterminate Forms

Indeterminate forms are expressions that appear when taking limits where direct substitution results in an ambiguous value, such as: \[\frac{0}{0}, \quad \frac{\infty}{\infty}, \quad 0 \cdot \infty, \quad \infty - \infty, \quad 1^{\infty}, \quad 0^0, \quad \infty^0\]

Example 1 (Eliminating Common Factors - \(\frac{0}{0}\) form): \[\lim_{x \to 1} \frac{x^2 + x - 2}{x^2 - x} = \lim_{x \to 1} \frac{(x-1)(x+2)}{x(x-1)} = \lim_{x \to 1} \frac{x+2}{x} = \frac{3}{1} = 3\]

Example 2 (Rationalization - \(\frac{0}{0}\) form): \[\lim_{x \to 0} \frac{\sqrt{x^2 + 100} - 10}{x} = \lim_{x \to 0} \frac{\sqrt{x^2 + 100} - 10}{x} \cdot \frac{\sqrt{x^2 + 100} + 10}{\sqrt{x^2 + 100} + 10}\] \[= \lim_{x \to 0} \frac{x^2 + 100 - 100}{x(\sqrt{x^2 + 100} + 10)} = \lim_{x \to 0} \frac{x^2}{x(\sqrt{x^2 + 100} + 10)} = \lim_{x \to 0} \frac{x}{\sqrt{x^2 + 100} + 10} = 0\]

1.2.4 The Sandwich Theorem

Theorem (Sandwich Theorem): Suppose that \(g(x) \leq f(x) \leq h(x)\) for all \(x\) in some open interval containing \(c\), except possibly at \(x = c\) itself. If \(\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L\), then \(\lim_{x \to c} f(x) = L\).

The function \(f\) is “sandwiched” between \(g\) and \(h\), and since both \(g\) and \(h\) approach the same limit \(L\), the function \(f\) must also approach \(L\).

Example 1: How does \(f(x) = x^2 \sin(1/x^2)\) behave near \(x = 0\)?

For any \(x \neq 0\): \[-1 \leq \sin(1/x^2) \leq 1 \implies -x^2 \leq x^2 \sin(1/x^2) \leq x^2\]

Since \(\lim_{x \to 0} (-x^2) = \lim_{x \to 0} x^2 = 0\), by the Sandwich Theorem: \[\lim_{x \to 0} x^2 \sin(1/x^2) = 0\]

Example 2: We can prove using the Sandwich Theorem that: \[\lim_{\theta \to 0} \sin \theta = 0, \quad \lim_{\theta \to 0} \cos \theta = 1\]

For all \(\theta \in \mathbb{R}\): \[-|\theta| \leq \sin \theta \leq |\theta|, \quad \text{and} \quad 0 \leq 1 - \cos \theta \leq |\theta|\]

Since \(\lim_{\theta \to 0} -|\theta| = \lim_{\theta \to 0} |\theta| = 0\), we get the desired limits by the Sandwich Theorem.

1.3 One-Sided Limits

One-sided limits extend the limit concept to situations where \(x\) approaches \(c\) from only one direction: from the left (\(x < c\)) or from the right (\(x > c\)).

Definition (One-Sided Limits): Let \(f: A \to \mathbb{R}\), and let \(c\) be a limit point of \(A\).

1. Right-hand limit: Assume \(A\) contains an interval \((c, d)\) to the right of \(c\). We say \(f(x)\) has right-hand limit \(L\) at \(c\), written: \[\lim_{x \to c^+} f(x) = L\] if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that: \[|f(x) - L| < \varepsilon \quad \text{whenever} \quad c < x < c + \delta\]
2. Left-hand limit: Assume \(A\) contains an interval \((b, c)\) to the left of \(c\). We say \(f\) has left-hand limit \(L\) at \(c\), written: \[\lim_{x \to c^-} f(x) = L\] if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that: \[|f(x) - L| < \varepsilon \quad \text{whenever} \quad c - \delta < x < c\]

Example: Show that \(\lim_{x \to 0^+} \sqrt{x} = 0\).

Let \(\varepsilon > 0\) be given. We want to find \(\delta > 0\) such that: \[|\sqrt{x} - 0| < \varepsilon \quad \text{whenever} \quad 0 < x < \delta\]

Squaring both sides: \(x < \varepsilon^2\) whenever \(0 < x < \delta\).

If we choose \(\delta = \varepsilon^2\), then: \[\sqrt{x} < \varepsilon \quad \text{whenever} \quad 0 < x < \delta = \varepsilon^2\]

This proves \(\lim_{x \to 0^+} \sqrt{x} = 0\).

Theorem (Relationship Between Two-Sided and One-Sided Limits): Suppose \(f\) is defined on an open interval containing \(c\), except perhaps at \(c\) itself. Then \(f(x)\) has a limit as \(x \to c\) if and only if it has both limits from the left and right at \(c\), and these limits are equal: \[\lim_{x \to c} f(x) = L \iff \lim_{x \to c^-} f(x) = L \quad \text{and} \quad \lim_{x \to c^+} f(x) = L\]

Example: Consider \(f(x) = \frac{x - c}{|x - c|}\), defined on \(\mathbb{R} \setminus \{c\}\).

• For \(x > c\): \(f(x) = \frac{x-c}{x-c} = 1\), so \(\lim_{x \to c^+} f(x) = 1\)
• For \(x < c\): \(f(x) = \frac{x-c}{-(x-c)} = -1\), so \(\lim_{x \to c^-} f(x) = -1\)

Since the one-sided limits are different, \(\lim_{x \to c} f(x)\) does not exist.

Example (finding parameter for limit existence): Consider: \[f(x) = \begin{cases} ax^2 + 1, & \text{if } x \leq 1 \\ -5x^3 + 3a, & \text{if } x > 1 \end{cases}\]

Find \(a \in \mathbb{R}\) such that \(\lim_{x \to 1} f(x)\) exists.

We have: \[\lim_{x \to 1^-} f(x) = a + 1, \quad \lim_{x \to 1^+} f(x) = 3a - 5\]

The limit exists if \(a + 1 = 3a - 5\), which gives \(2a = 6\), so \(a = 3\).

1.3.1 Important Limit: \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\)

Theorem: \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) (where \(\theta\) is in radians).

Proof: Consider a unit circle with angle \(\theta\) where \(0 < \theta < \frac{\pi}{2}\). From the geometry: \[\text{area}(\triangle OAP) < \text{area(sector } OAP) < \text{area}(\triangle OAT)\]

We can express these areas as: \[\text{area}(\triangle OAP) = \frac{1}{2}(1)(\sin \theta) = \frac{1}{2} \sin \theta\] \[\text{area(sector } OAP) = \frac{1}{2}r^2 \theta = \frac{1}{2}\theta \quad \text{(only for } \theta \text{ in radians)}\] \[\text{area}(\triangle OAT) = \frac{1}{2}(1)(\tan \theta) = \frac{1}{2} \tan \theta\]

Thus: \[\frac{1}{2} \sin \theta < \frac{1}{2} \theta < \frac{1}{2} \tan \theta\]

Since \(0 < \theta < \frac{\pi}{2}\), we have \(\sin \theta > 0\), so dividing by \(\frac{1}{2} \sin \theta\): \[1 < \frac{\theta}{\sin \theta} < \frac{1}{\cos \theta} \implies \cos \theta < \frac{\sin \theta}{\theta} < 1\]

Since \(\lim_{\theta \to 0^+} \cos \theta = 1\), by the Sandwich Theorem: \[\lim_{\theta \to 0^+} \frac{\sin \theta}{\theta} = 1\]

For the left-hand limit, note that \(\sin \theta\) and \(\theta\) are both odd functions, so \(f(\theta) = \frac{\sin \theta}{\theta}\) is an even function. By symmetry: \[\lim_{\theta \to 0^-} \frac{\sin \theta}{\theta} = \lim_{\theta \to 0^+} \frac{\sin \theta}{\theta} = 1\]

Therefore, \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\).

1.4 Limits at Infinity, Infinite Limits, and Asymptotes

1.4.1 Limits at Infinity

When studying function behavior as the magnitude of \(x\) becomes increasingly large, we use limits at infinity.

Definition:

1. We say \(f(x)\) has limit \(L\) as \(x\) approaches \(\infty\), written \(\lim_{x \to \infty} f(x) = L\), if for every \(\varepsilon > 0\), there exists \(M\) such that: \[|f(x) - L| < \varepsilon \quad \text{whenever} \quad x > M\]
2. We say \(f(x)\) has limit \(L\) as \(x\) approaches \(-\infty\), written \(\lim_{x \to -\infty} f(x) = L\), if for every \(\varepsilon > 0\), there exists \(N\) such that: \[|f(x) - L| < \varepsilon \quad \text{whenever} \quad x < N\]

Example 1: Let \(k \in \mathbb{R}\) be a constant. Then \(\lim_{x \to \pm\infty} k = k\).

For \(x \to \infty\): Let \(\varepsilon > 0\) be given. We need \(|k - k| < \varepsilon\) for all \(x > M\). Since \(|k - k| = 0 < \varepsilon\) always, any \(M\) works. Similarly for \(x \to -\infty\).

Example 2: Show that \(\lim_{x \to \pm\infty} \frac{1}{x} = 0\).

For \(x \to \infty\): Let \(\varepsilon > 0\) be given. We need: \[\left|\frac{1}{x} - 0\right| = \left|\frac{1}{x}\right| < \varepsilon \quad \text{whenever} \quad x > M\]

This holds if \(M = \frac{1}{\varepsilon}\) (or any larger number). Similarly, for \(x \to -\infty\), take \(N = -\frac{1}{\varepsilon}\).

Example 3: Show that \(\lim_{x \to -\infty} e^x = 0\).

Let \(\varepsilon > 0\) be given. We need \(|e^x - 0| < \varepsilon\) whenever \(x < N\). \[|e^x - 0| < \varepsilon \implies e^x < \varepsilon \implies x < \ln \varepsilon\]

Taking \(N = \ln \varepsilon\), we have \(|e^x - 0| < \varepsilon\) whenever \(x < \ln \varepsilon\).

Remark: All limit laws (sum, difference, product, quotient, etc.) apply when we replace \(\lim_{x \to c}\) by \(\lim_{x \to \infty}\) or \(\lim_{x \to -\infty}\). The Sandwich Theorem also holds for limits at infinity.

Examples:

1. \(\displaystyle \lim_{x \to \infty} \left(-1 + \frac{2}{x}\right) = -1 + 2 \cdot 0 = -1\)
2. \(\displaystyle \lim_{x \to \infty} \frac{3x^2 + 2x - 5}{-2x^2 + x - 1} = \lim_{x \to \infty} \frac{3 + \frac{2}{x} - \frac{5}{x^2}}{-2 + \frac{1}{x} - \frac{1}{x^2}} = \frac{3 + 0 - 0}{-2 + 0 - 0} = -\frac{3}{2}\)
3. \(\displaystyle \lim_{x \to -\infty} \frac{3x - 5}{2x^2 + x - 1} = \lim_{x \to -\infty} \frac{\frac{3}{x} - \frac{5}{x^2}}{2 + \frac{1}{x} - \frac{1}{x^2}} = \frac{0 - 0}{2 + 0 - 0} = 0\)
4. \(\displaystyle \lim_{x \to \infty} x \sin(1/x)\): Let \(t = 1/x\). As \(x \to \infty\), \(t \to 0^+\). Thus: \[\lim_{x \to \infty} x \sin(1/x) = \lim_{t \to 0^+} \frac{\sin t}{t} = 1\]
5. For \(L = \lim_{x \to \infty} (\sqrt{9x^2 - 13} - 3x)\) (indeterminate form \(\infty - \infty\)): \[L = \lim_{x \to \infty} \frac{(\sqrt{9x^2 - 13} - 3x)(\sqrt{9x^2 - 13} + 3x)}{\sqrt{9x^2 - 13} + 3x} = \lim_{x \to \infty} \frac{9x^2 - 13 - 9x^2}{\sqrt{9x^2 - 13} + 3x}\] \[= \lim_{x \to \infty} \frac{-13}{\sqrt{9x^2 - 13} + 3x} = \lim_{x \to \infty} \frac{-\frac{13}{x}}{\sqrt{9 - \frac{13}{x^2}} + 3} = \frac{0}{3 + 3} = 0\]

1.4.2 Infinite Limits

Definition:

1. We say \(f(x)\) approaches \(\infty\) as \(x\) approaches \(c\), written \(\lim_{x \to c} f(x) = \infty\), if for every positive real number \(B\) there exists \(\delta > 0\) such that: \[f(x) > B \quad \text{whenever} \quad 0 < |x - c| < \delta\]
2. We say \(f(x)\) approaches \(-\infty\) as \(x\) approaches \(c\), written \(\lim_{x \to c} f(x) = -\infty\), if for every negative real number \(-B\) (\(B > 0\)) there exists \(\delta > 0\) such that: \[f(x) < -B \quad \text{whenever} \quad 0 < |x - c| < \delta\]

Example 1: Prove that \(\lim_{x \to 0} \frac{1}{x^2} = \infty\).

