Calculus 1 Quiz: Squeeze Theorem

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Squeeze TheoremQuestion 1 of 20

If a function $f(x)$ satisfies $1 - \frac{1}{x^2} \le f(x) \le 1 + \frac{1}{x^2}$ for all $x \neq 0$ , what is $\lim_{x \to \infty} f(x)$ ?

x

The limit cannot be determined.

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Calculus 1 Quiz

Calculus 1 Quiz: Squeeze Theorem

Practice Squeeze Theorem 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 Squeeze Theorem, 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

If a function $f(x)$ satisfies $1 - \frac{1}{x^2} \le f(x) \le 1 + \frac{1}{x^2}$ for all $x \neq 0$ , what is $\lim_{x \to \infty} f(x)$ ?

0
1 (correct answer)
$x$
The limit cannot be determined.

Explanation: This is a direct application of the Squeeze Theorem for limits at infinity. We are given the inequality

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

. We need to find the limits of the bounding functions as

x \to \infty

. For the lower bound:

\lim_{x \to \infty} (1 - \frac{1}{x^2}) = 1 - 0 = 1

. For the upper bound:

\lim_{x \to \infty} (1 + \frac{1}{x^2}) = 1 + 0 = 1

. Since

f(x)

is between two functions that both approach 1, its limit as

x \to \infty

must also be 1.

Question 2

Evaluate the limit: $$ \lim_{x \to 1} (x-1)^2 \sin\left(\frac{\pi}{x-1}\right)

0 (correct answer)
1
$\pi$
The limit does not exist.

Explanation: The Squeeze Theorem is applicable here. We know that the sine function is bounded between -1 and 1. So, for any

x \neq 1

, we have

-1 \le \sin\left(\frac{\pi}{x-1}\right) \le 1

. Since

(x-1)^2

is always non-negative, we can multiply the inequality by it without changing the direction of the inequality signs:

-(x-1)^2 \le (x-1)^2 \sin\left(\frac{\pi}{x-1}\right) \le (x-1)^2

. Now, we take the limit of the bounding functions as

x \to 1

\lim_{x \to 1} -(x-1)^2 = 0

and

\lim_{x \to 1} (x-1)^2 = 0

. Since the function is squeezed between two functions that both approach 0, its limit must also be 0.

Question 3

Let $g(x) = -x^2$ and $h(x) = x^2$ . The Squeeze Theorem can be used to show $\lim_{x \to 0} f(x) = 0$ for which of the following functions $f(x)$ ?

$f(x) = x \sin(1/x)$
$f(x) = x^3 \cos(1/x)$
$f(x) = x^{1/3} \sin(x)$
$f(x) = x^2 \cos(x)$ (correct answer)

Explanation: For the Squeeze Theorem to apply with

g(x) = -x^2

and

h(x) = x^2

, we need

-x^2 \leq f(x) \leq x^2

near

x = 0

. Let's check each option: A)

f(x) = x\sin(1/x)

: We have

|x\sin(1/x)| \leq |x|

, so

-|x| \leq f(x) \leq |x|

. For small

x

|x| > x^2

, so the bounds

\pm x^2

are too restrictive. B)

f(x) = x^3\cos(1/x)

: We have

|f(x)| \leq |x^3|

, and for

|x| < 1

, we have

|x^3| < x^2

, so this works. C)

f(x) = x^{1/3}\sin(x)

: For small positive

x

x^{1/3} > x^2

, so the bounds are too restrictive. D)

f(x) = x^2\cos(x)

: Since

|\cos(x)| \leq 1

, we have

|f(x)| \leq x^2

, which gives

-x^2 \leq f(x) \leq x^2

. This satisfies the required inequality.

Question 4

Let $f(x) = x^2$ if $x$ is a rational number, and $f(x) = -x^2$ if $x$ is an irrational number. What is $\lim_{x \to 0} f(x)$ ?

0 (correct answer)
1
The limit exists but cannot be determined.
The limit does not exist.

Explanation: This is a classic application of the Squeeze Theorem. For any real number

x

, either

f(x) = x^2

f(x) = -x^2

. In all cases, the value of

f(x)

is bounded by

-x^2

and

x^2

. That is, for all

x

, we have the inequality

-x^2 \le f(x) \le x^2

. We can find the limit of the bounding functions as

x \to 0

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

and

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

. Since

f(x)

is squeezed between two functions that both approach 0, the limit of

f(x)

x \to 0

must also be 0.

Question 5

Evaluate $\lim_{x \to 0} \frac{x^2}{4 - \cos^2(1/x)}$ .

$0$ (correct answer)
$1/4$
$1/3$
The limit does not exist.

