Is the squeeze theorem the same as the sandwich theorem?

Yes, same theorem, different snack. It is also called the pinching theorem.

When do you use the squeeze theorem?

When the function has a bounded factor that oscillates, such as sin or cos of something that blows up, multiplied by a factor that tends to zero.

Does the squeeze theorem work for limits at infinity?

Yes. The inequality only has to hold for large enough x, and both bounds must tend to the same limit as x → ∞.

Why is lim[x → 0] (sin x)/x = 1 important?

It is the limit behind d/dx sin x = cos x, and every trig derivative is built from that one.

Where can I find squeeze theorem practice problems?

Right on this page: the practice problems section has three squeeze theorem problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 243 related problems after that, free and with no account: try a five-minute Sprint.

Blog · Calculus guides · 31 October 2026 · 4 min read

By the Asymptote team. Every answer on this page is checked by the same machine that grades the game.

Squeeze Theorem: 3 Examples & Practice Problems

The squeeze theorem says that if $g (x) \leq f (x) \leq h (x)$ near $a$ , and $g$ and $h$ both tend to the same limit $L$ as $x \to a$ , then $f$ tends to $L$ too. It is how you find limits of functions that wobble, like $x^{2} sin \frac{1}{x}$ as $x \to 0$ , where substitution fails and nothing cancels. Trap the function between two simpler ones that agree, and it has nowhere else to go.

The formula

g (x) \leq f (x) \leq h (x) near a and x \to a lim g (x) = x \to a lim h (x) = L ⟹ x \to a lim f (x) = L

Also sold as the sandwich theorem, and in Italy as the two carabinieri theorem: two police officers walking a suspect to the station, one on each side.

When to use it

Use the squeeze theorem when a limit involves something bounded but badly behaved, usually $sin$ or $cos$ of something that blows up, multiplied by something that goes to zero. The bounded part is the giveaway: $- 1 \leq sin (anything) \leq 1$ always, so you can bound the whole expression without knowing what the inside does. If plain substitution, factoring or L'Hôpital's rule works, use that instead; the squeeze is for the limits those methods cannot touch.

How to do it, step by step

Find the bounded piece. Usually $sin (\cdot)$ or $cos (\cdot)$ , which sits between $- 1$ and $1$ whatever is inside it.
Write the inequality for that piece. For example $- 1 \leq sin \frac{1}{x} \leq 1$ .
Build the rest of $f$ around it. Multiply or divide every part of the inequality by the other factor. If that factor is negative, flip the inequality signs; if it is positive (like $x^{2}$ ), they stay put.
Take the limit of both bounds. If they agree on $L$ , write the conclusion: by the squeeze theorem, $lim f = L$ . If they disagree, your bounds are too loose, and you need tighter ones.

Worked examples

Example 1: A wobble times something small

Easy

Find the limit:

x \to 0 lim x^{2} sin (\frac{1}{x})

Substitution gives $0 \cdot sin \frac{1}{0}$ , which is meaningless, and $sin \frac{1}{x}$ itself has no limit at $0$ : it oscillates faster and faster between $- 1$ and $1$ .

But it is bounded. Start there and multiply through by $x^{2}$ , which is positive, so the inequalities keep their direction:

- 1 \leq sin \frac{1}{x} \leq 1 ⟹ - x^{2} \leq x^{2} sin \frac{1}{x} \leq x^{2}

Both $- x^{2}$ and $x^{2}$ go to $0$ as $x \to 0$ . The function is pinned between them, so it goes to $0$ as well, however much it wiggles on the way.

On a graph, $x^{2} sin \frac{1}{x}$ is a wave trapped inside the two parabolas $y = x^{2}$ and $y = - x^{2}$ , wiggling infinitely often as it is crushed into the origin.

Answer

x \to 0 lim x^{2} sin (\frac{1}{x}) = 0

Example 2: A limit at infinity

Medium

Find the limit:

x \to \infty lim \frac{sin x}{x}

L'Hôpital's rule does not apply: the top does not go to $0$ or $\infty$ , it just keeps oscillating. The bound does the work instead.

For $x > 0$ , divide $- 1 \leq sin x \leq 1$ through by $x$ :

- \frac{1}{x} \leq \frac{sin x}{x} \leq \frac{1}{x}

Both bounds go to $0$ as $x \to \infty$ , so the function does too. On a graph, that is a wave whose height shrinks like $\frac{1}{x}$ , settling onto the horizontal asymptote $y = 0$ while crossing it infinitely often.

Answer

x \to \infty lim \frac{sin x}{x} = 0

Example 3: The famous one

Harder

Show that:

x \to 0 lim \frac{sin x}{x} = 1

Substitution gives $\frac{0}{0}$ , and L'Hôpital's rule would need the derivative of $sin x$ , which is itself proved using this limit. So it has to come from geometry.