Given \(B > 0\), we want \(\delta > 0\) such that: \[\frac{1}{x^2} > B \quad \text{whenever} \quad 0 < |x| < \delta\]

We have: \[\frac{1}{x^2} > B \iff x^2 < \frac{1}{B} \iff |x| < \frac{1}{\sqrt{B}}\]

Choosing \(\delta = \frac{1}{\sqrt{B}}\): if \(0 < |x| < \delta\), then \(\frac{1}{x^2} > \frac{1}{\delta^2} = B\).

More Examples:

• \(\displaystyle \lim_{x \to 2} \frac{(x-2)^2}{x^2-4} = \lim_{x \to 2} \frac{(x-2)^2}{(x-2)(x+2)} = \lim_{x \to 2} \frac{x-2}{x+2} = 0\)
• \(\displaystyle \lim_{x \to 2^+} \frac{x-3}{x^2-4} = \lim_{x \to 2^+} \frac{x-3}{(x-2)(x+2)} = -\infty\) (numerator approaches \(-1\), denominator approaches \(0^+\))
• \(\displaystyle \lim_{x \to 2^-} \frac{x-3}{x^2-4} = \lim_{x \to 2^-} \frac{x-3}{(x-2)(x+2)} = \infty\) (numerator approaches \(-1\), denominator approaches \(0^-\))
• \(\displaystyle \lim_{x \to 2} \frac{x-3}{x^2-4}\) does not exist (one-sided limits differ)
• \(\displaystyle \lim_{x \to \pm\infty} \frac{2x^5 - 6x^4 + 1}{3x^2 + x - 7} = \lim_{x \to \pm\infty} x^3 \frac{2 - \frac{6}{x} + \frac{1}{x^5}}{3 + \frac{1}{x} - \frac{7}{x^2}} = \pm\infty\)

1.4.3 Horizontal Asymptotes

If the distance between the graph of a function and a fixed line approaches zero as a point on the graph moves increasingly far from the origin, we say the line is an asymptote of the graph.

Definition (Horizontal Asymptote): A line \(y = b\) is a horizontal asymptote of the graph of \(y = f(x)\) if: \[\text{either } \lim_{x \to \infty} f(x) = b \quad \text{or } \lim_{x \to -\infty} f(x) = b\]

Example: Find the horizontal asymptotes of \(f(x) = \frac{x^3 - 2}{|x|^3 + 1}\).

For \(x > 0\): \[\lim_{x \to \infty} \frac{x^3 - 2}{|x|^3 + 1} = \lim_{x \to \infty} \frac{x^3 - 2}{x^3 + 1} = \lim_{x \to \infty} \frac{1 - \frac{2}{x^3}}{1 + \frac{1}{x^3}} = 1\]

For \(x < 0\): \[\lim_{x \to -\infty} \frac{x^3 - 2}{|x|^3 + 1} = \lim_{x \to -\infty} \frac{x^3 - 2}{(-x)^3 + 1} = \lim_{x \to -\infty} \frac{x^3 - 2}{-x^3 + 1} = \lim_{x \to -\infty} \frac{1 - \frac{2}{x^3}}{-1 + \frac{1}{x^3}} = -1\]

The horizontal asymptotes are \(y = -1\) and \(y = 1\).

1.4.4 Vertical Asymptotes

Definition (Vertical Asymptote): A line \(x = a\) is a vertical asymptote of the graph of \(y = f(x)\) if: \[\text{either } \lim_{x \to a^+} f(x) = \pm\infty \quad \text{or } \lim_{x \to a^-} f(x) = \pm\infty\]

Example 1: Find asymptotes of \(f(x) = \frac{x+3}{x+2}\).

Domain: \(D_f = \mathbb{R} \setminus \{-2\}\). We can write \(f(x) = 1 + \frac{1}{x+2}\).

• Horizontal: \(\lim_{x \to \pm\infty} f(x) = 1\), so \(y = 1\) is a horizontal asymptote
• Vertical: \(\lim_{x \to -2^+} f(x) = \infty\) and \(\lim_{x \to -2^-} f(x) = -\infty\), so \(x = -2\) is a vertical asymptote

Example 2: Find asymptotes of \(f(x) = -\frac{8}{x^2 - 4}\).

Domain: \(D_f = \mathbb{R} \setminus \{-2, 2\}\).

• Horizontal: \(\lim_{x \to \pm\infty} f(x) = 0\), so \(y = 0\) is a horizontal asymptote
• Vertical:
  • \(\lim_{x \to 2^+} f(x) = -\infty\), \(\lim_{x \to 2^-} f(x) = \infty\), so \(x = 2\) is a vertical asymptote
  • \(\lim_{x \to -2^+} f(x) = \infty\), \(\lim_{x \to -2^-} f(x) = -\infty\), so \(x = -2\) is a vertical asymptote

1.4.5 Oblique (Slant) Asymptotes

Definition (Oblique Asymptote): A function \(f(x)\) is asymptotic to the straight line \(y = \alpha x + \beta\) (\(\alpha \neq 0\)) if: \[\text{either } \lim_{x \to \infty} (f(x) - (\alpha x + \beta)) = 0 \quad \text{or } \lim_{x \to -\infty} (f(x) - (\alpha x + \beta)) = 0\]

Example: Find the oblique asymptote of \(f(x) = \frac{x^2 - 3}{2x - 4}\).

Using polynomial division: \[f(x) = \frac{x^2 - 3}{2x - 4} = \frac{1}{2}x + 1 + \frac{1}{2x - 4}\]

Therefore: \[\lim_{x \to \pm\infty} \left(f(x) - \left(\frac{1}{2}x + 1\right)\right) = \lim_{x \to \pm\infty} \frac{1}{2x - 4} = 0\]

The line \(y = \frac{1}{2}x + 1\) is an oblique asymptote.

1.5 Continuity

1.5.1 Continuity at a Point

The continuity is one of the most important properties of a function. Intuitively, a function is continuous if we can draw its graph without lifting the pen from paper.

Definition: Let \(c\) be either an interior point or an endpoint of an interval in the domain of \(f\).

1. The function \(f\) is continuous at \(c\) if: \[\lim_{x \to c} f(x) = f(c)\]
2. The function \(f\) is right-continuous at \(c\) if: \[\lim_{x \to c^+} f(x) = f(c)\]
3. The function \(f\) is left-continuous at \(c\) if: \[\lim_{x \to c^-} f(x) = f(c)\]

Note that for \(f\) to be continuous at \(c\), three conditions must hold:

• \(f(c)\) is defined
• \(\lim_{x \to c} f(x)\) exists
• \(\lim_{x \to c} f(x) = f(c)\)

A function is continuous over a closed interval \([a, b]\) if it is right-continuous at \(a\), left-continuous at \(b\), and continuous at all interior points.

If a function is not continuous at point \(c\) in its domain, we say \(f\) is discontinuous at \(c\).

1.5.2 Continuous Functions

A continuous function is one that is continuous at every point in its domain. If a function is discontinuous at one or more points, it is a discontinuous function.

Theorem (Properties of Continuous Functions): If functions \(f\) and \(g\) are continuous at \(x = c\), then the following are also continuous at \(x = c\):

1. Sums: \(f + g\)
2. Differences: \(f - g\)
3. Constant multiples: \(\alpha f\) for any \(\alpha \in \mathbb{R}\)
4. Products: \(fg\)
5. Quotients: \(f/g\), provided \(g(c) \neq 0\)
6. Powers: \(f^n\), \(n\) a positive integer
7. Roots: \(\sqrt[n]{f}\), provided it is defined on an interval containing \(c\)

Examples of continuous functions:

1. Identity and constant functions: \(f(x) = x\) and \(f(x) = k\) are continuous everywhere.
2. Reciprocal function: \(f(x) = \frac{1}{x}\) is continuous on its domain \(\mathbb{R} \setminus \{0\}\).
3. Polynomials: Every polynomial \(P(x) = a_n x^n + \cdots + a_1 x + a_0\) is continuous everywhere because \(\lim_{x \to c} P(x) = P(c)\).
4. Rational functions: If \(P(x)\) and \(Q(x)\) are polynomials, then \(\frac{P(x)}{Q(x)}\) is continuous wherever \(Q(x) \neq 0\).
5. Absolute value: \(f(x) = |x|\) is continuous everywhere. For \(x > 0\), \(f(x) = x\) (polynomial). For \(x < 0\), \(f(x) = -x\) (polynomial). At \(x = 0\), \(\lim_{x \to 0} |x| = 0 = |0|\).
6. Trigonometric functions: \(\sin x\) and \(\cos x\) are continuous everywhere. All six trigonometric functions are continuous wherever they are defined. For example, \(\tan x\) is continuous on \(\cdots \cup (-\frac{\pi}{2}, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{3\pi}{2}) \cup \cdots\).

1.6 Continuity of Compositions of Functions

Remark (Inverse Functions): When a continuous function defined on an interval has an inverse, the inverse function is itself continuous over its domain. This is because the graph of \(f^{-1}\) is the reflection of the graph of \(f\) across the line \(y = x\). As an example, all inverse trigonometric functions are continuous over their domains.

Theorem (Compositions of Continuous Functions): If \(f\) is continuous at \(c\) and \(g\) is continuous at \(f(c)\), then the composition \(g \circ f\) is continuous at \(c\), and: \[\lim_{x \to c} (g \circ f)(x) = g(f(c))\]

The continuity of compositions holds for any finite number of functions, provided each is continuous where it is applied.

Examples:

1. \(y = \sqrt{P_n(x)}\) where \(P_n(x)\) is a polynomial. The square root function is continuous on \([0, \infty)\). The given function is the composition of the polynomial with the square root function, so it’s continuous on its natural domain (where \(P_n(x) \geq 0\)).
2. \(y = \frac{x^{2/3}}{1 + x^4}\). The numerator is the cube root of the identity function squared; the denominator is an everywhere-positive polynomial. The quotient is continuous everywhere.
3. \(y = \left|\frac{x-2}{x^2-16}\right|\). The quotient \(\frac{x-2}{x^2-16}\) is continuous for \(x \in \mathbb{R} \setminus \{-4, 4\}\), and the function is the composition with the continuous absolute value function, so it’s continuous on \(\mathbb{R} \setminus \{-4, 4\}\).

Theorem (Limits of Continuous Functions): If \(\lim_{x \to c} f(x) = b\) and \(g\) is continuous at \(b\), then: \[\lim_{x \to c} g(f(x)) = g(b) = g\left(\lim_{x \to c} f(x)\right)\]

Proof: Let \(\varepsilon > 0\) be given. Since \(g\) is continuous at \(b\), \(\lim_{y \to b} g(y) = g(b)\), so there exists \(\delta_1 > 0\) such that: \[|g(y) - g(b)| < \varepsilon \quad \text{whenever} \quad |y - b| < \delta_1\]

Note this inequality also holds when \(y = b\).

Since \(\lim_{x \to c} f(x) = b\), setting \(y = f(x)\), there exists \(\delta > 0\) such that: \[|f(x) - b| < \delta_1 \quad \text{whenever} \quad 0 < |x - c| < \delta\]

This implies \(|g(f(x)) - g(b)| < \varepsilon\) whenever \(0 < |x - c| < \delta\).

By the definition of limit, \(\lim_{x \to c} g(f(x)) = g(b)\).

Examples:

1. \(\displaystyle \lim_{x \to 0} \sin\left(\frac{\pi}{2} \cos(\tan x)\right) = \sin\left(\frac{\pi}{2} \cos(0)\right) = \sin\left(\frac{\pi}{2}\right) = 1\)
2. \(\displaystyle \lim_{x \to 1} \cos^{-1}(\ln(\sqrt{x})) = \cos^{-1}(\ln(1)) = \cos^{-1}(0) = \frac{\pi}{2}\)
3. \(\displaystyle \lim_{x \to 0} \sec\left(e^x + \pi \tan\left(\frac{\pi}{4\sec x}\right) - 1\right) = \sec\left(1 + \pi \tan\left(\frac{\pi}{4}\right) - 1\right) = \sec(\pi) = -1\)

1.7 Intermediate Value Theorem

A function has the Intermediate Value Property if whenever it takes on two values, it also takes on all values in between.

Theorem (Intermediate Value Theorem): If \(f\) is a continuous function on a closed interval \([a, b]\), and if \(y_0\) is any value between \(f(a)\) and \(f(b)\), then \(y_0 = f(c)\) for some \(c\) in \([a, b]\).

Geometrically, any horizontal line \(y = y_0\) between \(f(a)\) and \(f(b)\) will cross the curve \(y = f(x)\) at least once over \([a, b]\).

The proof depends on the completeness property of real numbers. Continuity on the interval is essential—if \(f\) is discontinuous at even one point, the theorem may fail.

Consequence for Graphing (Connectedness): The graph of a function continuous on an interval cannot have any breaks—it will be a single, unbroken curve.

Consequence for Root Finding: A solution of \(f(x) = 0\) is called a root or zero of \(f\). The Intermediate Value Theorem tells us that if \(f\) is continuous and changes sign on an interval, then the interval contains a zero of the function.

Example: Show that there is a root of \(\sqrt{2x + 5} = 4 - x^2\) between 0 and 2.

