# The Squeeze Theorem: sin x/x, x² sin(1/x) and Classic Trig Limits

> Learn the squeeze theorem step by step, see why sin x/x approaches 1, and work through classic trig limits like x² sin(1/x) and (1 - cos x)/x.

Source: https://practicecalculus.com/blog/squeeze-theorem
Updated: 2026-11-05

## Key takeaways

- If $g(x) \le f(x) \le h(x)$ near $a$ and both $g$ and $h$ approach the same limit $L$, then $f(x)$ approaches $L$ too.
- The bound $-1 \le \sin(\text{anything}) \le 1$ is the starting point for almost every squeeze theorem problem.
- A unit circle area argument gives $\cos x \le \frac{\sin x}{x} \le 1$, which squeezes $\lim_{x \to 0} \frac{\sin x}{x}$ to exactly $1$.
- From $\frac{\sin x}{x} \to 1$ you get the other classic trig limits with algebra alone, such as $\frac{1 - \cos x}{x} \to 0$ and $\frac{\tan x}{x} \to 1$.
- The two bounding functions must approach the same number, or the squeeze tells you nothing.

You plug $x = 0$ into $\frac{\sin x}{x}$ and get $\frac{0}{0}$. You try $x^2 \sin(1/x)$ and the calculator shows a wiggle that never settles down. Factoring does nothing, and the limit laws say you cannot split the product because $\sin(1/x)$ has no limit. The **squeeze theorem** is the tool built for exactly these limits.

This guide explains what the squeeze theorem says, gives a reliable four-step method, and works through the classic trig limits every Calculus 1 course tests, including the famous proof that $\frac{\sin x}{x} \to 1$.

## What is the squeeze theorem?

In plain language: if a function is stuck between a floor and a ceiling, and the floor and ceiling meet at the same height, the function in the middle gets squeezed to that height too.

Here is the formal statement. Suppose that for all $x$ near $a$ (except possibly at $x = a$ itself),

$$g(x) \le f(x) \le h(x)$$

and that

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

Then

$$\lim_{x \to a} f(x) = L.$$

The same statement holds for one-sided limits and for limits as $x \to \infty$. You will also see it called the **sandwich theorem** or the **pinching theorem**. The [squeeze theorem reference page](/theorems/squeeze-theorem) has the statement in compact form.

Two details matter:

1. The inequality only has to hold **near** $a$, not everywhere. For $x \to 0$, it is enough that it holds on some small interval like $0 < \lvert x \rvert < 1$.
2. The two bounds must have the **same** limit. If the floor goes to $-1$ and the ceiling goes to $1$, you have learned only that $f$ stays between them, not where it goes.

If you want a refresher on what a limit is before going further, see [What Is a Limit in Calculus?](/blog/what-is-a-limit).

## The bound that powers almost every problem

Sine and cosine never leave the interval $[-1, 1]$, no matter what you feed them:

$$-1 \le \sin(\text{anything}) \le 1, \qquad -1 \le \cos(\text{anything}) \le 1.$$

That is the whole trick. The input can be $1/x$, $x^3$ or $e^x$, and the output is still trapped. Most squeeze problems start from this line and then multiply or divide every part of it by something that goes to $0$ or $\infty$. You can see the bounded wave on the [sine function page](/functions/sine-function).

## How to use the squeeze theorem in 4 steps

1. **Isolate the bounded piece.** Find the oscillating part, usually a sine or cosine, and write $-1 \le \sin(\dots) \le 1$.
2. **Build the rest of the function.** Multiply (or divide) all three parts by the remaining factor. If that factor can be negative, use absolute values so the inequality does not flip unexpectedly.
3. **Take the limit of both bounds.** Use direct substitution or the [limit laws](/rules/limit-laws).
4. **Check that they match.** If both bounds approach $L$, conclude the middle function does too.

## Worked example 1: x² sin(1/x) as x approaches 0

Find $\displaystyle \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$.

**Why other methods fail.** Substitution gives $0 \cdot \sin(1/0)$, which is undefined. The product law needs both limits to exist, but $\sin(1/x)$ oscillates between $-1$ and $1$ infinitely often near $0$ and has no limit.

