Calculus 1 Quiz: Infinite Limits And Vertical Asymptotes

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Infinite Limits And Vertical AsymptotesQuestion 1 of 20

For which of the following functions does $\lim_{x \to 2} f(x)$ exist and is finite, even though the expression for $f(x)$ has a denominator that is zero at $x=2$ ?

f(x) = \frac{x^2}{x-2}

f(x) = \frac{e^x}{(x-2)^2}

f(x) = \frac{x+2}{x^2-4}

f(x) = \frac{\sin(x-2)}{x^2-4}

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Calculus 1 Quiz

Calculus 1 Quiz: Infinite Limits And Vertical Asymptotes

Practice Infinite Limits And Vertical Asymptotes in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Infinite Limits And Vertical Asymptotes, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For which of the following functions does $\lim_{x \to 2} f(x)$ exist and is finite, even though the expression for $f(x)$ has a denominator that is zero at $x=2$ ?

$f(x) = \frac{x^2}{x-2}$
$f(x) = \frac{e^x}{(x-2)^2}$
$f(x) = \frac{x+2}{x^2-4}$
$f(x) = \frac{\sin(x-2)}{x^2-4}$ (correct answer)

Explanation: When you encounter a rational function where the denominator equals zero at a specific point, you're dealing with a potential removable discontinuity (hole) versus a vertical asymptote. The key is determining whether the numerator also approaches zero at that point, creating an indeterminate form that might have a finite limit. Let's examine each function at

x = 2

: For option D,

f(x) = \frac{\sin(x-2)}{x^2-4}

, both the numerator and denominator approach zero as

x \to 2

. Since

x^2-4 = (x-2)(x+2)

, we can rewrite this as

\frac{\sin(x-2)}{(x-2)(x+2)}

. Using the fundamental limit

\lim_{u \to 0} \frac{\sin u}{u} = 1

, we get:

\lim_{x \to 2} \frac{\sin(x-2)}{x^2-4} = \lim_{x \to 2} \frac{\sin(x-2)}{x-2} \cdot \frac{1}{x+2} = 1 \cdot \frac{1}{4} = \frac{1}{4}

Option A has

f(x) = \frac{x^2}{x-2}

. As

x \to 2

, the numerator approaches 4 while the denominator approaches 0, creating a vertical asymptote with an infinite limit. Option B has

f(x) = \frac{e^x}{(x-2)^2}

. The numerator approaches

e^2

while the denominator approaches 0, again yielding an infinite limit. Option C has

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

(after canceling

x+2

), which still has a vertical asymptote at

x = 2

. Remember: Look for cases where both numerator and denominator approach zero simultaneously—these often yield finite limits after applying L'Hôpital's rule or algebraic manipulation.

Question 2

Identify all vertical asymptotes of the function $g(x) = \frac{x^3 - 1}{x^2 + x - 2}$ .

$x=-2$ only (correct answer)
$x=1$ only
$x=-2$ and $x=1$
$x=-1$ and $x=2$

Explanation: To find vertical asymptotes, we first factor the numerator and the denominator. The numerator is a difference of cubes:

x^3 - 1 = (x-1)(x^2+x+1)

. The denominator is a quadratic:

x^2 + x - 2 = (x+2)(x-1)

. The function is

g(x) = \frac{(x-1)(x^2+x+1)}{(x+2)(x-1)}

. The factor

(x-1)

cancels, indicating a removable discontinuity (hole) at

x=1

. The simplified function is

g(x) = \frac{x^2+x+1}{x+2}

for

x \neq 1

. A vertical asymptote occurs where the denominator of the simplified function is zero, which is

x+2=0 \implies x=-2

Question 3

The function $f(x) = \frac{x^2+ax-6}{x^2-4}$ has a removable discontinuity at $x=2$ . What is the equation of its vertical asymptote?