The equation is equivalent to \(f(x) = 0\) where: \[f(x) = \sqrt{2x + 5} - 4 + x^2\]

The function \(f\) is continuous on \([0, 2]\). We have: \[f(0) = \sqrt{5} - 4 \approx 2.236 - 4 = -1.764 < 0\] \[f(2) = \sqrt{9} - 4 + 4 = 3 > 0\]

Since \(f(0) < 0 < f(2)\), by the Intermediate Value Theorem, there exists \(c \in [0, 2]\) such that \(f(c) = 0\), meaning there is a root in \([0, 2]\).

1.8 Continuous Extension to a Point

If \(f(c)\) is not defined but \(\lim_{x \to c} f(x) = L\) exists, we can define a new function: \[F(x) = \begin{cases} f(x), & \text{if } x \text{ is in the domain of } f \\ L, & \text{if } x = c \end{cases}\]

The function \(F\) is continuous at \(x = c\) and is called the continuous extension of \(f\) to \(x = c\).

Example: The function \(f(x) = \frac{x^2 + x - 6}{x^2 - 4}\), \(x \neq 2\), has a continuous extension to \(x = 2\).

We have: \[f(x) = \frac{x^2 + x - 6}{x^2 - 4} = \frac{(x-2)(x+3)}{(x-2)(x+2)} = \frac{x+3}{x+2} \quad \text{for } x \neq 2\]

Thus \(\lim_{x \to 2} f(x) = \frac{5}{4}\), and the continuous extension is: \[F(x) = \frac{x+3}{x+2}\]

1.9 The Number \(e\) and Related Limits

We have seen from sequences that: \[\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e = 2.71828\ldots\]

Theorem: For real \(x\): \[\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e\]

Corollary 1: This is equivalent to: \[\lim_{x \to 0} (1 + x)^{1/x} = e\]

Corollary 2: For any \(a \in \mathbb{R}\): \[\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = e^a\]

Proof: Let \(t = a/x\). When \(x \to \infty\), \(t \to 0\). Then: \[\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = \lim_{t \to 0} (1 + t)^{a/t} = \left(\lim_{t \to 0} (1 + t)^{1/t}\right)^a = e^a\]

Lemma: If \(\lim_{x \to c} g(x) = a\) and \(\lim_{x \to c} h(x) = b\), then: \[\lim_{x \to c} (g(x))^{h(x)} = a^b\]

Corollary (for indeterminate form \(1^{\infty}\)): Suppose \(\lim_{x \to c} g(x) = 1\) and \(\lim_{x \to c} h(x) = \infty\). Then: \[\lim_{x \to c} (g(x))^{h(x)} = \lim_{x \to c} \left([1 + (g(x) - 1)]^{\frac{1}{g(x)-1}}\right)^{(g(x)-1)h(x)} = e^{\lim_{x \to c} [h(x)(g(x)-1)]}\]

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2. Definitions

• Limit of a function (\(\varepsilon\)-\(\delta\) definition): \(\lim_{x \to c} f(x) = L\) means for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that \(|f(x) - L| < \varepsilon\) whenever \(0 < |x - c| < \delta\).
• Limit point: A point \(c\) is a limit point of a set \(A\) if each neighborhood of \(c\) contains at least one point of \(A\) different from \(c\).
• Heine’s definition of limit: \(f\) has limit \(L\) at \(c\) if for any sequence \(x_n \to c\) with \(x_n \neq c\), we have \(f(x_n) \to L\).
• Right-hand limit: \(\lim_{x \to c^+} f(x) = L\) means for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that \(|f(x) - L| < \varepsilon\) whenever \(c < x < c + \delta\).
• Left-hand limit: \(\lim_{x \to c^-} f(x) = L\) means for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that \(|f(x) - L| < \varepsilon\) whenever \(c - \delta < x < c\).
• Limit at infinity: \(\lim_{x \to \infty} f(x) = L\) means for every \(\varepsilon > 0\), there exists \(M\) such that \(|f(x) - L| < \varepsilon\) whenever \(x > M\).
• Infinite limit: \(\lim_{x \to c} f(x) = \infty\) means for every \(B > 0\), there exists \(\delta > 0\) such that \(f(x) > B\) whenever \(0 < |x - c| < \delta\).
• Horizontal asymptote: A line \(y = b\) is a horizontal asymptote of \(y = f(x)\) if \(\lim_{x \to \infty} f(x) = b\) or \(\lim_{x \to -\infty} f(x) = b\).
• Vertical asymptote: A line \(x = a\) is a vertical asymptote of \(y = f(x)\) if \(\lim_{x \to a^+} f(x) = \pm\infty\) or \(\lim_{x \to a^-} f(x) = \pm\infty\).
• Oblique (slant) asymptote: The line \(y = \alpha x + \beta\) (\(\alpha \neq 0\)) is an oblique asymptote if \(\lim_{x \to \pm\infty} (f(x) - (\alpha x + \beta)) = 0\).
• Continuous at a point: A function \(f\) is continuous at \(c\) if \(\lim_{x \to c} f(x) = f(c)\).
• Right-continuous: \(f\) is right-continuous at \(c\) if \(\lim_{x \to c^+} f(x) = f(c)\).
• Left-continuous: \(f\) is left-continuous at \(c\) if \(\lim_{x \to c^-} f(x) = f(c)\).
• Continuous function: A function continuous at every point in its domain.
• Continuous extension: If \(\lim_{x \to c} f(x) = L\) but \(f(c)\) is undefined, defining \(F(c) = L\) creates a continuous extension of \(f\) at \(c\).
• Intermediate Value Property: If \(f\) takes values \(f(a)\) and \(f(b)\), it takes all values between them.
• Indeterminate form: Expressions like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), \(\infty - \infty\), \(1^{\infty}\), \(0^0\), \(\infty^0\) that arise in limit evaluation where direct substitution is ambiguous.

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3. Formulas

• Basic limits: \(\lim_{x \to c} x = c\), \(\lim_{x \to c} k = k\) (constant)
• Sum rule: \(\lim_{x \to c} (f(x) + g(x)) = \lim_{x \to c} f(x) + \lim_{x \to c} g(x)\)
• Difference rule: \(\lim_{x \to c} (f(x) - g(x)) = \lim_{x \to c} f(x) - \lim_{x \to c} g(x)\)
• Product rule: \(\lim_{x \to c} (f(x) \cdot g(x)) = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x)\)
• Quotient rule: \(\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)}\), provided denominator \(\neq 0\)
• Power rule: \(\lim_{x \to c} [f(x)]^n = [\lim_{x \to c} f(x)]^n\)
• Root rule: \(\lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to c} f(x)}\) (with appropriate domain restrictions)
• Polynomial limit: If \(P(x)\) is a polynomial, then \(\lim_{x \to c} P(x) = P(c)\)
• Rational function limit: \(\lim_{x \to c} \frac{P(x)}{Q(x)} = \frac{P(c)}{Q(c)}\) if \(Q(c) \neq 0\)
• Important trigonometric limit: \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) (\(\theta\) in radians)
• Related trigonometric limits: \(\lim_{\theta \to 0} \sin \theta = 0\), \(\lim_{\theta \to 0} \cos \theta = 1\)
• Limit with exponential: \(\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e\)
• Equivalent form: \(\lim_{x \to 0} (1 + x)^{1/x} = e\)
• General exponential limit: \(\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = e^a\)
• Exponential-logarithmic limit: \(\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a\) (for \(a > 0\))
• Indeterminate form \(1^{\infty}\): If \(\lim_{x \to c} g(x) = 1\) and \(\lim_{x \to c} h(x) = \infty\), then \(\lim_{x \to c} (g(x))^{h(x)} = e^{\lim_{x \to c} [h(x)(g(x)-1)]}\)
• Limits at infinity: \(\lim_{x \to \pm\infty} \frac{1}{x} = 0\), \(\lim_{x \to -\infty} e^x = 0\)
• Composition of continuous functions: If \(f\) is continuous at \(c\) and \(g\) is continuous at \(f(c)\), then \(\lim_{x \to c} g(f(x)) = g(f(c))\)

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4. Examples

4.1. Limit with Cube Root (Lab 12, Task 1a)

Evaluate \(\displaystyle \lim_{x\to0} \frac{\sqrt[3]{x+1} - 1}{x}\).

Click to see the solution

Key Concept: Use the algebraic identity \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) by multiplying by the conjugate expression for cube roots.

1. Identify the conjugate factor: For \(\sqrt[3]{x+1} - 1\), we multiply by \(\frac{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}\)
2. Apply the identity: Using \(a - b = \frac{a^3 - b^3}{a^2 + ab + b^2}\) with \(a = \sqrt[3]{x+1}\) and \(b = 1\): \[\sqrt[3]{x+1} - 1 = \frac{(x+1) - 1}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1} = \frac{x}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}\]
3. Substitute into the limit: \[\lim_{x\to0} \frac{\sqrt[3]{x+1} - 1}{x} = \lim_{x\to0} \frac{x}{x \cdot \left[(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1\right]}\] \[= \lim_{x\to0} \frac{1}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}\]
4. Evaluate by direct substitution: \[= \frac{1}{(\sqrt[3]{1})^2 + \sqrt[3]{1} + 1} = \frac{1}{1 + 1 + 1} = \frac{1}{3}\]

Answer: \(\dfrac{1}{3}\)

4.2. Limit of Polynomial Rational Function (Lab 12, Task 1b)

Evaluate \(\displaystyle \lim_{x\to-1} \frac{x^3 - x^2 - 5x - 3}{(x+1)^2}\).

Click to see the solution

Key Concept: Since substituting \(x = -1\) gives \(\frac{0}{0}\), factor the numerator to cancel common factors.

1. Verify the indeterminate form: Numerator at \(x = -1\): \((-1)^3 - (-1)^2 - 5(-1) - 3 = -1 - 1 + 5 - 3 = 0\) Denominator at \(x = -1\): \((0)^2 = 0\) This is \(\frac{0}{0}\), so we need to factor.

2. Factor the numerator: Since \(x = -1\) is a root, \((x+1)\) is a factor. Perform polynomial division: \[x^3 - x^2 - 5x - 3 = (x+1)(x^2 - 2x - 3)\]

   Factor further: \(x^2 - 2x - 3 = (x-3)(x+1)\)

   So: \(x^3 - x^2 - 5x - 3 = (x+1)^2(x-3)\)

3. Simplify the expression: \[\frac{x^3 - x^2 - 5x - 3}{(x+1)^2} = \frac{(x+1)^2(x-3)}{(x+1)^2} = x - 3 \quad (x \neq -1)\]

4. Evaluate the limit: \[\lim_{x\to-1} (x - 3) = -1 - 3 = -4\]

Answer: \(-4\)

4.3. Trigonometric Limit with Sine and Cosine (Lab 12, Task 1c)

Evaluate \(\displaystyle \lim_{x\to0} \frac{x \sin x}{2 - 2 \cos x}\).

Click to see the solution

Key Concept: Use the identity \(1 - \cos x = 2\sin^2(x/2)\) and known limits.

1. Factor the denominator: \[\lim_{x\to0} \frac{x \sin x}{2(1 - \cos x)}\]
2. Use the identity \(1 - \cos x = 2\sin^2(x/2)\): \[= \lim_{x\to0} \frac{x \sin x}{2 \cdot 2\sin^2(x/2)} = \lim_{x\to0} \frac{x \sin x}{4\sin^2(x/2)}\]
3. Rewrite using standard limit forms: \[= \lim_{x\to0} \frac{x}{\sin(x/2)} \cdot \frac{\sin x}{4\sin(x/2)}\] \[= \lim_{x\to0} \frac{x}{x/2} \cdot \frac{x/2}{\sin(x/2)} \cdot \frac{\sin x}{x} \cdot \frac{x}{4\sin(x/2)}\] \[= \lim_{x\to0} 2 \cdot \frac{x/2}{\sin(x/2)} \cdot \frac{\sin x}{x} \cdot \frac{x/2}{\sin(x/2)} \cdot \frac{1}{2}\]
4. Alternatively, using L’Hôpital’s rule or direct substitution: \[= \lim_{x\to0} \frac{x \sin x}{4\sin^2(x/2)} = \lim_{x\to0} \frac{x \cdot 2\sin(x/2)\cos(x/2)}{4\sin^2(x/2)}\] \[= \lim_{x\to0} \frac{x \cos(x/2)}{2\sin(x/2)} = \lim_{x\to0} \frac{x/2}{\sin(x/2)} \cdot \cos(x/2) = 1 \cdot 1 = 1\]

Answer: \(1\)

4.4. One-Sided Limits of a Rational Function (Lab 12, Task 2)

Given \(f(x) = \frac{x^2 - 3x + 2}{x^3 - 4x}\), find the following limits:

1. \(\displaystyle \lim_{x\to-2^+} f(x)\)
2. \(\displaystyle \lim_{x\to0^-} f(x)\)
3. \(\displaystyle \lim_{x\to0^+} f(x)\)
4. \(\displaystyle \lim_{x\to1^+} f(x)\)
5. \(\displaystyle \lim_{x\to2^+} f(x)\)

Click to see the solution

Key Concept: First factor both numerator and denominator to identify removable discontinuities and vertical asymptotes.