**Step 1. Bound the oscillating piece.** For every $x \ne 0$,

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

**Step 2. Multiply by $x^2$.** Since $x^2 > 0$ for $x \ne 0$, multiplying does not flip the inequalities:

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

**Step 3. Take the limits of the bounds.**

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

**Step 4. Conclude.** Both bounds go to $0$, so

$$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0.$$

```graph
title: 'x² sin(1/x) squeezed between y = x² and y = -x² near 0'
caption: 'The curve $x^2 \sin(1/x)$ oscillates infinitely often near $0$, but it is trapped between $-x^2$ and $x^2$, which both shrink to $0$.'
x: [-0.5, 0.5]
y: [-0.25, 0.25]
functions:
  - expr: x^2*sin(1/x)
    label: y = x² sin(1/x)
  - expr: x^2
    label: y = x²
    dashed: true
  - expr: -x^2
    label: y = -x²
    dashed: true
```

## Worked example 2: sin x / x as x approaches infinity

Find $\displaystyle \lim_{x \to \infty} \frac{\sin x}{x}$.

**Step 1.** Start from $-1 \le \sin x \le 1$.

**Step 2.** We only care about large $x$, so take $x > 0$. Dividing by a positive number keeps the inequality direction:

$$-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}.$$

**Step 3.** Both bounds go to $0$ as $x \to \infty$ (see [limits at infinity](/blog/limits-at-infinity) if that step is unfamiliar).

**Step 4.** Therefore

$$\lim_{x \to \infty} \frac{\sin x}{x} = 0.$$

```graph
title: 'sin(x)/x trapped between 1/x and -1/x as x grows'
caption: 'As $x \to \infty$ the bounds $\pm\frac{1}{x}$ close in on $0$, and $\frac{\sin x}{x}$ is pinched to $0$ with them.'
x: [0, 20]
y: [-0.6, 1.1]
functions:
  - expr: sin(x)/x
    label: y = sin(x)/x
  - expr: 1/x
    label: y = 1/x
    dashed: true
    domain: [1, 20]
  - expr: -1/x
    label: y = -1/x
    dashed: true
    domain: [1, 20]
```

Notice the contrast with the next example. The same expression has limit $0$ at infinity but limit $1$ at zero. Always check where $x$ is heading.

## Worked example 3: why the limit of sin x / x as x approaches 0 is 1

This is the most important trig limit in Calculus 1:

$$\lim_{x \to 0} \frac{\sin x}{x} = 1.$$

Substitution gives $\frac{0}{0}$, and the simple $-1 \le \sin x \le 1$ bound is not good enough here (dividing by $x$ gives bounds $\pm\frac{1}{x}$, which blow up). We need sharper bounds, and they come from geometry.

**Step 1. Set up the unit circle.** Take $0 < x < \frac{\pi}{2}$, measured in radians. Let $O$ be the origin, $A = (1, 0)$, $P = (\cos x, \sin x)$ on the circle, and $T = (1, \tan x)$ where the line through $O$ and $P$ meets the vertical line $x = 1$.

**Step 2. Compare three areas.** The small triangle $OAP$ sits inside the sector $OAP$, which sits inside the big triangle $OAT$:

- Triangle $OAP$: base $1$, height $\sin x$, area $\frac{1}{2}\sin x$.
- Sector with angle $x$ in a circle of radius $1$: area $\frac{1}{2}x$.
- Triangle $OAT$: base $1$, height $\tan x$, area $\frac{1}{2}\tan x$.

So

$$\frac{1}{2}\sin x \le \frac{1}{2}x \le \frac{1}{2}\tan x \quad\Longrightarrow\quad \sin x \le x \le \frac{\sin x}{\cos x}.$$

**Step 3. Divide by $\sin x$.** It is positive on $0 < x < \frac{\pi}{2}$, so nothing flips:

$$1 \le \frac{x}{\sin x} \le \frac{1}{\cos x}.$$

**Step 4. Take reciprocals.** All three quantities are positive, and taking reciprocals of positive numbers reverses the inequalities:

$$\cos x \le \frac{\sin x}{x} \le 1.$$

**Step 5. Handle negative $x$.** Both $\cos x$ and $\frac{\sin x}{x}$ are even functions (replacing $x$ with $-x$ changes nothing), so the same inequality holds for $-\frac{\pi}{2} < x < 0$.