$x = -3$
$x = -2$ (correct answer)
$x = 2$
$x = 1$

Explanation: For a rational function to have a removable discontinuity at

x=2

(x-2)

must be a factor of both the numerator and the denominator. This means the numerator must be zero when

x=2

. So,

2^2 + a(2) - 6 = 0 \implies 4 + 2a - 6 = 0 \implies 2a = 2 \implies a=1

. Substituting

a=1

into the function gives

f(x) = \frac{x^2+x-6}{x^2-4} = \frac{(x+3)(x-2)}{(x+2)(x-2)}

. After canceling the

(x-2)

terms, the simplified function is

\frac{x+3}{x+2}

. The vertical asymptote occurs where the denominator of the simplified function is zero, which is

x+2=0

, or

x=-2

Question 4

How many vertical asymptotes does the function $f(x) = \frac{x^2-4}{\sin(\pi x)}$ have on the open interval $(-5, 5)$ ?

5
7 (correct answer)
9
11

Explanation: Vertical asymptotes can occur where the denominator is zero. The denominator

\sin(\pi x)

is zero when

\pi x = n\pi

for any integer

n

, which means

x=n

. The integers in the interval

(-5, 5)

are

\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}

. This gives 9 potential vertical asymptotes. However, if the numerator is also zero at one of these x-values, the limit might be finite, resulting in a hole instead of an asymptote. The numerator

x^2-4=(x-2)(x+2)

is zero at

x=2

and

x=-2

. Let's check the limits at these two points using L'Hôpital's Rule or the fundamental limit

\lim_{\theta \to 0} \frac{\sin \theta}{\theta}=1

. At

x=2

\lim_{x \to 2} \frac{x^2-4}{\sin(\pi x)}

is of the form

\frac{0}{0}

. Using L'Hôpital's Rule:

\lim_{x \to 2} \frac{2x}{\pi \cos(\pi x)} = \frac{2(2)}{\pi \cos(2\pi)} = \frac{4}{\pi}

. Since the limit is finite, there is a hole at

x=2

. At

x=-2

\lim_{x \to -2} \frac{x^2-4}{\sin(\pi x)}

is of the form

\frac{0}{0}

. Using L'Hôpital's Rule:

\lim_{x \to -2} \frac{2x}{\pi \cos(\pi x)} = \frac{2(-2)}{\pi \cos(-2\pi)} = \frac{-4}{\pi}

. Since the limit is finite, there is a hole at

x=-2

. For all other integers in the set (

\{-4, -3, -1, 0, 1, 3, 4\}

), the numerator is non-zero while the denominator is zero. Therefore, these 7 values correspond to vertical asymptotes. The total count is 7.

Question 5

Let $f(x) = \frac{1}{x-3}$ and $g(x) = x^2-1$ . What are the equations of the vertical asymptotes of the composite function $h(x) = (f \circ g)(x)$ ?

$x = 3$
$x = 1$ and $x = -1$
$x = 2$ and $x = -2$ (correct answer)
$x = 3$ , $x = 1$ , and $x = -1$

Explanation: First, we find the expression for the composite function

h(x) = (f \circ g)(x) = f(g(x))

. We substitute

g(x)

into

f(x)

h(x) = f(x^2-1) = \frac{1}{(x^2-1)-3} = \frac{1}{x^2-4}

. Vertical asymptotes occur where the denominator is zero. We set the denominator equal to zero and solve for

x

x^2-4 = 0

(x-2)(x+2) = 0

x = 2

x = -2

. Since the numerator is a non-zero constant, both of these values correspond to vertical asymptotes. Distractors are based on finding the vertical asymptote of

f(x)

alone (

x=3

) or the roots of

g(x)

(

x=\pm 1

Question 6

A function $f(x)$ satisfies the following conditions:

$\lim_{x \to 2^+} f(x) = +\infty$
$\lim_{x \to 2^-} f(x) = -\infty$
$\lim_{x \to -3} f(x) = +\infty$ Which of the following could be an expression for $f(x)$ ?