1. Factor the expressions:

   • Numerator: \(x^2 - 3x + 2 = (x-1)(x-2)\)
   • Denominator: \(x^3 - 4x = x(x^2-4) = x(x-2)(x+2)\)

2. Simplify (for \(x \neq 2\)): \[f(x) = \frac{(x-1)(x-2)}{x(x-2)(x+2)} = \frac{x-1}{x(x+2)} \quad (x \neq 2)\]

3. Evaluate each limit:

   a) \(\displaystyle \lim_{x\to-2^+} f(x)\): As \(x \to -2^+\): numerator \((x-1) \to -3\); denominator \(x(x+2) \to (-2)(0^+) = 0^-\) \[\lim_{x\to-2^+} \frac{x-1}{x(x+2)} = \frac{-3}{0^-} = +\infty\]

   b) \(\displaystyle \lim_{x\to0^-} f(x)\): As \(x \to 0^-\): numerator \((x-1) \to -1\); denominator \(x(x+2) \to (0^-)(2) = 0^-\) \[\lim_{x\to0^-} \frac{x-1}{x(x+2)} = \frac{-1}{0^-} = +\infty\]

   c) \(\displaystyle \lim_{x\to0^+} f(x)\): As \(x \to 0^+\): numerator \((x-1) \to -1\); denominator \(x(x+2) \to (0^+)(2) = 0^+\) \[\lim_{x\to0^+} \frac{x-1}{x(x+2)} = \frac{-1}{0^+} = -\infty\]

   d) \(\displaystyle \lim_{x\to1^+} f(x)\): Direct substitution works since \(x = 1\) is not a discontinuity of the simplified function: \[\lim_{x\to1^+} \frac{x-1}{x(x+2)} = \frac{1-1}{1(1+2)} = \frac{0}{3} = 0\]

   e) \(\displaystyle \lim_{x\to2^+} f(x)\): Although \(x = 2\) was cancelled, we use the simplified form: \[\lim_{x\to2^+} \frac{x-1}{x(x+2)} = \frac{2-1}{2(2+2)} = \frac{1}{8}\]

Answers:

• 1. \(+\infty\)
• 2. \(+\infty\)
• 3. \(-\infty\)
• 4. \(0\)
• 5. \(\frac{1}{8}\)

4.5. Find Asymptotes of Rational Functions (Lab 12, Task 3)

Find the equations of the asymptotes of the following functions:

1. \(f(x) = \dfrac{x^2 - 1}{2x + 4}\)

2. \(g(x) = \dfrac{x^3 + 1}{x^2}\)

Click to see the solution

Key Concept: There are three types of asymptotes:

• Vertical asymptotes occur where the denominator equals zero (and numerator doesn’t)
• Horizontal asymptotes are found by evaluating \(\lim_{x\to\pm\infty} f(x)\)
• Oblique (slant) asymptotes exist when the degree of numerator exceeds denominator by exactly 1

(a) \(f(x) = \dfrac{x^2 - 1}{2x + 4}\):

1. Find vertical asymptotes: Set denominator \(= 0\): \(2x + 4 = 0 \Rightarrow x = -2\)

   Check numerator at \(x = -2\): \((-2)^2 - 1 = 3 \neq 0\)

   Vertical asymptote: \(x = -2\)

2. Check for horizontal asymptote: Degree of numerator (2) > degree of denominator (1), so no horizontal asymptote.

3. Find oblique asymptote: Since degree difference = 1, perform polynomial long division: \[\frac{x^2 - 1}{2x + 4} = \frac{x^2 - 1}{2(x + 2)}\]

   Dividing \(x^2 - 1\) by \(x + 2\): \[x^2 - 1 = (x + 2)(x - 2) + 3\]

   So: \[f(x) = \frac{(x-2)(x+2) + 3}{2(x+2)} = \frac{x-2}{2} + \frac{3}{2(x+2)} = \frac{x}{2} - 1 + \frac{3}{2(x+2)}\]

   As \(x \to \pm\infty\), the term \(\frac{3}{2(x+2)} \to 0\)

   Oblique asymptote: \(y = \dfrac{x}{2} - 1\) (or equivalently \(y = \dfrac{x-2}{2}\))

Asymptotes for (a): Vertical: \(x = -2\); Oblique: \(y = \dfrac{x}{2} - 1\)

(b) \(g(x) = \dfrac{x^3 + 1}{x^2}\):

1. Find vertical asymptotes: Set denominator \(= 0\): \(x^2 = 0 \Rightarrow x = 0\)

   Check numerator at \(x = 0\): \(0^3 + 1 = 1 \neq 0\)

   Vertical asymptote: \(x = 0\)

2. Check for horizontal asymptote: Degree of numerator (3) > degree of denominator (2), so no horizontal asymptote.

3. Find oblique asymptote: Perform polynomial division: \[g(x) = \frac{x^3 + 1}{x^2} = \frac{x^3}{x^2} + \frac{1}{x^2} = x + \frac{1}{x^2}\]

   As \(x \to \pm\infty\), the term \(\frac{1}{x^2} \to 0\)

   Oblique asymptote: \(y = x\)

Asymptotes for (b): Vertical: \(x = 0\); Oblique: \(y = x\)

Answers:

• 1. Vertical asymptote: \(x = -2\); Oblique asymptote: \(y = \dfrac{x}{2} - 1\)
• 2. Vertical asymptote: \(x = 0\); Oblique asymptote: \(y = x\)

4.6. Find Asymptotes with Square Root and Parameters (Lab 13, Task 1)

Assume that constants \(a\) and \(b\) are positive. Find equations for all horizontal and vertical asymptotes for the graph of \(y = \frac{\sqrt{ax^2 + 4}}{x - b}\).

Click to see the solution

Key Concept: For functions involving square roots, we must carefully handle the sign of \(x\) when taking limits at \(\pm\infty\). Remember that \(\sqrt{x^2} = |x|\).

1. Find vertical asymptotes: Set denominator \(= 0\): \(x - b = 0 \Rightarrow x = b\)

   Check numerator at \(x = b\): \(\sqrt{ab^2 + 4} > 0\) (since \(a, b > 0\))

   Vertical asymptote: \(x = b\)

2. Find horizontal asymptotes as \(x \to \infty\): \[\lim_{x \to \infty} \frac{\sqrt{ax^2 + 4}}{x - b}\]

   For \(x > 0\), we have \(\sqrt{x^2} = x\). Factor out \(x\) from the square root: \[= \lim_{x \to \infty} \frac{\sqrt{x^2(a + \frac{4}{x^2})}}{x - b} = \lim_{x \to \infty} \frac{x\sqrt{a + \frac{4}{x^2}}}{x - b}\]

   Divide numerator and denominator by \(x\): \[= \lim_{x \to \infty} \frac{\sqrt{a + \frac{4}{x^2}}}{1 - \frac{b}{x}} = \frac{\sqrt{a}}{1} = \sqrt{a}\]

3. Find horizontal asymptotes as \(x \to -\infty\): \[\lim_{x \to -\infty} \frac{\sqrt{ax^2 + 4}}{x - b}\]

   For \(x < 0\), we have \(\sqrt{x^2} = |x| = -x\). Thus: \[= \lim_{x \to -\infty} \frac{-x\sqrt{a + \frac{4}{x^2}}}{x - b}\]

   Divide numerator and denominator by \(x\) (note: \(x < 0\)): \[= \lim_{x \to -\infty} \frac{-\sqrt{a + \frac{4}{x^2}}}{1 - \frac{b}{x}} = \frac{-\sqrt{a}}{1} = -\sqrt{a}\]

Answer:

• Vertical asymptote: \(x = b\)
• Horizontal asymptotes: \(y = \sqrt{a}\) (as \(x \to \infty\)) and \(y = -\sqrt{a}\) (as \(x \to -\infty\))

4.7. Indeterminate Form \(1^{\infty}\) with Squared Exponent (Lab 13, Task 2a)

Evaluate \(\displaystyle \lim_{x \to \infty} \left(\frac{x}{2x + 1}\right)^{x^2}\).

Click to see the solution

Key Concept: First determine if this is actually a \(1^{\infty}\) form by checking the base limit.

1. Check the base limit: \[\lim_{x \to \infty} \frac{x}{2x + 1} = \lim_{x \to \infty} \frac{1}{2 + \frac{1}{x}} = \frac{1}{2}\]

   The base approaches \(\frac{1}{2}\), not \(1\). This is NOT a \(1^{\infty}\) form.

2. Analyze the limit: Since the base approaches \(\frac{1}{2} < 1\) and the exponent approaches \(\infty\): \[\left(\frac{1}{2}\right)^{\infty} = 0\]

3. Verify rigorously: For large \(x\), \(\frac{x}{2x+1} < \frac{1}{2} + \varepsilon\) for any \(\varepsilon > 0\). Thus: \[\left(\frac{x}{2x+1}\right)^{x^2} \to 0 \text{ as } x \to \infty\]

Answer: \(0\)

4.8. Indeterminate Form \(1^{\infty}\) - Standard Type (Lab 13, Task 2b)

Evaluate \(\displaystyle \lim_{x \to \infty} \left(\frac{x + 1}{2x + 1}\right)^{x}\).

Click to see the solution

Key Concept: Check the base limit to determine the form type.

1. Check the base limit: \[\lim_{x \to \infty} \frac{x + 1}{2x + 1} = \lim_{x \to \infty} \frac{1 + \frac{1}{x}}{2 + \frac{1}{x}} = \frac{1}{2}\]

   The base approaches \(\frac{1}{2}\), not \(1\). This is NOT a \(1^{\infty}\) form.

2. Analyze the limit: Since \(\frac{1}{2} < 1\) and the exponent goes to \(\infty\): \[\left(\frac{1}{2}\right)^{\infty} = 0\]

3. Alternative verification using logarithms: \[L = \lim_{x \to \infty} \left(\frac{x + 1}{2x + 1}\right)^{x}\] \[\ln L = \lim_{x \to \infty} x \ln\left(\frac{x + 1}{2x + 1}\right) = \lim_{x \to \infty} x \ln\left(\frac{1 + \frac{1}{x}}{2 + \frac{1}{x}}\right)\]

   As \(x \to \infty\): \(\ln\left(\frac{1}{2}\right) = -\ln 2 < 0\)

   So \(\ln L = \lim_{x \to \infty} x \cdot (-\ln 2 + \text{terms} \to 0) = -\infty\)

   Therefore \(L = e^{-\infty} = 0\)

Answer: \(0\)

4.9. Indeterminate Form \(1^{\infty}\) with Trigonometric Functions (Lab 13, Task 2c)

Evaluate \(\displaystyle \lim_{x \to 0} (1 + x^2)^{\cot^2 x}\).

Click to see the solution

Key Concept: This IS a \(1^{\infty}\) form since the base approaches \(1\) and \(\cot^2 x \to \infty\) as \(x \to 0\). Use the corollary: if \(\lim g(x) = 1\) and \(\lim h(x) = \infty\), then \(\lim (g(x))^{h(x)} = e^{\lim h(x)(g(x)-1)}\).

1. Identify the components:

   • \(g(x) = 1 + x^2\), so \(g(x) - 1 = x^2\)
   • \(h(x) = \cot^2 x = \frac{\cos^2 x}{\sin^2 x}\)

2. Apply the formula: \[L = e^{\lim_{x \to 0} \cot^2 x \cdot x^2} = e^{\lim_{x \to 0} \frac{x^2 \cos^2 x}{\sin^2 x}}\]

3. Evaluate the exponent: \[\lim_{x \to 0} \frac{x^2 \cos^2 x}{\sin^2 x} = \lim_{x \to 0} \left(\frac{x}{\sin x}\right)^2 \cdot \cos^2 x\]

   Using \(\lim_{x \to 0} \frac{x}{\sin x} = 1\) and \(\lim_{x \to 0} \cos x = 1\): \[= 1^2 \cdot 1^2 = 1\]

4. Compute the final answer: \[L = e^1 = e\]

Answer: \(e\)

4.10. Continuous Extension - Removable Discontinuity (Lab 13, Task 3)

Define \(g(3)\) in a way that extends \(g(x) = \frac{x^2 - 9}{x - 3}\) to be continuous at \(x = 3\).

Click to see the solution

Key Concept: A function has a removable discontinuity at \(x = c\) if \(\lim_{x \to c} f(x)\) exists but differs from \(f(c)\) (or \(f(c)\) is undefined). To make it continuous, we define \(f(c)\) to equal the limit.

1. Simplify the function: \[g(x) = \frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3 \quad (x \neq 3)\]
2. Find the limit at \(x = 3\): \[\lim_{x \to 3} g(x) = \lim_{x \to 3} (x + 3) = 3 + 3 = 6\]
3. Define \(g(3)\) for continuity: For \(g\) to be continuous at \(x = 3\), we need: \[g(3) = \lim_{x \to 3} g(x) = 6\]

Answer: \(g(3) = 6\)

4.11. Continuous Extension - Another Example (Lab 13, Task 4)

Define \(h(2)\) in a way that extends \(h(t) = \frac{t^2 + 3t - 10}{t - 2}\) to be continuous at \(t = 2\).

