Use the squeeze theorem: lim_x to 0 x^4sin (3/x)

0, since -x^4 ≤ x^4sin (3/x) ≤ x^4

Use the squeeze theorem: lim_x to 0 x^4sin (1/x)

0, since -x^4 ≤ x^4sin (1/x) ≤ x^4

Use the squeeze theorem: lim_x to 0 x^2sin (1/x)

0, since -x^2 ≤ x^2sin (1/x) ≤ x^2

Use the squeeze theorem: lim_x to 0 xcos (1/x)

0, since -|x| ≤ xcos (1/x) ≤ |x|

Use the squeeze theorem: lim_x to 0 x^4sin (2/x)

0, since -x^4 ≤ x^4sin (2/x) ≤ x^4

Use the squeeze theorem: lim_x to 0 x^2cos (4/x)

0, since -x^2 ≤ x^2cos (4/x) ≤ x^2

Use the squeeze theorem: lim_x to 0 xsin (4/x)

0, since -|x| ≤ xsin (4/x) ≤ |x|

Use the squeeze theorem: lim_x to 0 x^2cos (3/x)

0, since -x^2 ≤ x^2cos (3/x) ≤ x^2

Worksheets · Calculus

Squeeze theorem worksheet

When a function is stuck between two others that head to the same value, it has nowhere else to go. The work is finding the two bounds, usually from the fact that sine and cosine never leave the interval from -1 to 1.

Select your difficulty

Pick more than one for a sheet that mixes them.



Sigma Prep · sigmaprep.io/worksheets Name The Squeeze Theorem Date Period Find each limit. Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{3}{x})$ If $24 - {(x - 4)}^{2} \leq f (x) \leq 24 + {(x - 4)}^{2}$ near $x = 4$ , find $\lim_{x \to 4} f (x)$ $\lim_{x \to \infty} \frac{6 + \cos x}{x^{2}}$ Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{1}{x})$ If $- 2 - {(x + 1)}^{2} \leq f (x) \leq - 2 + {(x + 1)}^{2}$ near $x = - 1$ , find $\lim_{x \to - 1} f (x)$ $\lim_{x \to \infty} \frac{\sin x}{x^{2}}$ Use the squeeze theorem: $\lim_{x \to 0} x^{2} \sin (\frac{1}{x})$ If $3 - 3 {(x + 1)}^{2} \leq f (x) \leq 3 + 3 {(x + 1)}^{2}$ near $x = - 1$ , find $\lim_{x \to - 1} f (x)$ $\lim_{x \to \infty} \frac{4 + \sin x}{x^{3}}$ Use the squeeze theorem: $\lim_{x \to 0} x \cos (\frac{1}{x})$ If $0 - {(x - 2)}^{2} \leq f (x) \leq 0 + {(x - 2)}^{2}$ near $x = 2$ , find $\lim_{x \to 2} f (x)$ $\lim_{x \to \infty} \frac{\sin x}{x}$ Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{2}{x})$ If $19 - 3 {(x + 4)}^{2} \leq f (x) \leq 19 + 3 {(x + 4)}^{2}$ near $x = - 4$ , find $\lim_{x \to - 4} f (x)$ $\lim_{x \to \infty} \frac{\cos x}{x^{3}}$ Use the squeeze theorem: $\lim_{x \to 0} x^{2} \cos (\frac{4}{x})$ If $20 - 3 {(x - 4)}^{2} \leq f (x) \leq 20 + 3 {(x - 4)}^{2}$ near $x = 4$ , find $\lim_{x \to 4} f (x)$ $\lim_{x \to \infty} \frac{2 + \cos x}{x}$ Use the squeeze theorem: $\lim_{x \to 0} x \sin (\frac{4}{x})$ If $8 - 3 x^{2} \leq f (x) \leq 8 + 3 x^{2}$ near $x = 0$ , find $\lim_{x \to 0} f (x)$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

The Squeeze Theorem

Date Period

Find each limit.

Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{3}{x})$
If $24 - {(x - 4)}^{2} \leq f (x) \leq 24 + {(x - 4)}^{2}$ near $x = 4$ , find $\lim_{x \to 4} f (x)$
$\lim_{x \to \infty} \frac{6 + \cos x}{x^{2}}$
Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{1}{x})$
If $- 2 - {(x + 1)}^{2} \leq f (x) \leq - 2 + {(x + 1)}^{2}$ near $x = - 1$ , find $\lim_{x \to - 1} f (x)$
$\lim_{x \to \infty} \frac{\sin x}{x^{2}}$
Use the squeeze theorem: $\lim_{x \to 0} x^{2} \sin (\frac{1}{x})$
If $3 - 3 {(x + 1)}^{2} \leq f (x) \leq 3 + 3 {(x + 1)}^{2}$ near $x = - 1$ , find $\lim_{x \to - 1} f (x)$
$\lim_{x \to \infty} \frac{4 + \sin x}{x^{3}}$
Use the squeeze theorem: $\lim_{x \to 0} x \cos (\frac{1}{x})$
If $0 - {(x - 2)}^{2} \leq f (x) \leq 0 + {(x - 2)}^{2}$ near $x = 2$ , find $\lim_{x \to 2} f (x)$
$\lim_{x \to \infty} \frac{\sin x}{x}$
Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{2}{x})$
If $19 - 3 {(x + 4)}^{2} \leq f (x) \leq 19 + 3 {(x + 4)}^{2}$ near $x = - 4$ , find $\lim_{x \to - 4} f (x)$
$\lim_{x \to \infty} \frac{\cos x}{x^{3}}$
Use the squeeze theorem: $\lim_{x \to 0} x^{2} \cos (\frac{4}{x})$
If $20 - 3 {(x - 4)}^{2} \leq f (x) \leq 20 + 3 {(x - 4)}^{2}$ near $x = 4$ , find $\lim_{x \to 4} f (x)$
$\lim_{x \to \infty} \frac{2 + \cos x}{x}$
Use the squeeze theorem: $\lim_{x \to 0} x \sin (\frac{4}{x})$
If $8 - 3 x^{2} \leq f (x) \leq 8 + 3 x^{2}$ near $x = 0$ , find $\lim_{x \to 0} f (x)$



Sigma Prep · sigmaprep.io/worksheets Name The Squeeze TheoremAnswer keyVersion 1 Date Period Find each limit. Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{3}{x})$ $0$ , since $- x^{4} \leq x^{4} \sin (\frac{3}{x}) \leq x^{4}$ If $24 - {(x - 4)}^{2} \leq f (x) \leq 24 + {(x - 4)}^{2}$ near $x = 4$ , find $\lim_{x \to 4} f (x)$ $24$ $\lim_{x \to \infty} \frac{6 + \cos x}{x^{2}}$ $0$ Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{1}{x})$ $0$ , since $- x^{4} \leq x^{4} \sin (\frac{1}{x}) \leq x^{4}$ If $- 2 - {(x + 1)}^{2} \leq f (x) \leq - 2 + {(x + 1)}^{2}$ near $x = - 1$ , find $\lim_{x \to - 1} f (x)$ $- 2$ $\lim_{x \to \infty} \frac{\sin x}{x^{2}}$ $0$ Use the squeeze theorem: $\lim_{x \to 0} x^{2} \sin (\frac{1}{x})$ $0$ , since $- x^{2} \leq x^{2} \sin (\frac{1}{x}) \leq x^{2}$ If $3 - 3 {(x + 1)}^{2} \leq f (x) \leq 3 + 3 {(x + 1)}^{2}$ near $x = - 1$ , find $\lim_{x \to - 1} f (x)$ $3$ $\lim_{x \to \infty} \frac{4 + \sin x}{x^{3}}$ $0$ Use the squeeze theorem: $\lim_{x \to 0} x \cos (\frac{1}{x})$ $0$ , since $- ∣ x ∣ \leq x \cos (\frac{1}{x}) \leq ∣ x ∣$ If $0 - {(x - 2)}^{2} \leq f (x) \leq 0 + {(x - 2)}^{2}$ near $x = 2$ , find $\lim_{x \to 2} f (x)$ $0$ $\lim_{x \to \infty} \frac{\sin x}{x}$ $0$ Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{2}{x})$ $0$ , since $- x^{4} \leq x^{4} \sin (\frac{2}{x}) \leq x^{4}$ If $19 - 3 {(x + 4)}^{2} \leq f (x) \leq 19 + 3 {(x + 4)}^{2}$ near $x = - 4$ , find $\lim_{x \to - 4} f (x)$ $19$ $\lim_{x \to \infty} \frac{\cos x}{x^{3}}$ $0$ Use the squeeze theorem: $\lim_{x \to 0} x^{2} \cos (\frac{4}{x})$ $0$ , since $- x^{2} \leq x^{2} \cos (\frac{4}{x}) \leq x^{2}$ If $20 - 3 {(x - 4)}^{2} \leq f (x) \leq 20 + 3 {(x - 4)}^{2}$ near $x = 4$ , find $\lim_{x \to 4} f (x)$ $20$ $\lim_{x \to \infty} \frac{2 + \cos x}{x}$ $0$ Use the squeeze theorem: $\lim_{x \to 0} x \sin (\frac{4}{x})$ $0$ , since $- ∣ x ∣ \leq x \sin (\frac{4}{x}) \leq ∣ x ∣$ If $8 - 3 x^{2} \leq f (x) \leq 8 + 3 x^{2}$ near $x = 0$ , find $\lim_{x \to 0} f (x)$ $8$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