$f(x) = \frac{1}{(x-2)(x+3)^2}$
$f(x) = \frac{x}{(x-2)^2(x+3)}$
$f(x) = \frac{x}{(x-2)(x+3)^2}$ (correct answer)
$f(x) = \frac{-x}{(x-2)(x+3)^2}$

Explanation: The conditions describe the behavior of

f(x)

near its vertical asymptotes. Conditions 1 and 2 imply a vertical asymptote at

x=2

. The fact that the sign of the limit is different from the left and right sides suggests that the factor

(x-2)

in the denominator has an odd power (like 1 or 3). Condition 3 implies a vertical asymptote at

x=-3

. The fact that the limit is

+\infty

(a two-sided limit) means the function approaches

+\infty

from both the left and right. This suggests the factor

(x+3)

in the denominator has an even power (like 2 or 4), which prevents a sign change. So, the denominator should look like

(x-2)^{odd}(x+3)^{even}

. Choices A, C, and D have denominators of the form

(x-2)(x+3)^2

. Let's test them. For choice C,

f(x) = \frac{x}{(x-2)(x+3)^2}

As $x \to 2^+$ , $f(x) \approx \frac{2}{(\text{small +})(\text{positive})} \to +\infty$ . (Matches condition 1)
As $x \to 2^-$ , $f(x) \approx \frac{2}{(\text{small -})(\text{positive})} \to -\infty$ . (Matches condition 2)
As $x \to -3$ , $f(x) \approx \frac{-3}{(\text{negative})(\text{small +})} \to +\infty$ . (Matches condition 3) All conditions are met by choice C. Choice A fails condition 3's sign check. Choice B has the wrong powers in the denominator. Choice D fails the sign checks for conditions 1 and 2.

Question 7

Let $f(x) = \frac{P(x)}{Q(x)}$ be a rational function where $P(x)$ and $Q(x)$ are non-zero polynomials. Which of the following statements is not always true?

If $Q(a)=0$ and $P(a) \neq 0$ , then $x=a$ is a vertical asymptote of $f(x)$ .
If $f(x)$ has a vertical asymptote at $x=a$ , then $Q(a)=0$ .
If $Q(a)=0$ and $P(a)=0$ , then $f(x)$ has a hole at $x=a$ . (correct answer)
The number of vertical asymptotes of $f(x)$ is less than or equal to the degree of $Q(x)$ .

Explanation: Let's analyze each statement. A) This is the fundamental condition for a vertical asymptote in a rational function. A non-zero numerator divided by a denominator approaching zero results in an infinite limit. This statement is always true. B) For a rational function to have an infinite limit at a finite value

x=a

, the denominator must approach zero. Since

Q(x)

is a polynomial, this means

Q(a)=0

. This statement is always true. C) If

Q(a)=0

and

P(a)=0

, it means

(x-a)

is a factor of both polynomials. This leads to an indeterminate form

\frac{0}{0}

. While this often results in a hole (removable discontinuity), it can also result in a vertical asymptote if the multiplicity of the root

a

is higher in the denominator

Q(x)

than in the numerator

P(x)

. For example, if

f(x) = \frac{(x-a)}{(x-a)^2} = \frac{1}{x-a}

, there is a vertical asymptote at

x=a

. Thus, this statement is not always true. D) The vertical asymptotes of a rational function can only occur at the real roots of the denominator polynomial

Q(x)

. A polynomial of degree

n

has at most

n

real roots. Therefore, the number of vertical asymptotes is at most the degree of

Q(x)

. This statement is always true.

Question 8

Let $f(x)$ be a polynomial. The graph of the function $g(x) = \frac{f(x)}{x^2-5x+6}$ is known to have vertical asymptotes at both $x=2$ and $x=3$ . Which of the following must be true about the polynomial $f(x)$ ?