Click to see the solution

Key Concept: Factor the numerator to cancel the problematic factor in the denominator.

1. Factor the numerator: We need to factor \(t^2 + 3t - 10\). Looking for two numbers that multiply to \(-10\) and add to \(3\): these are \(5\) and \(-2\). \[t^2 + 3t - 10 = (t + 5)(t - 2)\]
2. Simplify the function: \[h(t) = \frac{(t + 5)(t - 2)}{t - 2} = t + 5 \quad (t \neq 2)\]
3. Find the limit at \(t = 2\): \[\lim_{t \to 2} h(t) = \lim_{t \to 2} (t + 5) = 2 + 5 = 7\]
4. Define \(h(2)\) for continuity: \[h(2) = 7\]

Answer: \(h(2) = 7\)

4.12. Continuity of Piecewise Function - One Parameter (Lab 13, Task 5)

For what values of \(a\) is the function \[f(x) = \begin{cases} a^2x - 2a, & x \geq 2 \\ 12, & x < 2 \end{cases}\] continuous at every \(x\)?

Click to see the solution

Key Concept: A piecewise function is continuous everywhere if: (1) each piece is continuous on its domain, and (2) the pieces “connect” at the boundary points.

1. Analyze each piece:
   • For \(x < 2\): \(f(x) = 12\) (constant, always continuous)
   • For \(x \geq 2\): \(f(x) = a^2x - 2a\) (linear, always continuous)
2. Check continuity at the boundary \(x = 2\): For continuity at \(x = 2\), we need: \[\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)\]
3. Calculate the limits:
   • Left-hand limit: \(\lim_{x \to 2^-} f(x) = 12\)
   • Right-hand limit: \(\lim_{x \to 2^+} f(x) = a^2(2) - 2a = 2a^2 - 2a\)
   • Function value: \(f(2) = a^2(2) - 2a = 2a^2 - 2a\)
4. Set the limits equal: \[12 = 2a^2 - 2a\] \[2a^2 - 2a - 12 = 0\] \[a^2 - a - 6 = 0\] \[(a - 3)(a + 2) = 0\] \[a = 3 \quad \text{or} \quad a = -2\]

Answer: \(a = 3\) or \(a = -2\)

4.13. Continuity of Piecewise Function - Two Parameters (Lab 13, Task 6)

For what values of \(a\) and \(b\) is the function \[g(x) = \begin{cases} -2, & x \leq -1 \\ ax - b, & -1 < x < 1 \\ 3, & x \geq 1 \end{cases}\] continuous at every \(x\)?

Click to see the solution

Key Concept: We need to ensure continuity at both boundary points \(x = -1\) and \(x = 1\).

1. Analyze each piece:

   • For \(x \leq -1\): \(f(x) = -2\) (constant, continuous)
   • For \(-1 < x < 1\): \(f(x) = ax - b\) (linear, continuous)
   • For \(x \geq 1\): \(f(x) = 3\) (constant, continuous)

2. Check continuity at \(x = -1\):

   • Left-hand limit: \(\lim_{x \to -1^-} g(x) = -2\)
   • Right-hand limit: \(\lim_{x \to -1^+} g(x) = a(-1) - b = -a - b\)
   • Function value: \(g(-1) = -2\)

   For continuity: \(-a - b = -2\), which gives us: \[a + b = 2 \quad \text{...(Equation 1)}\]

3. Check continuity at \(x = 1\):

   • Left-hand limit: \(\lim_{x \to 1^-} g(x) = a(1) - b = a - b\)
   • Right-hand limit: \(\lim_{x \to 1^+} g(x) = 3\)
   • Function value: \(g(1) = 3\)

   For continuity: \(a - b = 3 \quad \text{...(Equation 2)}\)

4. Solve the system of equations: From Equation 1: \(a + b = 2\) From Equation 2: \(a - b = 3\)

   Adding these equations: \(2a = 5 \Rightarrow a = \frac{5}{2}\)

   Substituting back: \(\frac{5}{2} + b = 2 \Rightarrow b = 2 - \frac{5}{2} = -\frac{1}{2}\)

Answer: \(a = \frac{5}{2}\) and \(b = -\frac{1}{2}\)

4.14. Continuity of Piecewise Function - Three Pieces with Two Parameters (Lab 13, Task 7)

For what values of \(a\) and \(b\) is the function \[h(x) = \begin{cases} ax + 2b, & x \leq 0 \\ x^2 + 3a - b, & 0 < x \leq 2 \\ 3x - 5, & x > 2 \end{cases}\] continuous at every \(x\)?

Click to see the solution

Key Concept: We need continuity at both boundary points \(x = 0\) and \(x = 2\).

1. Analyze each piece:

   • For \(x \leq 0\): \(h(x) = ax + 2b\) (linear, continuous)
   • For \(0 < x \leq 2\): \(h(x) = x^2 + 3a - b\) (polynomial, continuous)
   • For \(x > 2\): \(h(x) = 3x - 5\) (linear, continuous)

2. Check continuity at \(x = 0\):

   • Left-hand limit and function value: \(\lim_{x \to 0^-} h(x) = h(0) = a(0) + 2b = 2b\)
   • Right-hand limit: \(\lim_{x \to 0^+} h(x) = 0^2 + 3a - b = 3a - b\)

   For continuity: \(2b = 3a - b\), which gives us: \[3b = 3a \Rightarrow a = b \quad \text{...(Equation 1)}\]

3. Check continuity at \(x = 2\):

   • Left-hand limit and function value: \(\lim_{x \to 2^-} h(x) = h(2) = 2^2 + 3a - b = 4 + 3a - b\)
   • Right-hand limit: \(\lim_{x \to 2^+} h(x) = 3(2) - 5 = 1\)

   For continuity: \(4 + 3a - b = 1\), which gives us: \[3a - b = -3 \quad \text{...(Equation 2)}\]

4. Solve the system of equations: Substituting Equation 1 (\(a = b\)) into Equation 2: \[3a - a = -3\] \[2a = -3\] \[a = -\frac{3}{2}\]

   From Equation 1: \(b = a = -\frac{3}{2}\)

Answer: \(a = -\frac{3}{2}\) and \(b = -\frac{3}{2}\)

4.15. Evaluating a Limit Using Factorization (Chapter 5, Example 1)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{5x^3 + 8x^2}{3x^4 - 16x^2}\).

Click to see the solution

Key Concept: Factor out common terms from numerator and denominator to cancel and remove the \(\frac{0}{0}\) indeterminate form.

1. Factor the numerator and denominator: \[\frac{5x^3 + 8x^2}{3x^4 - 16x^2} = \frac{x^2(5x + 8)}{x^2(3x^2 - 16)}\]
2. Cancel common factors (valid since \(x \neq 0\)): \[= \frac{5x + 8}{3x^2 - 16}\]
3. Evaluate the limit by direct substitution: \[\lim_{x \to 0} \frac{5x + 8}{3x^2 - 16} = \frac{5(0) + 8}{3(0)^2 - 16} = \frac{8}{-16} = -\frac{1}{2}\]

Answer: \(-\frac{1}{2}\)

4.16. Simplifying Before Taking the Limit (Chapter 5, Example 2)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{\frac{1}{x-1} + \frac{1}{x+1}}{x}\).

Click to see the solution

Key Concept: Combine fractions in the numerator, then simplify.

1. Combine the fractions in the numerator: \[\frac{1}{x-1} + \frac{1}{x+1} = \frac{(x+1) + (x-1)}{(x-1)(x+1)} = \frac{2x}{x^2-1}\]
2. Rewrite the original limit: \[\lim_{x \to 0} \frac{\frac{2x}{x^2-1}}{x} = \lim_{x \to 0} \frac{2x}{x(x^2-1)} = \lim_{x \to 0} \frac{2}{x^2-1}\]
3. Evaluate by direct substitution: \[= \frac{2}{0^2-1} = \frac{2}{-1} = -2\]

Answer: \(-2\)

4.17. Factoring and Canceling (Chapter 5, Example 3)

Evaluate \(\displaystyle \lim_{x \to 2} \frac{x^3 - 8}{x^4 - 16}\).

Click to see the solution

1. Factor using difference formulas:
   • Numerator: \(x^3 - 8 = (x-2)(x^2 + 2x + 4)\)
   • Denominator: \(x^4 - 16 = (x^2-4)(x^2+4) = (x-2)(x+2)(x^2+4)\)
2. Cancel the common factor \((x-2)\): \[\frac{x^3 - 8}{x^4 - 16} = \frac{(x-2)(x^2 + 2x + 4)}{(x-2)(x+2)(x^2+4)} = \frac{x^2 + 2x + 4}{(x+2)(x^2+4)}\]
3. Evaluate by substitution: \[\lim_{x \to 2} \frac{x^2 + 2x + 4}{(x+2)(x^2+4)} = \frac{4 + 4 + 4}{(4)(8)} = \frac{12}{32} = \frac{3}{8}\]

Answer: \(\frac{3}{8}\)

4.18. Polynomial Division (Chapter 5, Example 4)

Evaluate \(\displaystyle \lim_{x \to -1} \frac{x^3 - x^2 - 5x - 3}{(x+1)^2}\).

Click to see the solution

Key Concept: Since \(x = -1\) is a root of the numerator, factor out \((x+1)\).

1. Check if \(x = -1\) is a root of the numerator: \[(-1)^3 - (-1)^2 - 5(-1) - 3 = -1 - 1 + 5 - 3 = 0\]

   Yes, so \((x+1)\) is a factor.

2. Factor the numerator by polynomial division or synthetic division: \[x^3 - x^2 - 5x - 3 = (x+1)(x^2 - 2x - 3) = (x+1)(x-3)(x+1) = (x+1)^2(x-3)\]

3. Simplify and evaluate: \[\lim_{x \to -1} \frac{(x+1)^2(x-3)}{(x+1)^2} = \lim_{x \to -1} (x-3) = -1 - 3 = -4\]

Answer: \(-4\)

4.19. Rationalizing the Denominator (Chapter 5, Example 5)

Evaluate \(\displaystyle \lim_{x \to 1} \frac{x-1}{\sqrt{x+3} - 2}\).

Click to see the solution

Key Concept: Multiply by the conjugate of the denominator.

1. Multiply numerator and denominator by the conjugate: \[\lim_{x \to 1} \frac{x-1}{\sqrt{x+3} - 2} \cdot \frac{\sqrt{x+3} + 2}{\sqrt{x+3} + 2}\]
2. Simplify: \[= \lim_{x \to 1} \frac{(x-1)(\sqrt{x+3} + 2)}{(x+3) - 4} = \lim_{x \to 1} \frac{(x-1)(\sqrt{x+3} + 2)}{x-1}\]
3. Cancel \((x-1)\) and evaluate: \[= \lim_{x \to 1} (\sqrt{x+3} + 2) = \sqrt{4} + 2 = 2 + 2 = 4\]

Answer: \(4\)

4.20. Rationalizing the Numerator (Chapter 5, Example 6)

Evaluate \(\displaystyle \lim_{x \to -3} \frac{2 - \sqrt{x^2-5}}{x+3}\).

Click to see the solution

1. Multiply by the conjugate of the numerator: \[\lim_{x \to -3} \frac{2 - \sqrt{x^2-5}}{x+3} \cdot \frac{2 + \sqrt{x^2-5}}{2 + \sqrt{x^2-5}}\]
2. Simplify: \[= \lim_{x \to -3} \frac{4 - (x^2-5)}{(x+3)(2 + \sqrt{x^2-5})} = \lim_{x \to -3} \frac{9 - x^2}{(x+3)(2 + \sqrt{x^2-5})}\]
3. Factor the numerator: \[= \lim_{x \to -3} \frac{(3-x)(3+x)}{(x+3)(2 + \sqrt{x^2-5})} = \lim_{x \to -3} \frac{-(x-3)(x+3)}{(x+3)(2 + \sqrt{x^2-5})}\]
4. Cancel \((x+3)\) and evaluate: \[= \lim_{x \to -3} \frac{-(x-3)}{2 + \sqrt{x^2-5}} = \frac{-(-3-3)}{2 + \sqrt{9-5}} = \frac{6}{2 + 2} = \frac{6}{4} = \frac{3}{2}\]

Answer: \(\frac{3}{2}\)

4.21. Another Rationalization Problem (Chapter 5, Example 7)

Evaluate \(\displaystyle \lim_{x \to 3} \frac{x^2-9}{\sqrt{x^2+7} - 4}\).

Click to see the solution

1. Multiply by the conjugate of the denominator: \[\lim_{x \to 3} \frac{x^2-9}{\sqrt{x^2+7} - 4} \cdot \frac{\sqrt{x^2+7} + 4}{\sqrt{x^2+7} + 4}\]
2. Simplify: \[= \lim_{x \to 3} \frac{(x^2-9)(\sqrt{x^2+7} + 4)}{(x^2+7) - 16} = \lim_{x \to 3} \frac{(x^2-9)(\sqrt{x^2+7} + 4)}{x^2-9}\]
3. Cancel \((x^2-9)\) and evaluate: \[= \lim_{x \to 3} (\sqrt{x^2+7} + 4) = \sqrt{9+7} + 4 = 4 + 4 = 8\]

Answer: \(8\)

4.22. Cube Root Rationalization (Chapter 5, Example 8)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{\sqrt[3]{x+1} - 1}{x}\).