Take $0 < x < \frac{π}{2}$ on the unit circle. A small triangle, a sector of angle $x$ and a larger triangle sit inside each other, with areas $\frac{1}{2} sin x$ , $\frac{1}{2} x$ and $\frac{1}{2} tan x$ :

sin x \leq x \leq tan x ⟹ 1 \leq \frac{x}{sin x} \leq \frac{1}{cos x} ⟹ cos x \leq \frac{sin x}{x} \leq 1

The middle step divides by $sin x > 0$ ; the last one flips everything upside down, which reverses the inequalities. Both bounds tend to $1$ as $x \to 0$ . For negative $x$ , $\frac{s i n x}{x}$ takes the same values, since both top and bottom change sign, so the limit from the left is $1$ as well.

Answer

x \to 0 lim \frac{sin x}{x} = 1

Squeeze theorem practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

Find the limit:

x \to 0 lim x cos (\frac{1}{x})

Stuck? Show the answer

Answer

0

How to start

Start from $- 1 \leq cos \frac{1}{x} \leq 1$ and multiply by $∣ x ∣$ : the function is trapped between $- ∣ x ∣$ and $∣ x ∣$ .

Problem 2

Medium

Find the limit:

x \to \infty lim \frac{3 x + sin x}{x}

Stuck? Show the answer

Answer

3

How to start

Since $- 1 \leq sin x \leq 1$ , the top is between $3 x - 1$ and $3 x + 1$ . Divide those bounds by $x$ and take their limits.

Problem 3

Harder

Find the limit:

x \to 0 lim \frac{1 - cos ( 4 x )}{2 x ^{2}}

Stuck? Show the answer

Answer

4

How to start

Multiply top and bottom by $1 + cos (4 x)$ , so the top becomes $sin^{2} (4 x)$ . Then build $(\frac{s i n 4 x}{4 x})^{2}$ , the squeezed limit, which tends to $1$ .

Want more? Asymptote has 243 related problems, free, with no account.

Common mistakes

Bounds that do not agree. $- 1 \leq sin \frac{1}{x} \leq 1$ is true, but the bounds tend to $- 1$ and $1$ , so it proves nothing about $sin \frac{1}{x}$ alone. The squeeze only works when both sides meet.
Forgetting to flip the inequality. Multiplying by a negative number reverses both signs. If $x < 0$ is in play, check which bound is really on top.
Writing $lim sin \frac{1}{x} = 0$ along the way. It has no limit at all. The product has a limit because the $x^{2}$ crushes it, not because the sine settles down.
Splitting the limit of a product. $lim x^{2} \cdot lim sin \frac{1}{x}$ is not allowed when the second limit does not exist. That is precisely the case the squeeze theorem exists for.

Practice it now

Five minutes, no account

Asymptote has no squeeze-theorem problems as such, since a proof is hard to type into a box. The closest thing is the standard trig limits, $\frac{s i n ( a x )}{b x}$ and friends as $x \to 0$ , which run on the limit in the third example and turn up in a Sprint once the difficulty reaches 4.

Play a Sprint See it on the formula sheet

FAQ

Is the squeeze theorem the same as the sandwich theorem?: Yes, same theorem, different snack. It is also called the pinching theorem.
When do you use the squeeze theorem?: When the function has a bounded factor that oscillates, such as $sin$ or $cos$ of something that blows up, multiplied by a factor that tends to zero.
Does the squeeze theorem work for limits at infinity?: Yes. The inequality only has to hold for large enough $x$ , and both bounds must tend to the same limit as $x \to \infty$ .
Why is $lim_{x \to 0} \frac{s i n x}{x} = 1$ important?: It is the limit behind $\frac{d}{d x} sin x = cos x$ , and every trig derivative is built from that one.
Where can I find squeeze theorem practice problems?: Right on this page: the practice problems section has three squeeze theorem problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 243 related problems after that, free and with no account: try a five-minute Sprint.

Blog · Calculus guides · 31 October 2026 · 4 min read

By the Asymptote team. Every answer on this page is checked by the same machine that grades the game.

Squeeze Theorem: 3 Examples & Practice Problems

The formula

g (x) \leq f (x) \leq h (x) near a and x \to a lim g (x) = x \to a lim h (x) = L ⟹ x \to a lim f (x) = L

Also sold as the sandwich theorem, and in Italy as the two carabinieri theorem: two police officers walking a suspect to the station, one on each side.

When to use it

How to do it, step by step

Find the bounded piece. Usually $sin (\cdot)$ or $cos (\cdot)$ , which sits between $- 1$ and $1$ whatever is inside it.
Write the inequality for that piece. For example $- 1 \leq sin \frac{1}{x} \leq 1$ .
Build the rest of $f$ around it. Multiply or divide every part of the inequality by the other factor. If that factor is negative, flip the inequality signs; if it is positive (like $x^{2}$ ), they stay put.
Take the limit of both bounds. If they agree on $L$ , write the conclusion: by the squeeze theorem, $lim f = L$ . If they disagree, your bounds are too loose, and you need tighter ones.

Worked examples

Example 1: A wobble times something small

Easy

Find the limit:

x \to 0 lim x^{2} sin (\frac{1}{x})

Substitution gives $0 \cdot sin \frac{1}{0}$ , which is meaningless, and $sin \frac{1}{x}$ itself has no limit at $0$ : it oscillates faster and faster between $- 1$ and $1$ .