$f(2) \neq 0$ and $f(3) \neq 0$ (correct answer)
$f(x)$ has no real roots between $x=2$ and $x=3$
The degree of $f(x)$ must be less than 2
$f(2) = 0$ and $f(3) = 0$

Explanation: The function is

g(x) = \frac{f(x)}{x^2-5x+6} = \frac{f(x)}{(x-2)(x-3)}

. The denominator is zero at

x=2

and

x=3

. The problem states that vertical asymptotes exist at both of these locations. A vertical asymptote exists at

x=a

for a rational function if the denominator is zero at

x=a

and the numerator is non-zero at

x=a

. If the numerator were also zero, there would be a hole (removable discontinuity) or potentially still an asymptote, but the existence of the asymptote requires that the factor in the denominator is not completely canceled out by a factor in the numerator. The most direct conclusion from the given information is that the numerator is not zero at these points, preventing cancellation. Therefore, it must be that

f(2) \neq 0

and

f(3) \neq 0

Question 9

Suppose $\lim_{x \to 0} f(x) = \infty$ . Which of the following limits must also be infinite?

I. $\lim_{x \to 0} (f(x) + \sin(x))$

II. $\lim_{x \to 0} (x \cdot f(x))$

III. $\lim_{x \to 0} (f(x))^2$

I only
III only
I and III only (correct answer)
I, II, and III

Explanation: I. As

x \to 0

f(x) \to \infty

. The function

\sin(x)

is bounded between -1 and 1. Adding a bounded function to a function that grows to infinity will still result in a function that grows to infinity. So,

\lim_{x \to 0} (f(x) + \sin(x)) = \infty

. Statement I is true. II. This limit is of the indeterminate form

0 \cdot \infty

. Its value is not necessarily infinite. For example, if

f(x) = 1/x

, the limit is 1. If

f(x) = 1/x^2

, the limit is infinite. Since it is not guaranteed to be infinite, statement II is false. III. If

f(x) \to \infty

, then squaring it will also result in a limit of

\infty

. So,

\lim_{x \to 0} (f(x))^2 = \infty

. Statement III is true.

Question 10

A function $f(x)$ has a single vertical asymptote at $x=-1$ . Furthermore, $\lim_{x \to -1^+} f(x) = -\infty$ and $\lim_{x \to -1^-} f(x) = -\infty$ . Which of the following could be an equation for $f(x)$ ?

$f(x) = \frac{1}{x+1}$
$f(x) = \frac{x}{x+1}$
$f(x) = \frac{-1}{(x+1)^2}$ (correct answer)
$f(x) = \frac{1}{(x+1)^2}$

Explanation: The condition that the limit is the same (

-\infty

) from both the left and the right suggests that the term in the denominator that goes to zero should be squared, as this makes its sign positive regardless of the direction of approach. A denominator of

(x+1)^2

approaches

0^+

x \to -1

from either side. For the limit to be

-\infty

, the numerator must be negative. In option C,

f(x) = \frac{-1}{(x+1)^2}

, the numerator is

-1

and the denominator approaches

0^+

. Thus,

\lim_{x \to -1} f(x) = \frac{-1}{0^+} = -\infty

, which matches the given conditions.

Question 11

Let P be the statement "The function $f$ has a vertical asymptote at $x=c$ " and Q be the statement " $f(c)$ is undefined." Which of the following correctly describes the logical relationship between P and Q?

P is a necessary but not sufficient condition for Q.
P is a sufficient but not necessary condition for Q.
P is a necessary and sufficient condition for Q.
Neither statement is a necessary or sufficient condition for the other. (correct answer)

Explanation: P does not imply Q: A function can have a vertical asymptote at

x=c

even if

f(c)

is defined. For example,

f(x) = \begin{cases} 1/x & x \neq 0 \\ 5 & x=0 \end{cases}

has a vertical asymptote at

x=0

, but

f(0)=5

. Q does not imply P: A function can be undefined at

x=c

without having a vertical asymptote. For example,

f(x) = \frac{x^2-1}{x-1}

is undefined at

x=1

, but it has a hole (a removable discontinuity), not a vertical asymptote. Therefore, neither statement implies the other.