Explanation: We can use the Squeeze Theorem by first establishing bounds for the denominator. We know that

0 \le \cos^2(1/x) \le 1

. Multiplying by

-1

reverses the inequality:

-1 \le -\cos^2(1/x) \le 0

. Adding

4

to all parts gives

3 \le 4 - \cos^2(1/x) \le 4

. Since the denominator is always positive, we can take the reciprocal, which reverses the inequality again:

\frac{1}{4} \le \frac{1}{4 - \cos^2(1/x)} \le \frac{1}{3}

. Finally, we multiply the entire inequality by

x^2

. Since

x^2 \ge 0

, the inequality direction is preserved:

\frac{x^2}{4} \le \frac{x^2}{4 - \cos^2(1/x)} \le \frac{x^2}{3}

. Taking the limit as

x \to 0

, we have

\lim_{x \to 0} \frac{x^2}{4} = 0

and

\lim_{x \to 0} \frac{x^2}{3} = 0

. By the Squeeze Theorem, the limit of the original function is

0

Question 6

A function $f(x)$ satisfies the inequality $4x - x^2 \le f(x) \le x^2 - 4x + 8$ for all real numbers $x$ . For which of the following values of $c$ can the Squeeze Theorem be used to determine $\lim_{x \to c} f(x)$ ?

$c=0$ only
$c=2$ only (correct answer)
$c=4$ only
No value of $c$

Explanation: The Squeeze Theorem can be used to determine the limit of

f(x)

at a point

x=c

if the limits of the lower and upper bounding functions are equal at that point. Let

g(x) = 4x - x^2

and

h(x) = x^2 - 4x + 8

. We need to find

c

such that

\lim_{x \to c} g(x) = \lim_{x \to c} h(x)

. Since polynomials are continuous, we can set

g(c) = h(c)

4c - c^2 = c^2 - 4c + 8

. Rearranging gives

2c^2 - 8c + 8 = 0

, which simplifies to

c^2 - 4c + 4 = 0

. Factoring gives

(c-2)^2 = 0

, so

c=2

Question 7

Let the function $f$ be defined by $f(x) = \begin{cases} x^2 \cos(\frac{\pi}{x}) + k & \text{if } x \ne 0 \\ 5 & \text{if } x = 0 \end{cases}$ . For what value of the constant $k$ is the function $f$ continuous at $x=0$ ?

$k = 0$
$k = 4$
$k = 5$ (correct answer)
No such value of $k$ exists.

Explanation: For

f

to be continuous at

x=0

, we must have

\lim_{x \to 0} f(x) = f(0)

. We are given

f(0) = 5

. We need to evaluate

\lim_{x \to 0} (x^2 \cos(\frac{\pi}{x}) + k)

. We can evaluate

\lim_{x \to 0} x^2 \cos(\frac{\pi}{x})

using the Squeeze Theorem. Since

-1 \le \cos(\frac{\pi}{x}) \le 1

for all

x \ne 0

, we can multiply by

x^2

(which is non-negative) to get

-x^2 \le x^2 \cos(\frac{\pi}{x}) \le x^2

. Since

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

and

\lim_{x \to 0} x^2 = 0

, the Squeeze Theorem implies

\lim_{x \to 0} x^2 \cos(\frac{\pi}{x}) = 0

. Therefore,

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

. For continuity, we set this limit equal to

f(0)

, so

k=5

Question 8

Suppose $g(x) \le f(x) \le h(x)$ for all $x$ in an open interval containing $c$ , except possibly at $c$ . If $\lim_{x \to c} g(x) = L$ and $\lim_{x \to c} h(x) = M$ with $L < M$ , which of the following statements is necessarily true?

$\lim_{x \to c} f(x)$ does not exist.
$\lim_{x \to c} f(x)$ exists and is equal to $\frac{L+M}{2}$ .
If $\lim_{x \to c} f(x)$ exists, its value must be in the interval $[L, M]$ . (correct answer)
The Squeeze Theorem cannot be applied, so no conclusion can be drawn about $f(x)$ .

Explanation: The Squeeze Theorem requires that the limits of the upper and lower bounds are equal (

L=M

). Since

L \ne M

, the theorem cannot be used to determine the value of

\lim_{x \to c} f(x)

. However, we can still draw a conclusion using limit properties. If

\lim_{x \to c} f(x)

exists, then from

g(x) \le f(x)

we must have

\lim_{x \to c} g(x) \le \lim_{x \to c} f(x)

, which means

L \le \lim_{x \to c} f(x)

. Similarly, from

f(x) \le h(x)

, we have

\lim_{x \to c} f(x) \le \lim_{x \to c} h(x)

, which means

\lim_{x \to c} f(x) \le M

. Combining these, if the limit exists, it must lie in the closed interval

[L, M]

. We cannot conclude that the limit must exist; for example,

f(x)

could oscillate between

L

and

M

Question 9

To evaluate $\lim_{x \to \infty} \frac{e^{-x}}{2-\cos(x)}$ using the Squeeze Theorem, which of the following inequalities provides the correct bounding functions, $g(x)$ and $h(x)$ , for $f(x) = \frac{e^{-x}}{2-\cos(x)}$ ?