The Squeeze TheoremAnswer keyVersion 1

Date Period

Find each limit.

Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{3}{x})$ $0$ , since $- x^{4} \leq x^{4} \sin (\frac{3}{x}) \leq x^{4}$
If $24 - {(x - 4)}^{2} \leq f (x) \leq 24 + {(x - 4)}^{2}$ near $x = 4$ , find $\lim_{x \to 4} f (x)$ $24$
$\lim_{x \to \infty} \frac{6 + \cos x}{x^{2}}$ $0$
Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{1}{x})$ $0$ , since $- x^{4} \leq x^{4} \sin (\frac{1}{x}) \leq x^{4}$
If $- 2 - {(x + 1)}^{2} \leq f (x) \leq - 2 + {(x + 1)}^{2}$ near $x = - 1$ , find $\lim_{x \to - 1} f (x)$ $- 2$
$\lim_{x \to \infty} \frac{\sin x}{x^{2}}$ $0$
Use the squeeze theorem: $\lim_{x \to 0} x^{2} \sin (\frac{1}{x})$ $0$ , since $- x^{2} \leq x^{2} \sin (\frac{1}{x}) \leq x^{2}$
If $3 - 3 {(x + 1)}^{2} \leq f (x) \leq 3 + 3 {(x + 1)}^{2}$ near $x = - 1$ , find $\lim_{x \to - 1} f (x)$ $3$
$\lim_{x \to \infty} \frac{4 + \sin x}{x^{3}}$ $0$
Use the squeeze theorem: $\lim_{x \to 0} x \cos (\frac{1}{x})$ $0$ , since $- ∣ x ∣ \leq x \cos (\frac{1}{x}) \leq ∣ x ∣$
If $0 - {(x - 2)}^{2} \leq f (x) \leq 0 + {(x - 2)}^{2}$ near $x = 2$ , find $\lim_{x \to 2} f (x)$ $0$
$\lim_{x \to \infty} \frac{\sin x}{x}$ $0$
Use the squeeze theorem: $\lim_{x \to 0} x^{4} \sin (\frac{2}{x})$ $0$ , since $- x^{4} \leq x^{4} \sin (\frac{2}{x}) \leq x^{4}$
If $19 - 3 {(x + 4)}^{2} \leq f (x) \leq 19 + 3 {(x + 4)}^{2}$ near $x = - 4$ , find $\lim_{x \to - 4} f (x)$ $19$
$\lim_{x \to \infty} \frac{\cos x}{x^{3}}$ $0$
Use the squeeze theorem: $\lim_{x \to 0} x^{2} \cos (\frac{4}{x})$ $0$ , since $- x^{2} \leq x^{2} \cos (\frac{4}{x}) \leq x^{2}$
If $20 - 3 {(x - 4)}^{2} \leq f (x) \leq 20 + 3 {(x - 4)}^{2}$ near $x = 4$ , find $\lim_{x \to 4} f (x)$ $20$
$\lim_{x \to \infty} \frac{2 + \cos x}{x}$ $0$
Use the squeeze theorem: $\lim_{x \to 0} x \sin (\frac{4}{x})$ $0$ , since $- ∣ x ∣ \leq x \sin (\frac{4}{x}) \leq ∣ x ∣$
If $8 - 3 x^{2} \leq f (x) \leq 8 + 3 x^{2}$ near $x = 0$ , find $\lim_{x \to 0} f (x)$ $8$