Question 12

Suppose $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = L$ , where $L$ is a non-zero finite number. Let $h(x) = \frac{g(x)}{f(x)}$ . Which of the following statements about $h(x)$ at $x=a$ must be true?

$h(x)$ has a vertical asymptote at $x=a$ .
$h(x)$ has a jump discontinuity at $x=a$ .
$\lim_{x \to a} h(x) = 0$ . (correct answer)
$\lim_{x \to a} h(x)$ does not exist.

Explanation: We can evaluate the limit of

h(x)

using the properties of limits.

\lim_{x \to a} h(x) = \lim_{x \to a} \frac{g(x)}{f(x)} = \frac{\lim_{x \to a} g(x)}{\lim_{x \to a} f(x)}

. Given that

\lim_{x \to a} g(x) = L

(a finite non-zero number) and

\lim_{x \to a} f(x) = \infty

, the limit of the quotient is

\frac{L}{\infty}

, which is 0. Therefore,

\lim_{x \to a} h(x) = 0

. This means there is no vertical asymptote or jump discontinuity; rather, the function approaches a finite value.

Question 13

What are the equations of all vertical asymptotes of the function $h(x) = \ln(|x^2-4|)$ ?

$x = 2$ only
$x = 2$ and $x = -2$ (correct answer)
$x = 0$ and $x=4$
The function has no vertical asymptotes.

Explanation: The natural logarithm function

\ln(u)

has a vertical asymptote when its argument

u

approaches 0 from the positive side. For the function

h(x) = \ln(|x^2-4|)

, the argument is

u = |x^2-4|

. We need to find the values of

x

for which the argument approaches 0. This happens when

x^2-4 = 0

, which gives

x=2

and

x=-2

. As

x

approaches 2 from either side,

x^2-4

approaches 0, so

|x^2-4|

approaches

0^+

. Therefore,

\lim_{x \to 2} \ln(|x^2-4|) = -\infty

. This means

x=2

is a vertical asymptote. Similarly, as

x

approaches -2 from either side,

x^2-4

approaches 0, so

|x^2-4|

approaches

0^+

. Therefore,

\lim_{x \to -2} \ln(|x^2-4|) = -\infty

. This means

x=-2

is a vertical asymptote. Thus, the function has two vertical asymptotes:

x=2

and

x=-2

Question 14

Which of the following describes the vertical asymptotes of the function $f(x) = \frac{1}{x-1} - \frac{2}{x^2-1}$ ?

A vertical asymptote at $x = -1$ only (correct answer)
A vertical asymptote at $x = 1$ only
Vertical asymptotes at $x = 1$ and $x = -1$
The function has no vertical asymptotes.

Explanation: To find the vertical asymptotes, we first combine the two terms into a single rational expression. The common denominator is

x^2-1 = (x-1)(x+1)

f(x) = \frac{1}{x-1} - \frac{2}{(x-1)(x+1)}

f(x) = \frac{1(x+1)}{(x-1)(x+1)} - \frac{2}{(x-1)(x+1)}

f(x) = \frac{x+1-2}{(x-1)(x+1)}

f(x) = \frac{x-1}{(x-1)(x+1)}

Potential vertical asymptotes are at the zeros of the denominator,

x=1

and

x=-1

. At

x=1

, both the numerator and denominator are zero. We can simplify the expression for

x \neq 1

f(x) = \frac{1}{x+1}

. The limit as

x \to 1

\lim_{x \to 1} \frac{1}{x+1} = \frac{1}{2}

. Since the limit is finite, there is a removable discontinuity (a hole) at

x=1

, not a vertical asymptote. At

x=-1

, the simplified function is

f(x)=\frac{1}{x+1}

. The limit is

\lim_{x \to -1} \frac{1}{x+1}

, which is infinite. Therefore, there is a vertical asymptote at

x=-1

Question 15

What are the equations of the vertical asymptotes of the function $f(x)$ defined as $f(x) = \begin{cases} \frac{1}{x+2} & \text{if } x < 0 \\ \frac{x-1}{x^2-1} & \text{if } x \ge 0 \end{cases}$ ?