$e^{-x} \le f(x) \le \frac{e^{-x}}{3}$
$\frac{e^{-x}}{3} \le f(x) \le e^{-x}$ (correct answer)
$-e^{-x} \le f(x) \le e^{-x}$
$\frac{e^{-x}}{2} \le f(x) \le e^{-x}$

Explanation: First, we must bound the denominator,

2-\cos(x)

. We know that

-1 \le \cos(x) \le 1

. Multiplying by -1 reverses the inequalities:

1 \ge -\cos(x) \ge -1

. Adding 2 to all parts gives

2+1 \ge 2-\cos(x) \ge 2-1

, which is

3 \ge 2-\cos(x) \ge 1

. Since the denominator is always positive, we can take the reciprocal, which reverses the inequalities again:

\frac{1}{3} \le \frac{1}{2-\cos(x)} \le 1

. The numerator

e^{-x}

is positive for all real

x

. Multiplying the inequality by

e^{-x}

gives the correct bounds:

\frac{e^{-x}}{3} \le \frac{e^{-x}}{2-\cos(x)} \le e^{-x}

Question 10

Evaluate the limit: $$ \lim_{x \to \infty} \frac{\arctan(x^2)}{x}

0 (correct answer)
$\pi/2$
$1$
The limit does not exist.

Explanation: The range of the arctangent function is

(-\pi/2, \pi/2)

. This means that for any input, the output of

\arctan(x^2)

is bounded. Specifically, as

x \to \infty

x^2 \to \infty

, and

\arctan(x^2)

approaches

\pi/2

. More formally, for all

x

, we have

-\frac{\pi}{2} < \arctan(x^2) < \frac{\pi}{2}

. For

x > 0

, we can divide by

x

to get

\frac{-\pi/2}{x} < \frac{\arctan(x^2)}{x} < \frac{\pi/2}{x}

. Now, we apply the Squeeze Theorem as

x \to \infty

\lim_{x \to \infty} \frac{-\pi/2}{x} = 0

and

\lim_{x \to \infty} \frac{\pi/2}{x} = 0

. Therefore, the limit of the function squeezed between them must also be 0.

Question 11

Let $f(x)$ be a function such that for all $x>5$ , $\frac{4x-1}{x} < f(x) < \frac{4x^2+3x}{x^2}$ . What is $\lim_{x \to \infty} f(x)$ ?

0
4 (correct answer)
3
The limit cannot be determined.

Explanation: We use the Squeeze Theorem to find the limit at infinity. First, find the limit of the lower bound:

\lim_{x \to \infty} \frac{4x-1}{x} = \lim_{x \to \infty} (4 - \frac{1}{x}) = 4 - 0 = 4

. Next, find the limit of the upper bound:

\lim_{x \to \infty} \frac{4x^2+3x}{x^2} = \lim_{x \to \infty} (4 + \frac{3}{x}) = 4 + 0 = 4

. Since

f(x)

is squeezed between two functions that both approach 4 as

x \to \infty

, the limit of

f(x)

must also be 4.

Question 12

Evaluate the limit: $$ \lim_{x \to 0} (x^2 + x^4) \sin(\pi/x)

0 (correct answer)
$1$
$\pi$
The limit does not exist.

Explanation: The Squeeze Theorem can be applied. The term

\sin(\pi/x)

is bounded between -1 and 1. The term

(x^2 + x^4)

is non-negative and approaches 0 as

x \to 0

. We can establish the inequality:

-1 \le \sin(\pi/x) \le 1

. Multiplying by

(x^2+x^4)

(which is

\ge 0

) gives

-(x^2+x^4) \le (x^2+x^4)\sin(\pi/x) \le (x^2+x^4)

. Now we take the limits of the bounding functions:

\lim_{x \to 0} -(x^2+x^4) = 0

and

\lim_{x \to 0} (x^2+x^4) = 0

. By the Squeeze Theorem, the limit of the original function must be 0.

Question 13

If $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$ and $g(x) < f(x) < h(x)$ for all $x$ in an open interval containing $c$ (except possibly at $c$ ), what can be said about $\lim_{x \to c} f(x)$ ?