Click to see the solution

Key Concept: Use the identity \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). Here, let \(a = \sqrt[3]{x+1}\) and \(b = 1\).

1. Multiply by the conjugate factor: \[\lim_{x \to 0} \frac{\sqrt[3]{x+1} - 1}{x} \cdot \frac{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}\]
2. Use the difference of cubes in the numerator: \[= \lim_{x \to 0} \frac{(\sqrt[3]{x+1})^3 - 1^3}{x((\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1)} = \lim_{x \to 0} \frac{(x+1) - 1}{x((\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1)}\]
3. Simplify and evaluate: \[= \lim_{x \to 0} \frac{x}{x((\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1)} = \lim_{x \to 0} \frac{1}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1} + 1}\] \[= \frac{1}{1 + 1 + 1} = \frac{1}{3}\]

Answer: \(\frac{1}{3}\)

4.23. One-Sided Limit from the Right (Chapter 5, Example 9)

Evaluate \(\displaystyle \lim_{x \to -2^+} \frac{(x+3)|x+2|}{x+2}\).

Click to see the solution

Key Concept: As \(x \to -2^+\), we have \(x > -2\), so \(x + 2 > 0\), which means \(|x+2| = x+2\).

1. Simplify using \(|x+2| = x+2\) when \(x > -2\): \[\frac{(x+3)|x+2|}{x+2} = \frac{(x+3)(x+2)}{x+2} = x+3\]
2. Evaluate the limit: \[\lim_{x \to -2^+} (x+3) = -2 + 3 = 1\]

Answer: \(1\)

4.24. One-Sided Limit from the Left (Chapter 5, Example 10)

Evaluate \(\displaystyle \lim_{x \to -2^-} \frac{(x+3)|x+2|}{x+2}\).

Click to see the solution

Key Concept: As \(x \to -2^-\), we have \(x < -2\), so \(x + 2 < 0\), which means \(|x+2| = -(x+2)\).

1. Simplify using \(|x+2| = -(x+2)\) when \(x < -2\): \[\frac{(x+3)|x+2|}{x+2} = \frac{(x+3)(-(x+2))}{x+2} = -(x+3)\]
2. Evaluate the limit: \[\lim_{x \to -2^-} (-(x+3)) = -(-2+3) = -1\]

Answer: \(-1\)

4.25. Left-Hand Limit with Nested Radicals (Chapter 5, Example 11)

Evaluate \(\displaystyle \lim_{h \to 0^-} \frac{\sqrt{6} - \sqrt{5h^2+11h+6}}{h}\).

Click to see the solution

1. Multiply by the conjugate: \[\lim_{h \to 0^-} \frac{\sqrt{6} - \sqrt{5h^2+11h+6}}{h} \cdot \frac{\sqrt{6} + \sqrt{5h^2+11h+6}}{\sqrt{6} + \sqrt{5h^2+11h+6}}\]
2. Simplify: \[= \lim_{h \to 0^-} \frac{6 - (5h^2+11h+6)}{h(\sqrt{6} + \sqrt{5h^2+11h+6})} = \lim_{h \to 0^-} \frac{-5h^2-11h}{h(\sqrt{6} + \sqrt{5h^2+11h+6})}\]
3. Factor and cancel \(h\): \[= \lim_{h \to 0^-} \frac{h(-5h-11)}{h(\sqrt{6} + \sqrt{5h^2+11h+6})} = \lim_{h \to 0^-} \frac{-5h-11}{\sqrt{6} + \sqrt{5h^2+11h+6}}\]
4. Evaluate: \[= \frac{-11}{\sqrt{6} + \sqrt{6}} = \frac{-11}{2\sqrt{6}} = -\frac{11\sqrt{6}}{12}\]

Answer: \(-\frac{11\sqrt{6}}{12}\)

4.26. Tangent Limit (Chapter 5, Example 12)

Evaluate \(\displaystyle \lim_{\theta \to 0} \frac{\tan \theta}{\theta}\).

Click to see the solution

Key Concept: Use \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and known limits.

1. Rewrite: \[\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = \lim_{\theta \to 0} \frac{\sin \theta}{\theta \cos \theta} = \lim_{\theta \to 0} \frac{\sin \theta}{\theta} \cdot \frac{1}{\cos \theta}\]
2. Apply limit laws: \[= \left(\lim_{\theta \to 0} \frac{\sin \theta}{\theta}\right) \cdot \left(\lim_{\theta \to 0} \frac{1}{\cos \theta}\right) = 1 \cdot \frac{1}{1} = 1\]

Answer: \(1\)

4.27. Ratio of Sine Functions (Chapter 5, Example 13)

Evaluate \(\displaystyle \lim_{y \to 0} \frac{\sin 3y}{\sin 5y}\).

Click to see the solution

Key Concept: Multiply and divide to create the standard limit form.

1. Rewrite: \[\lim_{y \to 0} \frac{\sin 3y}{\sin 5y} = \lim_{y \to 0} \frac{\sin 3y}{3y} \cdot \frac{5y}{\sin 5y} \cdot \frac{3y}{5y}\]
2. Use substitutions: Let \(u = 3y\) and \(v = 5y\). As \(y \to 0\), both \(u \to 0\) and \(v \to 0\). \[= \lim_{u \to 0} \frac{\sin u}{u} \cdot \lim_{v \to 0} \frac{v}{\sin v} \cdot \frac{3}{5} = 1 \cdot 1 \cdot \frac{3}{5} = \frac{3}{5}\]

Answer: \(\frac{3}{5}\)

4.28. Tangent and Sine Combined (Chapter 5, Example 14)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{\tan 3x}{\sin 8x}\).

Click to see the solution

1. Rewrite tangent: \[\lim_{x \to 0} \frac{\tan 3x}{\sin 8x} = \lim_{x \to 0} \frac{\sin 3x}{\cos 3x \cdot \sin 8x}\]
2. Rearrange: \[= \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot \frac{8x}{\sin 8x} \cdot \frac{3x}{8x} \cdot \frac{1}{\cos 3x}\]
3. Evaluate: \[= 1 \cdot 1 \cdot \frac{3}{8} \cdot \frac{1}{1} = \frac{3}{8}\]

Answer: \(\frac{3}{8}\)

4.29. Nested Trigonometric Functions (Chapter 5, Example 15)

Evaluate \(\displaystyle \lim_{t \to 0} \frac{\sin(1-\cos t)}{2 - 2\cos t}\).

Click to see the solution

1. Factor the denominator: \[\lim_{t \to 0} \frac{\sin(1-\cos t)}{2(1 - \cos t)}\]
2. Let \(u = 1 - \cos t\). As \(t \to 0\), \(u \to 0\): \[= \lim_{u \to 0} \frac{\sin u}{2u} = \frac{1}{2} \lim_{u \to 0} \frac{\sin u}{u} = \frac{1}{2} \cdot 1 = \frac{1}{2}\]

Answer: \(\frac{1}{2}\)

4.30. Sine of Hyperbolic Sine (Chapter 5, Example 16)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{\sin(\sinh x)}{2\sinh x}\).

Click to see the solution

Key Concept: Use the fact that \(\lim_{x \to 0} \frac{\sinh x}{x} = 1\).

1. Let \(u = \sinh x\). As \(x \to 0\), \(u \to 0\): \[\lim_{x \to 0} \frac{\sin(\sinh x)}{2\sinh x} = \lim_{u \to 0} \frac{\sin u}{2u} = \frac{1}{2} \lim_{u \to 0} \frac{\sin u}{u} = \frac{1}{2}\]

Answer: \(\frac{1}{2}\)

4.31. Product of Trigonometric Functions (Chapter 5, Example 17)

Evaluate \(\displaystyle \lim_{x \to 0} 8x^2 (\cot x)(\csc x)\).

Click to see the solution

1. Rewrite in terms of sine and cosine: \[\lim_{x \to 0} 8x^2 \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = \lim_{x \to 0} \frac{8x^2 \cos x}{\sin^2 x}\]
2. Rearrange: \[= \lim_{x \to 0} 8 \left(\frac{x}{\sin x}\right)^2 \cos x = 8 \cdot 1^2 \cdot 1 = 8\]

Answer: \(8\)

4.32. Sine Squared in Denominator (Chapter 5, Example 18)

Evaluate \(\displaystyle \lim_{x \to 0} \frac{x - x\cos x}{\sin^2 3x}\).

Click to see the solution

1. Factor the numerator: \[\lim_{x \to 0} \frac{x(1 - \cos x)}{\sin^2 3x}\]
2. Use the identity \(1 - \cos x = 2\sin^2(x/2)\): \[= \lim_{x \to 0} \frac{2x\sin^2(x/2)}{\sin^2 3x} = \lim_{x \to 0} 2x \cdot \frac{\sin^2(x/2)}{\sin^2 3x}\]
3. Rearrange: \[= \lim_{x \to 0} 2x \cdot \left(\frac{\sin(x/2)}{x/2}\right)^2 \cdot \left(\frac{3x}{\sin 3x}\right)^2 \cdot \frac{(x/2)^2}{(3x)^2}\] \[= \lim_{x \to 0} 2x \cdot 1^2 \cdot 1^2 \cdot \frac{1/4}{9} = \lim_{x \to 0} 2x \cdot \frac{1}{36} = \lim_{x \to 0} \frac{x}{18} = 0\]

Answer: \(0\)

4.33. Product of Trigonometric Ratios (Chapter 5, Example 19)

Evaluate \(\displaystyle \lim_{z \to 0} \frac{\sin(3z) \cot(5z)}{z \cot(4z)}\).

Click to see the solution

1. Rewrite cotangents: \[\lim_{z \to 0} \frac{\sin(3z) \cdot \frac{\cos 5z}{\sin 5z}}{z \cdot \frac{\cos 4z}{\sin 4z}} = \lim_{z \to 0} \frac{\sin(3z) \cos(5z) \sin(4z)}{z \sin(5z) \cos(4z)}\]
2. Rearrange: \[= \lim_{z \to 0} \frac{\sin 3z}{3z} \cdot \frac{4z}{\sin 4z} \cdot \frac{5z}{\sin 5z} \cdot \frac{3z \cdot 4z}{z \cdot 5z} \cdot \frac{\cos 5z}{\cos 4z}\] \[= 1 \cdot 1 \cdot 1 \cdot \frac{12}{5} \cdot \frac{1}{1} = \frac{12}{5}\]

Answer: \(\frac{12}{5}\)

4.34. Limit of Rational Function at Infinity (Chapter 5, Example 20)

Find \(\displaystyle \lim_{x \to \infty} \frac{9x^4 + x}{6 - x + 5x^2 + 2x^4}\) and \(\displaystyle \lim_{x \to -\infty} \frac{9x^4 + x}{6 - x + 5x^2 + 2x^4}\).

Click to see the solution

Key Concept: Divide numerator and denominator by the highest power of \(x\).

1. Divide by \(x^4\): \[\lim_{x \to \infty} \frac{9x^4 + x}{6 - x + 5x^2 + 2x^4} = \lim_{x \to \infty} \frac{9 + \frac{1}{x^3}}{\frac{6}{x^4} - \frac{1}{x^3} + \frac{5}{x^2} + 2}\]
2. Evaluate as \(x \to \infty\): \[= \frac{9 + 0}{0 - 0 + 0 + 2} = \frac{9}{2}\]
3. For \(x \to -\infty\), the same calculation applies: \[\lim_{x \to -\infty} \frac{9x^4 + x}{6 - x + 5x^2 + 2x^4} = \frac{9}{2}\]

Answer: Both limits equal \(\frac{9}{2}\)

4.35. Limit at Infinity (Higher Degree in Denominator) (Chapter 5, Example 21)

Find \(\displaystyle \lim_{x \to \infty} \frac{10x^5 + x^4 + 3}{x^6 - 2x^2 + 5}\) and \(\displaystyle \lim_{x \to -\infty} \frac{10x^5 + x^4 + 3}{x^6 - 2x^2 + 5}\).

Click to see the solution

1. Divide by \(x^6\): \[\lim_{x \to \infty} \frac{10x^5 + x^4 + 3}{x^6 - 2x^2 + 5} = \lim_{x \to \infty} \frac{\frac{10}{x} + \frac{1}{x^2} + \frac{3}{x^6}}{1 - \frac{2}{x^4} + \frac{5}{x^6}} = \frac{0}{1} = 0\]
2. For \(x \to -\infty\): \[\lim_{x \to -\infty} \frac{10x^5 + x^4 + 3}{x^6 - 2x^2 + 5} = 0\]

Answer: Both limits equal \(0\)

4.36. Limit at Infinity (Higher Degree in Numerator) (Chapter 5, Example 22)

Find \(\displaystyle \lim_{x \to \infty} \frac{5x^8 - 2x^3 + 9}{3 + x - 4x^5}\) and \(\displaystyle \lim_{x \to -\infty} \frac{5x^8 - 2x^3 + 9}{3 + x - 4x^5}\).