$x = -2$ only (correct answer)
$x = -2$ and $x = 1$
$x = 1$ and $x = -1$
$x = -2$ , $x = -1$ , and $x = 1$

Explanation: We analyze each piece of the function separately. For

x < 0

, the function is

f(x) = \frac{1}{x+2}

. A vertical asymptote occurs when the denominator is zero, which is at

x=-2

. Since

-2 < 0

, this is within the domain for this piece. Thus,

x=-2

is a vertical asymptote. For

x \ge 0

, the function is

f(x) = \frac{x-1}{x^2-1} = \frac{x-1}{(x-1)(x+1)}

. Potential vertical asymptotes are at

x=1

and

x=-1

. The value

x=-1

is not in the domain

x \ge 0

, so we disregard it. For

x=1

, we check the limit:

\lim_{x \to 1} \frac{x-1}{(x-1)(x+1)} = \lim_{x \to 1} \frac{1}{x+1} = \frac{1}{2}

. Since the limit is a finite value, there is a hole at

x=1

, not a vertical asymptote. Finally, we check the boundary point

x=0

. The left-hand and right-hand limits are finite, so there is no asymptote there. Therefore, the only vertical asymptote is at

x=-2

Question 16

Which of the following statements is true about the function $f(x) = \frac{x-3}{|x-2|(x-1)}$ ?

$\lim_{x \to 1^+} f(x) = +\infty$ and $\lim_{x \to 2} f(x) = -\infty$
$\lim_{x \to 1^-} f(x) = +\infty$ and $\lim_{x \to 2} f(x) = -\infty$ (correct answer)
$f(x)$ has a vertical asymptote at $x=1$ and a hole at $x=2$
$\lim_{x \to 1} f(x)$ exists and $\lim_{x \to 2} f(x)$ does not exist

Explanation: The function has potential vertical asymptotes at

x=1

and

x=2

. Analysis at

x=1

: As

x \to 1^+

x-3 \approx -2

|x-2| \approx 1

, and

x-1

is a small positive number. So,

f(x) \approx \frac{-2}{1 \cdot (\text{small positive})} \to -\infty

. As

x \to 1^-

x-3 \approx -2

|x-2| \approx 1

, and

x-1

is a small negative number. So,

f(x) \approx \frac{-2}{1 \cdot (\text{small negative})} \to +\infty

. Thus,

\lim_{x \to 1^-} f(x) = +\infty

. Analysis at

x=2

: As

x \to 2

(from either side),

x-3 \approx -1

|x-2|

is a small positive number, and

x-1 \approx 1

. So,

f(x) \approx \frac{-1}{(\text{small positive}) \cdot 1} \to -\infty

. Since the limits from both the left and right are

-\infty

, we have

\lim_{x \to 2} f(x) = -\infty

. Comparing these results with the options, choice B is the only true statement.

Question 17

Given the function $f(x) = e^{1/(x-3)}$ . Which of the following statements correctly describes the behavior of $f(x)$ at $x=3$ ?

$\lim_{x \to 3} f(x) = \infty$ , so $x=3$ is a vertical asymptote.
$\lim_{x \to 3^+} f(x) = \infty$ , so $x=3$ is a vertical asymptote. (correct answer)
The function has a jump discontinuity at $x=3$ because the one-sided limits are unequal and finite.
The function is continuous at $x=3$ because the left-hand limit is 0.