How to do these

$\lim_{x \to 0} x^{2} \cos (\frac{1}{x})$

Cosine stays between -1 and 1, so the product stays between -x² and x².
Both -x² and x² go to 0 as x goes to 0.
The function is trapped between them.

The answer is $0$ .

Where students go wrong

Treating cos(1/x) as if it had a limit at 0. It does not: it swings between -1 and 1 faster and faster. Only the product with x² settles down, and only because of the bound.

Also called sandwich theorem or pinching theorem.

What you can put on this worksheet

A power of x times sine or cosine: Use the squeeze theorem: $\lim_{x \to 0} x^{2} \cos (\frac{3}{x})$ → $0$ , since $- x^{2} \leq x^{2} \cos (\frac{3}{x}) \leq x^{2}$
From two bounding functions: If $2 - 3 x^{2} \leq f (x) \leq 2 + 3 x^{2}$ near $x = 0$ , find $\lim_{x \to 0} f (x)$ → $2$
Limits at infinity: $\lim_{x \to \infty} \frac{\cos x}{x^{3}}$ → $0$

Questions about these worksheets

Yes. Every squeeze theorem sheet is free to print and free to download as a PDF. There is no account to make, no email to hand over and no limit on how many you take.

Yes. The answer key prints on a second page, and you can choose whether it shows each question beside its answer or just the answers in a list.

Yes. Print straight from the page, or download the PDF and print that. The sheet is laid out for paper rather than squeezed off a screen, so there is room to work under each question, and the answer key comes out on its own page.

Calculus is usually taken around 12th grade or a first college course, though schools vary and plenty of students meet it earlier or later. Pick the difficulty rather than the grade: the easy level suits a first lesson on it, the hard level suits review before a test.

Yes, as many as you like. The questions are built fresh each time rather than picked from a fixed set of files, so pressing Generate gives a new sheet. That is what you want for a second class, a retake, or two students sitting next to each other.

That is the closest comparison, and Kuta Software is good software. It is also paid software you install on a computer. These worksheets run in a browser tab for free, with no account and nothing to install. Like Kuta, the squeeze theorem questions are generated when you ask for them rather than pulled from a fixed set of files, so two classes never get the same sheet, and the answer key comes with it.

Yes. There are three levels, and you can tick more than one for a sheet that mixes them, in which case the questions come out easiest first. The harder levels are not just larger numbers: they ask for something the easy ones do not.

You pick from 3: a power of x times sine or cosine, from two bounding functions and limits at infinity. Tick as many as you want and set how many of each, or let it spread them evenly.

Treating cos(1/x) as if it had a limit at 0. It does not: it swings between -1 and 1 faster and faster. Only the product with x² settles down, and only because of the bound.

Squeeze theorem worksheet

Select your difficulty

The Squeeze Theorem

The Squeeze TheoremAnswer keyVersion 1

How to do these

Where students go wrong

What you can put on this worksheet

Questions about these worksheets

Are these squeeze theorem worksheets free?+

Do the squeeze theorem worksheets come with an answer key?+

Are these squeeze theorem worksheets printable?+

What grade are these squeeze theorem worksheets for?+

Can I get a different version of the same squeeze theorem worksheet?+

Is this a free alternative to Kuta Software for squeeze theorem?+

Can I choose how hard the squeeze theorem questions are?+

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What do students get wrong with squeeze theorem?+