Click to see the solution

1. Divide by \(x^8\): \[\lim_{x \to \infty} \frac{5x^8 - 2x^3 + 9}{3 + x - 4x^5} = \lim_{x \to \infty} \frac{5 - \frac{2}{x^5} + \frac{9}{x^8}}{\frac{3}{x^8} + \frac{1}{x^7} - \frac{4}{x^3}}\]
2. As \(x \to \infty\), the numerator approaches \(5\) and denominator approaches \(0^-\): \[= -\infty\]
3. For \(x \to -\infty\), numerator approaches \(5\) and denominator approaches \(0^+\): \[= \infty\]

Answer: \(\lim_{x \to \infty} = -\infty\), \(\lim_{x \to -\infty} = \infty\)

4.37. Square Root Limit at Infinity (Chapter 5, Example 23)

Evaluate \(\displaystyle \lim_{x \to \infty} \sqrt{\frac{8x^2 - 3}{2x^2 + x}}\).

Click to see the solution

1. Move the limit inside (since square root is continuous): \[\lim_{x \to \infty} \sqrt{\frac{8x^2 - 3}{2x^2 + x}} = \sqrt{\lim_{x \to \infty} \frac{8x^2 - 3}{2x^2 + x}}\]
2. Divide by \(x^2\): \[= \sqrt{\lim_{x \to \infty} \frac{8 - \frac{3}{x^2}}{2 + \frac{1}{x}}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2\]

Answer: \(2\)

4.38. Cube Root Limit as \(x \to -\infty\) (Chapter 5, Example 24)

Evaluate \(\displaystyle \lim_{x \to -\infty} \left(\frac{x^2 + x - 1}{8x^2 - 3}\right)^{1/3}\).

Click to see the solution

1. Move limit inside: \[\lim_{x \to -\infty} \left(\frac{x^2 + x - 1}{8x^2 - 3}\right)^{1/3} = \left(\lim_{x \to -\infty} \frac{x^2 + x - 1}{8x^2 - 3}\right)^{1/3}\]
2. Divide by \(x^2\): \[= \left(\lim_{x \to -\infty} \frac{1 + \frac{1}{x} - \frac{1}{x^2}}{8 - \frac{3}{x^2}}\right)^{1/3} = \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2}\]

Answer: \(\frac{1}{2}\)

4.39. Fifth Power Limit (Chapter 5, Example 25)

Evaluate \(\displaystyle \lim_{x \to -\infty} \left(\frac{1 - x^3}{x^2 + 7x}\right)^5\).

Click to see the solution

1. Move limit inside: \[\left(\lim_{x \to -\infty} \frac{1 - x^3}{x^2 + 7x}\right)^5\]

2. Divide by \(x^3\): \[= \left(\lim_{x \to -\infty} \frac{\frac{1}{x^3} - 1}{\frac{1}{x} + \frac{7}{x^2}}\right)^5 = \left(\frac{0 - 1}{0 + 0}\right)^5\]

   The denominator approaches 0 while numerator approaches \(-1\), so the fraction goes to \(-\infty\).

Answer: \(-\infty\) (or more precisely, the limit does not exist as a finite number)

4.40. Mixed Root and Polynomial (Chapter 5, Example 26)

Evaluate \(\displaystyle \lim_{x \to \infty} \frac{2\sqrt{x} + x^{-1}}{3x - 7}\).

Click to see the solution

1. Divide by \(x\): \[\lim_{x \to \infty} \frac{2\sqrt{x} + x^{-1}}{3x - 7} = \lim_{x \to \infty} \frac{\frac{2}{\sqrt{x}} + \frac{1}{x^2}}{3 - \frac{7}{x}} = \frac{0 + 0}{3 - 0} = 0\]

Answer: \(0\)

4.41. Square Root Ratio (Chapter 5, Example 27)

Evaluate \(\displaystyle \lim_{x \to \infty} \frac{2 + \sqrt{x}}{2 - \sqrt{x}}\).

Click to see the solution

1. Divide numerator and denominator by \(\sqrt{x}\): \[\lim_{x \to \infty} \frac{2 + \sqrt{x}}{2 - \sqrt{x}} = \lim_{x \to \infty} \frac{\frac{2}{\sqrt{x}} + 1}{\frac{2}{\sqrt{x}} - 1} = \frac{0 + 1}{0 - 1} = -1\]

Answer: \(-1\)

4.42. Cube and Fifth Root (Chapter 5, Example 28)

Evaluate \(\displaystyle \lim_{x \to -\infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}\).

Click to see the solution

1. For \(x < 0\), write \(x = -|x|\). As \(x \to -\infty\), \(|x| \to \infty\): \[\sqrt[3]{x} = -\sqrt[3]{|x|}, \quad \sqrt[5]{x} = -\sqrt[5]{|x|}\]
2. Substitute: \[\lim_{x \to -\infty} \frac{-\sqrt[3]{|x|} - (-\sqrt[5]{|x|})}{-\sqrt[3]{|x|} + (-\sqrt[5]{|x|})} = \lim_{|x| \to \infty} \frac{-\sqrt[3]{|x|} + \sqrt[5]{|x|]}{-\sqrt[3]{|x|} - \sqrt[5]{|x|}}\]
3. Divide by \(\sqrt[3]{|x|}\) (higher order): \[= \lim_{|x| \to \infty} \frac{-1 + \frac{1}{|x|^{3/5 - 1/3}}}{-1 - \frac{1}{|x|^{1/3 - 1/5}}} = \lim_{|x| \to \infty} \frac{-1 + \frac{1}{|x|^{2/15}}}{-1 - \frac{1}{|x|^{2/15}}} = \frac{-1 + 0}{-1 - 0} = 1\]

Answer: \(1\)

4.43. Negative Powers (Chapter 5, Example 29)

Evaluate \(\displaystyle \lim_{x \to \infty} \frac{x^{-1} + x^{-4}}{x^{-2} - x^{-3}}\).

Click to see the solution

1. Multiply numerator and denominator by \(x^4\): \[\lim_{x \to \infty} \frac{x^{-1} + x^{-4}}{x^{-2} - x^{-3}} = \lim_{x \to \infty} \frac{x^3 + 1}{x^2 - x}\]
2. Divide by \(x^3\): \[= \lim_{x \to \infty} \frac{1 + \frac{1}{x^3}}{\frac{1}{x} - \frac{1}{x^2}} = \frac{1}{0} = \infty\]

Answer: \(\infty\)

4.44. Fractional Powers (Chapter 5, Example 30)

Evaluate \(\displaystyle \lim_{x \to \infty} \frac{2x^{5/3} - x^{1/3} + 7}{x^{8/5} + 3x + \sqrt{x}}\).

Click to see the solution

1. Identify the highest power in each:

   • Numerator: \(x^{5/3} \approx x^{1.667}\) is highest
   • Denominator: \(x^{8/5} = x^{1.6}\) is highest (since \(x^{8/5} > x^1 = x\) and \(x^{8/5} > x^{1/2} = \sqrt{x}\))

2. Divide by \(x^{5/3}\): \[= \lim_{x \to \infty} \frac{2 - x^{-4/3} + 7x^{-5/3}}{x^{8/5 - 5/3} + 3x^{1 - 5/3} + x^{1/2 - 5/3}}\] \[= \lim_{x \to \infty} \frac{2 - x^{-4/3} + 7x^{-5/3}}{x^{-1/15} + 3x^{-2/3} + x^{-7/6}}\]

   Since numerator \(\to 2\) and denominator \(\to 0^+\):

Answer: \(\infty\)

4.45. Polynomial Under Square Root (Chapter 5, Example 31)

Evaluate \(\displaystyle \lim_{x \to -\infty} \frac{4 - 3x^3}{\sqrt{x^6 + 9}}\).

Click to see the solution

1. For \(x < 0\), \(\sqrt{x^6} = |x^3| = -x^3\): \[\lim_{x \to -\infty} \frac{4 - 3x^3}{\sqrt{x^6 + 9}} = \lim_{x \to -\infty} \frac{4 - 3x^3}{-x^3\sqrt{1 + \frac{9}{x^6}}}\]
2. Divide numerator by \(x^3\): \[= \lim_{x \to -\infty} \frac{\frac{4}{x^3} - 3}{-\sqrt{1 + \frac{9}{x^6}}} = \frac{0 - 3}{-\sqrt{1 + 0}} = \frac{-3}{-1} = 3\]

Answer: \(3\)

4.46. One-Sided Limits at a Point (Chapter 5, Example 32)

Calculate \(\displaystyle \lim_{x \to ?} \frac{x^2 - 3x + 2}{x^3 - 2x^2}\) as \(x \to 2^+\), \(x \to 2^-\), \(x \to 2\), \(x \to 0^+\), \(x \to 0\).

Click to see the solution

1. Factor: \[\frac{x^2 - 3x + 2}{x^3 - 2x^2} = \frac{(x-1)(x-2)}{x^2(x-2)} = \frac{x-1}{x^2} \quad (x \neq 2)\]
2. For \(x \to 2^+\), \(x \to 2^-\), and \(x \to 2\): \[\lim_{x \to 2} \frac{x-1}{x^2} = \frac{2-1}{4} = \frac{1}{4}\]
3. For \(x \to 0^+\): \[\lim_{x \to 0^+} \frac{x-1}{x^2} = \frac{-1}{0^+} = -\infty\]
4. For \(x \to 0\): Since both one-sided limits are \(-\infty\): \[\lim_{x \to 0} \frac{x-1}{x^2} = -\infty\]

Answer:

• \(x \to 2^+\), \(x \to 2^-\), \(x \to 2\): all equal \(\frac{1}{4}\)
• \(x \to 0^+\): \(-\infty\)
• \(x \to 0\): \(-\infty\)

4.47. Difference of Square Roots (Chapter 5, Example 33)

Calculate \(\displaystyle \lim_{x \to \infty} (\sqrt{x+9} - \sqrt{x+4})\).

Click to see the solution

Key Concept: Multiply by the conjugate to rationalize.

1. Multiply by conjugate: \[\lim_{x \to \infty} (\sqrt{x+9} - \sqrt{x+4}) \cdot \frac{\sqrt{x+9} + \sqrt{x+4}}{\sqrt{x+9} + \sqrt{x+4}}\]
2. Simplify: \[= \lim_{x \to \infty} \frac{(x+9) - (x+4)}{\sqrt{x+9} + \sqrt{x+4}} = \lim_{x \to \infty} \frac{5}{\sqrt{x+9} + \sqrt{x+4}}\]
3. As \(x \to \infty\), denominator \(\to \infty\): \[= 0\]

Answer: \(0\)

4.48. Difference of Square Roots with Same Leading Term (Chapter 5, Example 34)

Calculate \(\displaystyle \lim_{x \to \infty} (\sqrt{x^2+25} - \sqrt{x^2-1})\).

Click to see the solution

1. Multiply by conjugate: \[\lim_{x \to \infty} \frac{(\sqrt{x^2+25} - \sqrt{x^2-1})(\sqrt{x^2+25} + \sqrt{x^2-1})}{\sqrt{x^2+25} + \sqrt{x^2-1}}\]
2. Simplify: \[= \lim_{x \to \infty} \frac{(x^2+25) - (x^2-1)}{\sqrt{x^2+25} + \sqrt{x^2-1}} = \lim_{x \to \infty} \frac{26}{\sqrt{x^2+25} + \sqrt{x^2-1}} = 0\]

Answer: \(0\)

4.49. Sum with Square Root (Negative Infinity) (Chapter 5, Example 35)

Calculate \(\displaystyle \lim_{x \to -\infty} (\sqrt{x^2+3} + x)\).

Click to see the solution

1. For \(x < 0\), write \(\sqrt{x^2} = |x| = -x\). Multiply by conjugate: \[\lim_{x \to -\infty} (\sqrt{x^2+3} + x) \cdot \frac{\sqrt{x^2+3} - x}{\sqrt{x^2+3} - x}\]
2. Simplify: \[= \lim_{x \to -\infty} \frac{(x^2+3) - x^2}{\sqrt{x^2+3} - x} = \lim_{x \to -\infty} \frac{3}{\sqrt{x^2+3} - x}\]
3. For \(x \to -\infty\), write \(\sqrt{x^2+3} = -x\sqrt{1 + \frac{3}{x^2}}\) (since \(x < 0\)): \[= \lim_{x \to -\infty} \frac{3}{-x\sqrt{1 + \frac{3}{x^2}} - x} = \lim_{x \to -\infty} \frac{3}{-x(\sqrt{1 + \frac{3}{x^2}} + 1)}\]
4. As \(x \to -\infty\) (so \(x\) is large negative), \(-x\) is large positive: \[= \lim_{x \to -\infty} \frac{3}{-x \cdot 2} = 0\]

Answer: \(0\)

4.50. Linear Plus Square Root (Chapter 5, Example 36)

Calculate \(\displaystyle \lim_{x \to -\infty} (2x + \sqrt{4x^2+3x-2})\).