**Step 6. Squeeze.** $\lim_{x \to 0} \cos x = 1$ and $\lim_{x \to 0} 1 = 1$, so

$$\lim_{x \to 0} \frac{\sin x}{x} = 1.$$

```graph
title: 'sin(x)/x squeezed between cos(x) and 1 near x = 0'
caption: 'For $0 < \lvert x \rvert < \frac{\pi}{2}$, the graph of $\frac{\sin x}{x}$ lies between $\cos x$ and $1$. Both bounds equal $1$ at $x = 0$, so the limit is $1$.'
x: [-1.6, 1.6]
y: [-0.1, 1.3]
functions:
  - expr: sin(x)/x
    label: y = sin(x)/x
  - expr: cos(x)
    label: y = cos(x)
    dashed: true
  - expr: '1'
    label: y = 1
    dashed: true
points:
  - [0, 1, 'hole at (0, 1)', open]
```

> **Tip:** this limit is the reason the derivative of $\sin x$ is $\cos x$. The proof in [Derivatives of Trig Functions](/blog/derivatives-of-trig-functions) uses it directly. That is also why using [L'Hôpital's rule](/blog/lhopitals-rule) on $\frac{\sin x}{x}$ is circular: L'Hôpital needs that derivative first.

## Worked example 4: the limit of (1 - cos x)/x

Find $\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x}$.

Once you know $\frac{\sin x}{x} \to 1$, you do not need a new squeeze. Algebra turns this into the limit you already have.

**Step 1. Multiply by the conjugate** $1 + \cos x$:

$$\frac{1 - \cos x}{x} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{x(1 + \cos x)}.$$

**Step 2. Use $1 - \cos^2 x = \sin^2 x$:**

$$\frac{\sin^2 x}{x(1 + \cos x)} = \frac{\sin x}{x} \cdot \frac{\sin x}{1 + \cos x}.$$

**Step 3. Take each limit.** The first factor goes to $1$. The second is continuous at $0$, so substitute: $\frac{\sin 0}{1 + \cos 0} = \frac{0}{2} = 0$.

$$\lim_{x \to 0} \frac{1 - \cos x}{x} = 1 \cdot 0 = 0.$$

The same steps give a close cousin. For $\frac{1 - \cos x}{x^2}$, you get $\left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1 + \cos x} \to 1 \cdot \frac{1}{2} = \frac{1}{2}$.

## Worked example 5: rescaling sin(kx)/x and tan x/x

Many exam problems disguise $\frac{\sin x}{x}$. The fix is to make the denominator match the inside of the sine.

**Find $\displaystyle \lim_{x \to 0} \frac{\sin(5x)}{3x}$.**

The inside is $5x$, so we want $5x$ in the denominator. Multiply and divide by $5$:

$$\frac{\sin(5x)}{3x} = \frac{5}{3} \cdot \frac{\sin(5x)}{5x}.$$

As $x \to 0$, $u = 5x \to 0$ too, so $\frac{\sin u}{u} \to 1$. The answer is $\frac{5}{3}$.

**Find $\displaystyle \lim_{x \to 0} \frac{\tan x}{x}$.**

Write tangent as sine over cosine:

$$\frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x} \to 1 \cdot \frac{1}{1} = 1.$$

## The absolute value version

A handy shortcut follows from the squeeze theorem: if $\lvert f(x) \rvert \le h(x)$ and $h(x) \to 0$, then $f(x) \to 0$. That is because $-h(x) \le f(x) \le h(x)$, and both sides go to $0$.

For example, $\lvert x \sin(1/x) \rvert \le \lvert x \rvert$, and $\lvert x \rvert \to 0$, so $\lim_{x \to 0} x \sin(1/x) = 0$. This version only works when the target limit is $0$.