But it is bounded. Start there and multiply through by $x^{2}$ , which is positive, so the inequalities keep their direction:

- 1 \leq sin \frac{1}{x} \leq 1 ⟹ - x^{2} \leq x^{2} sin \frac{1}{x} \leq x^{2}

Both $- x^{2}$ and $x^{2}$ go to $0$ as $x \to 0$ . The function is pinned between them, so it goes to $0$ as well, however much it wiggles on the way.

On a graph, $x^{2} sin \frac{1}{x}$ is a wave trapped inside the two parabolas $y = x^{2}$ and $y = - x^{2}$ , wiggling infinitely often as it is crushed into the origin.

Answer

x \to 0 lim x^{2} sin (\frac{1}{x}) = 0

Example 2: A limit at infinity

Medium

Find the limit:

x \to \infty lim \frac{sin x}{x}

L'Hôpital's rule does not apply: the top does not go to $0$ or $\infty$ , it just keeps oscillating. The bound does the work instead.

For $x > 0$ , divide $- 1 \leq sin x \leq 1$ through by $x$ :

- \frac{1}{x} \leq \frac{sin x}{x} \leq \frac{1}{x}

Answer

x \to \infty lim \frac{sin x}{x} = 0

Example 3: The famous one

Harder

Show that:

x \to 0 lim \frac{sin x}{x} = 1

Substitution gives $\frac{0}{0}$ , and L'Hôpital's rule would need the derivative of $sin x$ , which is itself proved using this limit. So it has to come from geometry.

sin x \leq x \leq tan x ⟹ 1 \leq \frac{x}{sin x} \leq \frac{1}{cos x} ⟹ cos x \leq \frac{sin x}{x} \leq 1

Answer

x \to 0 lim \frac{sin x}{x} = 1

Squeeze theorem practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

Find the limit:

x \to 0 lim x cos (\frac{1}{x})

Stuck? Show the answer

Answer

0

How to start

Start from $- 1 \leq cos \frac{1}{x} \leq 1$ and multiply by $∣ x ∣$ : the function is trapped between $- ∣ x ∣$ and $∣ x ∣$ .

Problem 2

Medium

Find the limit:

x \to \infty lim \frac{3 x + sin x}{x}

Stuck? Show the answer

Answer

3

How to start

Since $- 1 \leq sin x \leq 1$ , the top is between $3 x - 1$ and $3 x + 1$ . Divide those bounds by $x$ and take their limits.

Problem 3

Harder

Find the limit:

x \to 0 lim \frac{1 - cos ( 4 x )}{2 x ^{2}}

Stuck? Show the answer

Answer

4

How to start

Multiply top and bottom by $1 + cos (4 x)$ , so the top becomes $sin^{2} (4 x)$ . Then build $(\frac{s i n 4 x}{4 x})^{2}$ , the squeezed limit, which tends to $1$ .

Want more? Asymptote has 243 related problems, free, with no account.

Common mistakes

Bounds that do not agree. $- 1 \leq sin \frac{1}{x} \leq 1$ is true, but the bounds tend to $- 1$ and $1$ , so it proves nothing about $sin \frac{1}{x}$ alone. The squeeze only works when both sides meet.
Forgetting to flip the inequality. Multiplying by a negative number reverses both signs. If $x < 0$ is in play, check which bound is really on top.
Writing $lim sin \frac{1}{x} = 0$ along the way. It has no limit at all. The product has a limit because the $x^{2}$ crushes it, not because the sine settles down.
Splitting the limit of a product. $lim x^{2} \cdot lim sin \frac{1}{x}$ is not allowed when the second limit does not exist. That is precisely the case the squeeze theorem exists for.

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

Is the squeeze theorem the same as the sandwich theorem?: Yes, same theorem, different snack. It is also called the pinching theorem.
When do you use the squeeze theorem?: When the function has a bounded factor that oscillates, such as $sin$ or $cos$ of something that blows up, multiplied by a factor that tends to zero.
Does the squeeze theorem work for limits at infinity?: Yes. The inequality only has to hold for large enough $x$ , and both bounds must tend to the same limit as $x \to \infty$ .
Why is $lim_{x \to 0} \frac{s i n x}{x} = 1$ important?: It is the limit behind $\frac{d}{d x} sin x = cos x$ , and every trig derivative is built from that one.
Where can I find squeeze theorem practice problems?: Right on this page: the practice problems section has three squeeze theorem problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 243 related problems after that, free and with no account: try a five-minute Sprint.

Squeeze Theorem: 3 Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A wobble times something small

Example 2: A limit at infinity

Example 3: The famous one

Squeeze theorem practice problems

Common mistakes

FAQ

Keep going

Squeeze Theorem: 3 Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A wobble times something small

Example 2: A limit at infinity

Example 3: The famous one

Squeeze theorem practice problems

Common mistakes

FAQ

Keep going