Click to see the solution

1. Multiply by conjugate: \[\lim_{x \to -\infty} (2x + \sqrt{4x^2+3x-2}) \cdot \frac{2x - \sqrt{4x^2+3x-2}}{2x - \sqrt{4x^2+3x-2}}\]
2. Simplify: \[= \lim_{x \to -\infty} \frac{4x^2 - (4x^2+3x-2)}{2x - \sqrt{4x^2+3x-2}} = \lim_{x \to -\infty} \frac{-3x+2}{2x - \sqrt{4x^2+3x-2}}\]
3. For \(x \to -\infty\), \(\sqrt{4x^2} = 2|x| = -2x\): \[= \lim_{x \to -\infty} \frac{-3x+2}{2x - (-2x)\sqrt{1 + \frac{3}{4x} - \frac{2}{4x^2}}}\]
4. Divide by \(x\) (noting \(x < 0\)): \[= \lim_{x \to -\infty} \frac{-3 + \frac{2}{x}}{2 + 2\sqrt{1 + \frac{3}{4x} - \frac{1}{2x^2}}} = \frac{-3}{2 + 2} = -\frac{3}{4}\]

Answer: \(-\frac{3}{4}\)

4.51. Quadratic Minus Linear Under Root (Chapter 5, Example 37)

Calculate \(\displaystyle \lim_{x \to \infty} (\sqrt{9x^2-x} - 3x)\).

Click to see the solution

1. Multiply by conjugate: \[\lim_{x \to \infty} \frac{(\sqrt{9x^2-x} - 3x)(\sqrt{9x^2-x} + 3x)}{\sqrt{9x^2-x} + 3x}\]
2. Simplify: \[= \lim_{x \to \infty} \frac{9x^2 - x - 9x^2}{\sqrt{9x^2-x} + 3x} = \lim_{x \to \infty} \frac{-x}{\sqrt{9x^2-x} + 3x}\]
3. Divide by \(x\) (for \(x > 0\), \(\sqrt{x^2} = x\)): \[= \lim_{x \to \infty} \frac{-1}{\sqrt{9 - \frac{1}{x}} + 3} = \frac{-1}{3 + 3} = -\frac{1}{6}\]

Answer: \(-\frac{1}{6}\)

4.52. Difference of Square Roots with Different Powers (Chapter 5, Example 38)

Calculate \(\displaystyle \lim_{x \to \infty} (\sqrt{x^2+3x} - \sqrt{x^2-2x})\).

Click to see the solution

1. Multiply by conjugate: \[\lim_{x \to \infty} \frac{(x^2+3x) - (x^2-2x)}{\sqrt{x^2+3x} + \sqrt{x^2-2x}} = \lim_{x \to \infty} \frac{5x}{\sqrt{x^2+3x} + \sqrt{x^2-2x}}\]
2. Divide by \(x\): \[= \lim_{x \to \infty} \frac{5}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 - \frac{2}{x}}} = \frac{5}{1 + 1} = \frac{5}{2}\]

Answer: \(\frac{5}{2}\)

4.53. Find All Asymptotes (Chapter 5, Example 39)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^2}{x-1}\).

Click to see the solution

1. Vertical asymptote: Domain excludes \(x = 1\). Check limits: \[\lim_{x \to 1^+} \frac{x^2}{x-1} = \frac{1}{0^+} = \infty\] \[\lim_{x \to 1^-} \frac{x^2}{x-1} = \frac{1}{0^-} = -\infty\]

   Vertical asymptote: \(x = 1\)

2. Horizontal asymptote: Check limits at infinity: \[\lim_{x \to \pm\infty} \frac{x^2}{x-1} = \lim_{x \to \pm\infty} \frac{x}{1 - \frac{1}{x}} = \pm\infty\]

   No horizontal asymptote.

3. Oblique asymptote: Perform polynomial division: \[\frac{x^2}{x-1} = x + 1 + \frac{1}{x-1}\]

   As \(x \to \pm\infty\), \(\frac{1}{x-1} \to 0\), so the oblique asymptote is \(y = x + 1\).

Answer: Vertical: \(x = 1\); Oblique: \(y = x + 1\)

4.54. Asymptotes of Rational Function (Chapter 5, Example 40)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^2+1}{x-1}\).

Click to see the solution

1. Vertical asymptote: \(x = 1\) (same analysis as previous problem)

2. Oblique asymptote: Polynomial division: \[\frac{x^2+1}{x-1} = x + 1 + \frac{2}{x-1}\]

   Oblique asymptote: \(y = x + 1\)

Answer: Vertical: \(x = 1\); Oblique: \(y = x + 1\)

4.55. Another Asymptote Problem (Chapter 5, Example 41)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^2-4}{x-1}\).

Click to see the solution

1. Vertical asymptote: \(x = 1\)

2. Oblique asymptote: Polynomial division: \[\frac{x^2-4}{x-1} = x + 1 - \frac{3}{x-1}\]

   Oblique asymptote: \(y = x + 1\)

Answer: Vertical: \(x = 1\); Oblique: \(y = x + 1\)

4.56. Asymptotes with Factored Denominator (Chapter 5, Example 42)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^2-1}{2x+4}\).

Click to see the solution

1. Vertical asymptote: \(2x + 4 = 0 \implies x = -2\)

2. Oblique asymptote: Polynomial division: \[\frac{x^2-1}{2x+4} = \frac{x^2-1}{2(x+2)} = \frac{1}{2}x - 1 + \frac{1}{2(x+2)}\]

   Oblique asymptote: \(y = \frac{1}{2}x - 1\)

Answer: Vertical: \(x = -2\); Oblique: \(y = \frac{x}{2} - 1\)

4.57. Asymptotes of \(\frac{x^2-1}{x}\) (Chapter 5, Example 43)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^2-1}{x}\).

Click to see the solution

1. Vertical asymptote: \(x = 0\) \[\lim_{x \to 0^+} \frac{x^2-1}{x} = \frac{-1}{0^+} = -\infty\]

2. Oblique asymptote: Simplify: \[\frac{x^2-1}{x} = x - \frac{1}{x}\]

   As \(x \to \pm\infty\), \(\frac{1}{x} \to 0\), so oblique asymptote: \(y = x\)

Answer: Vertical: \(x = 0\); Oblique: \(y = x\)

4.58. Asymptotes with Cubic Numerator (Chapter 5, Example 44)

Find the equations of all asymptotes of \(\displaystyle f(x) = \frac{x^3+1}{x^2}\).

Click to see the solution

1. Vertical asymptote: \(x = 0\)

2. Oblique asymptote: Polynomial division: \[\frac{x^3+1}{x^2} = x + \frac{1}{x^2}\]

   As \(x \to \pm\infty\), \(\frac{1}{x^2} \to 0\), so oblique asymptote: \(y = x\)

Answer: Vertical: \(x = 0\); Oblique: \(y = x\)

4.59. Limit with Exponential Form \((1+\frac{1}{x})^x\) (Chapter 5, Example 45)

Find \(\displaystyle L_1 = \lim_{x \to \infty} \left(\frac{x+2}{x+1}\right)^x\).

Click to see the solution

Key Concept: Rewrite in the form \((1 + \frac{a}{t})^t\).

1. Rewrite: \[L_1 = \lim_{x \to \infty} \left(\frac{x+2}{x+1}\right)^x = \lim_{x \to \infty} \left(1 + \frac{1}{x+1}\right)^x\]
2. Substitute \(t = x + 1\), so \(x = t - 1\). As \(x \to \infty\), \(t \to \infty\): \[L_1 = \lim_{t \to \infty} \left(1 + \frac{1}{t}\right)^{t-1} = \lim_{t \to \infty} \frac{\left(1 + \frac{1}{t}\right)^t}{\left(1 + \frac{1}{t}\right)}\]
3. Evaluate: \[= \frac{\lim_{t \to \infty} \left(1 + \frac{1}{t}\right)^t}{\lim_{t \to \infty} \left(1 + \frac{1}{t}\right)} = \frac{e}{1} = e\]

Answer: \(e\)

4.60. Limit with \((1 + \frac{1}{x^2})^x\) (Chapter 5, Example 46)

Find \(\displaystyle L_2 = \lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^x\).

Click to see the solution

Key Concept: Use the relationship \(a^b = e^{b \ln a}\).

1. Rewrite: \[L_2 = \lim_{x \to \infty} \left[\left(1 + \frac{1}{x^2}\right)^{x^2}\right]^{1/x}\]
2. Note that \(\lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^{x^2} = e\): \[L_2 = \lim_{x \to \infty} e^{1/x} = e^0 = 1\]

Answer: \(1\)

4.61. Complex Exponential Limit (Chapter 5, Example 47)

Find \(\displaystyle L_3 = \lim_{x \to \infty} \left(\frac{5x-2}{2x+1}\right)^{2x-1}\).

Click to see the solution

1. Use the exponential form: \[L_3 = \lim_{x \to \infty} e^{(2x-1) \ln\left(\frac{5x-2}{2x+1}\right)}\]
2. Evaluate the limit in the exponent: \[\lim_{x \to \infty} (2x-1) \cdot \ln\left(\frac{5x-2}{2x+1}\right) = \lim_{x \to \infty} (2x-1) \cdot \ln\left(\frac{5 - \frac{2}{x}}{2 + \frac{1}{x}}\right)\] \[= \lim_{x \to \infty} (2x-1) \cdot \ln\left(\frac{5}{2}\right) = \infty \cdot \ln\left(\frac{5}{2}\right) = \infty\]
3. Since \(\ln(5/2) > 0\): \[L_3 = e^{\infty} = \infty\]

Answer: \(\infty\)

4.62. Indeterminate Form \(1^{\infty}\) Example 1 (Chapter 5, Example 48)

Find \(\displaystyle L_4 = \lim_{x \to \infty} \left(\frac{x^2+1}{x^2-2}\right)^{x^2}\).

Click to see the solution

Key Concept: For \(1^{\infty}\) form, use \(\lim_{x \to c} (g(x))^{h(x)} = e^{\lim_{x \to c} h(x)(g(x)-1)}\).

1. Let \(g(x) = \frac{x^2+1}{x^2-2}\) and \(h(x) = x^2\): \[L_4 = e^{\lim_{x \to \infty} x^2 \left(\frac{x^2+1}{x^2-2} - 1\right)}\]
2. Simplify the expression: \[\lim_{x \to \infty} x^2 \left(\frac{x^2+1 - x^2+2}{x^2-2}\right) = \lim_{x \to \infty} x^2 \cdot \frac{3}{x^2-2} = \lim_{x \to \infty} \frac{3x^2}{x^2-2}\] \[= \lim_{x \to \infty} \frac{3}{1 - \frac{2}{x^2}} = 3\]
3. Therefore: \[L_4 = e^3\]

Answer: \(e^3\)

4.63. Indeterminate Form \(1^{\infty}\) Example 2 (Chapter 5, Example 49)

Find \(\displaystyle L_5 = \lim_{x \to 0} \left(\frac{1+\tan x}{1+\sin x}\right)^{1/\sin x}\).

Click to see the solution

1. Let \(g(x) = \frac{1+\tan x}{1+\sin x}\) and \(h(x) = \frac{1}{\sin x}\): \[L_5 = e^{\lim_{x \to 0} \frac{1}{\sin x}\left(\frac{1+\tan x}{1+\sin x} - 1\right)}\]
2. Simplify: \[h(x)(g(x) - 1) = \frac{1}{\sin x} \cdot \frac{1+\tan x - 1 - \sin x}{1+\sin x} = \frac{\tan x - \sin x}{\sin x(1+\sin x)}\]
3. Rewrite tangent: \[= \frac{\frac{\sin x}{\cos x} - \sin x}{\sin x(1+\sin x)} = \frac{\sin x(1 - \cos x)}{\cos x \cdot \sin x(1+\sin x)} = \frac{1-\cos x}{\cos x(1+\sin x)}\]
4. Use \(1 - \cos x = 2\sin^2(x/2)\): \[\lim_{x \to 0} \frac{2\sin^2(x/2)}{\cos x(1+\sin x)} = \frac{0}{1 \cdot 1} = 0\]
5. Therefore: \[L_5 = e^0 = 1\]

Answer: \(1\)

4.64. Limit of \(\frac{a^x-1}{x}\) (Chapter 5, Example 50)

Find \(\displaystyle L_6 = \lim_{x \to 0} \frac{a^x-1}{x}\) for \(a > 0\).

Click to see the solution

Key Concept: Use the fact that \(a^x = e^{x \ln a}\) and known limits.

1. Note that: \[\frac{x}{a^x-1} = \frac{x}{e^{x\ln a} - 1}\]
2. Let \(t = x \ln a\). As \(x \to 0\), \(t \to 0\): \[\frac{x}{a^x-1} = \frac{t/\ln a}{e^t - 1} = \frac{1}{\ln a} \cdot \frac{t}{e^t-1}\]
3. We know that \(\lim_{t \to 0} \frac{e^t-1}{t} = 1\), so: \[\lim_{x \to 0} \frac{x}{a^x-1} = \frac{1}{\ln a} \cdot 1 = \frac{1}{\ln a}\]
4. Therefore: \[L_6 = \lim_{x \to 0} \frac{a^x-1}{x} = \ln a\]

Answer: \(\ln a\)