The limit is $L$ . (correct answer)
The limit is some value strictly between $L$ and $L$ , which is impossible.
The limit does not exist because the inequalities are strict.
The limit cannot be determined without more information about $f(x)$ .

Explanation: The Squeeze Theorem holds even if the inequalities are strict (i.e.,

g(x) < f(x) < h(x)

). The conclusion is the same. As

x

gets arbitrarily close to

c

, the values of

g(x)

and

h(x)

get arbitrarily close to

L

. Since

f(x)

is trapped between them, its value must also get arbitrarily close to

L

. Therefore,

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

. The strictness of the inequality does not affect the limit value.

Question 14

Evaluate the limit: $$ \lim_{x \to \infty} \frac{\ln(x) + \sin(x)}{\ln(x)}

0
1 (correct answer)
$\infty$
The limit does not exist.

Explanation: We can rewrite the expression as

\lim_{x \to \infty} \left( \frac{\ln(x)}{\ln(x)} + \frac{\sin(x)}{\ln(x)} \right) = \lim_{x \to \infty} \left( 1 + \frac{\sin(x)}{\ln(x)} \right)

. To find the limit of the second term,

\frac{\sin(x)}{\ln(x)}

, we use the Squeeze Theorem. We know

-1 \le \sin(x) \le 1

. For

x > 1

\ln(x) > 0

, so we can divide by

\ln(x)

without changing the inequality direction:

\frac{-1}{\ln(x)} \le \frac{\sin(x)}{\ln(x)} \le \frac{1}{\ln(x)}

. As

x \to \infty

\ln(x) \to \infty

, so both

\frac{-1}{\ln(x)}

and

\frac{1}{\ln(x)}

approach 0. By the Squeeze Theorem,

\lim_{x \to \infty} \frac{\sin(x)}{\ln(x)} = 0

. Therefore, the original limit is

1 + 0 = 1

Question 15

Suppose that for $x$ in $(-1, 1)$ , we have $g(x) \le f(x) \le h(x)$ , where $\lim_{x \to 0} g(x) = 0$ and $\lim_{x \to 0} h(x) = 0$ . What additional condition is necessary to prove that $f(x)$ is continuous at $x=0$ ?

$f(0)$ is defined.
$f(0)=0$ . (correct answer)
$g(0) = h(0) = 0$ .
No additional condition is needed; continuity is already guaranteed.

Explanation: The Squeeze Theorem, with the given conditions, guarantees that

\lim_{x \to 0} f(x) = 0

. For a function to be continuous at a point

c

, three conditions must be met: 1)

f(c)

is defined, 2)

\lim_{x \to c} f(x)

exists, and 3)

\lim_{x \to c} f(x) = f(c)

. We have established that the limit exists and is 0. To satisfy the third condition for continuity at

x=0

, we must have

f(0)

equal to this limit. Therefore, the necessary additional condition is

f(0)=0

. Option A is not sufficient, as

f(0)

could be defined but not equal to 0. Option C is about the bounding functions, not

f(x)

itself. Option D is incorrect because the limit existing does not guarantee continuity.

Question 16

Let $f(x)$ be a function defined on $(-1, 1)$ such that $x^4 \le f(x) \le 2x^2$ for all $x$ in its domain. Based on this information and the Squeeze Theorem, what is $\lim_{x \to 0} \frac{f(x)}{x}$ ?

0 (correct answer)
1
2
The limit cannot be determined.

Explanation: We are given

x^4 \le f(x) \le 2x^2

. To find the limit of

\frac{f(x)}{x}

, we need to divide the inequality by

x

. This requires considering two cases. Case 1:

x > 0

. Dividing by

x

gives

x^3 \le \frac{f(x)}{x} \le 2x

. Taking the limit as

x \to 0^+

, we get

\lim_{x \to 0^+} x^3 = 0

and

\lim_{x \to 0^+} 2x = 0

. By the Squeeze Theorem,

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

. Case 2:

x < 0

. Dividing by

x

reverses the inequalities:

x^3 \ge \frac{f(x)}{x} \ge 2x

. Taking the limit as

x \to 0^-\, we get \(\lim_{x \to 0^-} x^3 = 0

and

\lim_{x \to 0^-} 2x = 0

. By the Squeeze Theorem,

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

. Since both one-sided limits are 0, the two-sided limit is 0.