## Common mistakes with the squeeze theorem

1. **Bounds that do not match.** Writing $-1 \le \cos x \le 1$ and stopping there proves nothing about the limit. The floor and ceiling must approach the same number.
2. **Flipping signs carelessly.** For $x \sin(1/x)$, writing $-x \le x\sin(1/x) \le x$ is wrong when $x < 0$, because multiplying by a negative number reverses the inequalities. Use $-\lvert x \rvert \le x\sin(1/x) \le \lvert x \rvert$ instead.
3. **Splitting the limit with the product law.** Writing $\lim x^2 \cdot \lim \sin(1/x) = 0 \cdot (\text{something})$ is invalid, because $\lim_{x \to 0}\sin(1/x)$ does not exist. The limit laws need every piece to have a limit.
4. **Confusing $x \to 0$ with $x \to \infty$.** $\frac{\sin x}{x}$ goes to $1$ at $0$ but to $0$ at infinity.
5. **Mismatched inside and denominator.** $\frac{\sin(5x)}{x}$ is not $1$. It is $5 \cdot \frac{\sin(5x)}{5x} \to 5$.
6. **Using degrees.** The area of a sector is $\frac{1}{2}x$ only in radians. In degrees, $\frac{\sin x}{x}$ approaches $\frac{\pi}{180}$, not $1$.

For more traps like these across all of Calculus 1, see [12 Common Calculus Mistakes](/blog/common-calculus-mistakes).

## Quick reference: classic trig limits

| Limit | Value | How to get it |
| --- | --- | --- |
| $\lim_{x \to 0} \frac{\sin x}{x}$ | $1$ | Squeeze with $\cos x$ and $1$ |
| $\lim_{x \to 0} \frac{\sin(kx)}{x}$ | $k$ | Rewrite as $k \cdot \frac{\sin(kx)}{kx}$ |
| $\lim_{x \to 0} \frac{\tan x}{x}$ | $1$ | $\frac{\sin x}{x} \cdot \frac{1}{\cos x}$ |
| $\lim_{x \to 0} \frac{1 - \cos x}{x}$ | $0$ | Multiply by conjugate $1 + \cos x$ |
| $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$ | $\frac{1}{2}$ | Conjugate, then $\left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1 + \cos x}$ |
| $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$ | $0$ | Squeeze with $\pm x^2$ |
| $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$ | $0$ | Squeeze with $\pm\lvert x \rvert$ |
| $\lim_{x \to \infty} \frac{\sin x}{x}$ | $0$ | Squeeze with $\pm\frac{1}{x}$ |

The squeeze theorem is one of five standard limit strategies. For the full toolkit, including when to reach for substitution, factoring or conjugates first, read [How to Evaluate Limits](/blog/how-to-evaluate-limits).

## Practice until the bounds come naturally

The hard part of the squeeze theorem is not the theorem. It is choosing the bounds and keeping the inequality signs straight. That becomes automatic after a dozen problems. Work through the interactive limits practice below: each question gives instant feedback, flags a mismatched bound or a flipped inequality right away, and shows the full solution so you can see exactly where the squeeze happens.

## Frequently asked questions

### What is the squeeze theorem in simple terms?

If a function is trapped between two other functions, and those two functions head to the same value, the trapped function has nowhere else to go. It must approach that same value. That is why it is also called the sandwich theorem or the pinching theorem.

### When should I use the squeeze theorem?

Use it when a limit involves a bounded but wildly oscillating piece, like $\sin(1/x)$ or $\cos x$ as $x \to \infty$, multiplied or divided by something that goes to zero or infinity. Direct substitution and factoring fail on these, and the limit laws do not apply because the oscillating piece has no limit of its own.

### Why is the limit of sin x over x equal to 1?

Comparing the areas of two triangles and a circular sector on the unit circle shows that $\cos x \le \frac{\sin x}{x} \le 1$ for $x$ near $0$. Since $\cos x \to 1$, the squeeze theorem forces $\frac{\sin x}{x} \to 1$. This only works when $x$ is measured in radians.

### Can I use L'Hôpital's rule for sin x over x instead?

You will get the right number, but the argument is circular. L'Hôpital's rule needs the derivative of $\sin x$, and that derivative is proved using $\lim_{x \to 0} \frac{\sin x}{x} = 1$. Many instructors require the squeeze theorem proof for exactly this reason.

### What is the limit of sin x over x as x goes to infinity?

It is $0$, not $1$. For $x > 0$ we have $-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$, and both bounds go to $0$, so the squeeze theorem gives $0$.