Explanation: To determine the behavior at

x=3

, we must evaluate the one-sided limits. As

x \to 3^+

x-3

is a small positive number, so

1/(x-3) \to +\infty

. Thus,

\lim_{x \to 3^+} e^{1/(x-3)} = e^{+\infty} = +\infty

. As

x \to 3^-

x-3

is a small negative number, so

1/(x-3) \to -\infty

. Thus,

\lim_{x \to 3^-} e^{1/(x-3)} = e^{-\infty} = 0

. A vertical asymptote exists if at least one of the one-sided limits is infinite. Since

\lim_{x \to 3^+} f(x) = \infty

, the line

x=3

is a vertical asymptote.

Question 18

Let $f(x)$ be an odd function. If $\lim_{x \to 2^+} f(x) = -\infty$ , which of the following limits must be true?

$\lim_{x \to -2^+} f(x) = -\infty$
$\lim_{x \to -2^-} f(x) = -\infty$
$\lim_{x \to -2^-} f(x) = +\infty$ (correct answer)
$f(x)$ has a removable discontinuity at $x=-2$ .

Explanation: By definition, an odd function satisfies

f(-x) = -f(x)

. We are given

\lim_{x \to 2^+} f(x) = -\infty

. To find the limit as

x

approaches

-2

, let

y = -x

. As

x \to 2^+

y \to -2^-

. We can write:

\lim_{y \to -2^-} f(y) = \lim_{x \to 2^+} f(-x)

. Using the odd function property, this becomes

\lim_{x \to 2^+} -f(x) = -\lim_{x \to 2^+} f(x)

. Substituting the given limit, we get

-(-\infty) = +\infty

. Therefore,

\lim_{x \to -2^-} f(x) = +\infty

Question 19

Find all vertical asymptotes of $f(x) = \frac{x-3}{|x^2-9|}$ .

$x=-3$ only (correct answer)
$x=3$ only
$x=-3$ and $x=3$
The function has no vertical asymptotes.

Explanation: Potential vertical asymptotes occur where the denominator is zero, i.e.,

|x^2-9|=0

, which means

x=3

and

x=-3

. We must check the limits at these points. At

x=-3

: The numerator approaches

-6

and the denominator approaches

0^+

. Thus,

\lim_{x \to -3} f(x) = \frac{-6}{0^+} = -\infty

. So,

x=-3

is a vertical asymptote. At

x=3

: This is an indeterminate form. We rewrite

f(x) = \frac{x-3}{|x-3||x+3|}

. For

x \to 3^+

|x-3|=x-3

, so

f(x) = \frac{1}{|x+3|}

, and the limit is

1/6

. For

x \to 3^-

|x-3|=-(x-3)

, so

f(x) = \frac{-1}{|x+3|}

, and the limit is

-1/6

. Since both one-sided limits are finite, there is no vertical asymptote at

x=3

Question 20

Let $f(x) = \frac{x^2 - \pi^2}{\sin(x)}$ . Which of the following is a vertical asymptote for $f(x)$ ?

$x = \pi$
$x = -\pi$
$x = 2\pi$ (correct answer)
The function has no vertical asymptotes.

Explanation: Vertical asymptotes can occur where the denominator is zero, which is

\sin(x) = 0

, or

x=n\pi

for any integer

n

. We must check if the numerator is also zero at these points. The numerator is

x^2 - \pi^2 = (x-\pi)(x+\pi)

, which is zero at

x=\pi

and

x=-\pi

. At

x=\pi

: The limit is of the form

0/0

. Using L'Hôpital's Rule:

\lim_{x \to \pi} \frac{2x}{\cos(x)} = \frac{2\pi}{-1} = -2\pi

. Since the limit is finite,

x=\pi

is a hole. At

x=-\pi

: The limit is

0/0

. Using L'Hôpital's Rule:

\lim_{x \to -\pi} \frac{2x}{\cos(x)} = \frac{-2\pi}{-1} = 2\pi

. This is also a hole. At

x=2\pi

: The numerator is

(2\pi)^2 - \pi^2 = 3\pi^2 \neq 0

. The denominator is

\sin(2\pi)=0

. Since the limit is of the form

k/0

with

k \neq 0

, the limit is infinite, and

x=2\pi

is a vertical asymptote.