Question 17

A student attempts to evaluate $\lim_{x \to 0} \frac{1}{x^2} \cos(x)$ and reasons that since $-1 \le \cos(x) \le 1$ , then $\frac{-1}{x^2} \le \frac{\cos(x)}{x^2} \le \frac{1}{x^2}$ . Since $\lim_{x \to 0} \frac{-1}{x^2} = -\infty$ and $\lim_{x \to 0} \frac{1}{x^2} = \infty$ , the student concludes the Squeeze Theorem is inconclusive. What is the correct evaluation of the limit?

The student is correct; the limit cannot be determined.
The limit is 0, because $1/x^2$ is multiplied by a bounded function.
The limit is 1, as $\lim_{x \to 0} \cos(x) = 1$ .
The limit is $\infty$ , as $\cos(x)$ approaches 1 while $1/x^2$ approaches $\infty$ . (correct answer)

Explanation: The student's application of the Squeeze Theorem is technically correct in that it yields an inconclusive result. However, the Squeeze Theorem is the wrong tool here. This is not a '0 times bounded' form. As

x \to 0

, the term

\cos(x)

approaches

\cos(0) = 1

, a positive constant. The term

1/x^2

approaches

+\infty

. Therefore, the limit is of the form

\infty \cdot 1

, which is

\infty

. The student misidentified the problem type, leading to an unnecessarily complex and inconclusive analysis.

Question 18

Let $f, g,$ and $h$ be functions defined for all real numbers. Suppose that for all $x \neq 2$ , we have $g(x) \le f(x) \le h(x)$ . If $\lim_{x \to 2} g(x) = -1$ and $\lim_{x \to 2} h(x) = 1$ , what can be concluded about $\lim_{x \to 2} f(x)$ ?

The limit must be 0.
The limit must be a value $L$ such that $-1 \le L \le 1$ .
The limit must not exist.
The Squeeze Theorem does not provide sufficient information to determine the limit. (correct answer)

Explanation: The Squeeze Theorem requires that the limits of the lower and upper bounding functions are equal. In this case,

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

and

\lim_{x \to 2} h(x) = 1

. Since these limits are not equal, the conditions for the Squeeze Theorem are not met. Therefore, we cannot draw any conclusion about

\lim_{x \to 2} f(x)

. The limit of

f(x)

may or may not exist. For example, if

f(x) = 0

, the limit is 0. If

f(x) = (x-2) \sin(1/(x-2))

, the limit is 0. If

f(x) = \sin(\pi/(x-2))

, the limit does not exist. The theorem is inconclusive.

Question 19

Consider the function $f(x) = (x - |x|) \cos(1/x^2)$ . Evaluate $\lim_{x \to 0} f(x)$ .

0 (correct answer)
-2
1
The limit does not exist.

Explanation: Because of the absolute value function

|x|

, we should analyze the left-hand and right-hand limits separately. For the right-hand limit (

x \to 0^+

x > 0

|x| = x

. The function becomes

f(x) = (x - x) \cos(1/x^2) = 0 \cdot \cos(1/x^2) = 0

. So,

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

. For the left-hand limit (

x \to 0^-

x < 0

|x| = -x

. The function becomes

f(x) = (x - (-x)) \cos(1/x^2) = 2x \cos(1/x^2)

. To evaluate

\lim_{x \to 0^-} 2x \cos(1/x^2)

, we use the Squeeze Theorem. We know

-1 \le \cos(1/x^2) \le 1

. Multiplying by

2x

(which is negative) reverses the inequalities:

-2x \ge 2x \cos(1/x^2) \ge 2x

. As

x \to 0^-\, \lim_{x \to 0^-} (-2x) = 0

and

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

. Thus,

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

. Since both one-sided limits are 0, the overall limit is 0.

Question 20

Evaluate the limit: $$ \lim_{x \to \infty} (2 + \frac{\cos(x^2)}{x+1})

0
1
2 (correct answer)
The limit does not exist.

Explanation: We can evaluate this limit by considering its parts. The limit of a sum is the sum of the limits, provided they exist. So,

\lim_{x \to \infty} (2 + \frac{\cos(x^2)}{x+1}) = \lim_{x \to \infty} 2 + \lim_{x \to \infty} \frac{\cos(x^2)}{x+1}

. The first limit is clearly 2. For the second limit, we use the Squeeze Theorem. We know

-1 \le \cos(x^2) \le 1

. For

x > -1

, we can divide by

x+1

(which is positive) to get

\frac{-1}{x+1} \le \frac{\cos(x^2)}{x+1} \le \frac{1}{x+1}

. As

x \to \infty

, both

\frac{-1}{x+1}

and

\frac{1}{x+1}

approach 0. So, by the Squeeze Theorem,

\lim_{x \to \infty} \frac{\cos(x^2)}{x+1} = 0

. Therefore, the original limit is

2 + 0 = 2