Squeeze Theorem: Statement & Examples

The squeeze theorem finds a limit by trapping a function between two others. If $g(x) \le f(x) \le h(x)$ for x near a, and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$ , then $\lim_{x \to a} f(x) = L$ too. It is also called the sandwich theorem.

Before this lesson: Limit Laws, Limits at Infinity and Graphs of Sine and Cosine

The idea behind the squeeze theorem

Picture two friends walking on either side of you down a narrow hallway, and both of them end up at the same door. Since you are always between them, you must end up at that door too.

That is the squeeze theorem. When a function is too wild to handle directly, such as one that wiggles forever near a point, you can often trap it between two simple functions. If the simple functions have the same limit, the wild one is forced to have it as well.

The most common trap comes from sine and cosine: $-1 \le \sin(\text{anything}) \le 1$ and $-1 \le \cos(\text{anything}) \le 1$ .

Graph of y = x² sin(1/x) for x from −0.4 to 0.4, oscillating faster near 0 while staying between the dashed parabolas y = x² above and y = −x² below. Both parabolas meet at the origin, so the squeezed function is forced to approach 0 as x approaches 0. — The curve x² sin(1/x) wiggles infinitely often near 0, but it is trapped between −x² and x², so its limit is 0.

The squeeze theorem

Suppose $g(x) \le f(x) \le h(x)$ for all x in an open interval around a (except possibly at a itself), and

\lim_{x \to a} g(x) = L \quad \text{and} \quad \lim_{x \to a} h(x) = L

Then $\lim_{x \to a} f(x) = L$ . The theorem also works for one-sided limits and for limits as $x \to \infty$ .

Step	What to do
1	Start from a known inequality, usually $-1 \le \sin(\cdot) \le 1$ or $-1 \le \cos(\cdot) \le 1$ .
2	Multiply or divide all three parts to build f(x) in the middle. Multiplying by a positive quantity keeps the inequality signs.
3	Find the limits of the outside functions.
4	If they are equal, the middle function has that limit.

Squeeze theorem examples

Example 1: Given bounds

Suppose $4 - x^2 \le f(x) \le 4 + x^2$ for all x near 0. Find $\lim_{x \to 0} f(x)$ .

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

Find the limit of the lower function by direct substitution.

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

Find the limit of the upper function.

\lim_{x \to 0} f(x) = 4

The bounds have the same limit, so the squeeze theorem applies.

Answer: 4.

Example 2: The classic wiggle

Find $\lim_{x \to 0} x^2 \sin\frac{1}{x}$ .

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

Sine is always between −1 and 1, for every x ≠ 0.

-x^2 \le x^2 \sin\frac{1}{x} \le x^2

Multiply all three parts by $x^2$ . Since $x^2$ is positive, the signs stay the same.

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

Find the limit of the lower bound.

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

Find the limit of the upper bound.

\lim_{x \to 0} x^2 \sin\frac{1}{x} = 0

Squeeze theorem.

Answer: 0, as the graph above shows. You cannot use the product law here, because $\lim_{x \to 0} \sin\frac{1}{x}$ does not exist.

Example 3: A limit at infinity

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

-1 \le \sin x \le 1

Start with the bounds on sine.

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

Divide all three parts by x. For x greater than 0, the signs stay the same.

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

A constant over x goes to 0 (see limits at infinity).

\lim_{x \to \infty} \frac{1}{x} = 0

Same for the upper bound.

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

Squeeze theorem.

Answer: 0. The graph wiggles across the x-axis with smaller and smaller swings, so y = 0 is a horizontal asymptote even though the curve crosses it again and again.

Example 4: The special limit sin x / x as x approaches 0

Use the inequality $\cos x \le \frac{\sin x}{x} \le 1$ , which holds for x near 0 (x ≠ 0), to find $\lim_{x \to 0} \frac{\sin x}{x}$ .

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

Comparing areas of two triangles and a sector of the unit circle gives this inequality, for x in radians.

\lim_{x \to 0} \cos x = \cos 0 = 1

Cosine is continuous, so substitute.

\lim_{x \to 0} 1 = 1

The upper bound is the constant 1.

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

Squeeze theorem.

Answer: 1. This confirms the value suggested by the table in Introduction to Limits, and it is the key fact behind the derivative of sine later in Calculus 1.

Example 5: Word problem with a damped spring

A weight on a spring moves so that its position after t seconds is $d(t) = 3e^{-t}\cos(5t)$ centimeters from rest. Show that $\lim_{t \to \infty} d(t) = 0$ .

-1 \le \cos(5t) \le 1

Cosine is between −1 and 1.

-3e^{-t} \le 3e^{-t}\cos(5t) \le 3e^{-t}

Multiply all parts by $3e^{-t}$ , which is always positive.

\lim_{t \to \infty} 3e^{-t} = 3(0) = 0

$e^{-t} = \frac{1}{e^t} \to 0$ as t grows (see exponential decay).

\lim_{t \to \infty} (-3e^{-t}) = 0

The lower bound also goes to 0.

\lim_{t \to \infty} d(t) = 0

Squeeze theorem.

Answer: the position approaches 0, so the weight settles at rest. The bounds $\pm 3e^{-t}$ form an envelope that closes in as the bouncing fades.

Common mistakes

Mistake	Example of the error	Fix
Using the product law with a limit that doesn’t exist	$\lim x^2 \cdot \lim \sin\frac{1}{x} = 0 \cdot (\text{something})$	$\lim \sin\frac{1}{x}$ does not exist, so the product law fails. Squeeze instead.
Bounds with different limits	Using $-1 \le \sin\frac{1}{x} \le 1$ to find $\lim \sin\frac{1}{x}$	The bounds go to −1 and 1, which differ, so the theorem tells you nothing.
Multiplying by a negative without flipping	Multiplying $-1 \le \sin x \le 1$ by x when x is negative	For x less than 0, the inequality signs reverse. Use \|x\| or check each side.
Forgetting the bounds must hold near a	Using an inequality that is only true far from a	The trap must hold on an interval around a (or for large x).
Skipping the inequality step	Writing “the limit is 0 by the squeeze theorem” with no bounds	Show the inequality and the two outside limits.
Calculator in degree mode	Checking $\frac{\sin x}{x}$ near 0 and getting about 0.01745	Use radians; the limit 1 only holds in radians.

Related methods

Using absolute values

A handy version: if $|f(x)| \le h(x)$ and $h(x) \to 0$ , then $f(x) \to 0$ , because $-h(x) \le f(x) \le h(x)$ . This avoids sign trouble when x can be negative, as in $x \sin\frac{1}{x}$ .

A companion limit

Using $\frac{\sin x}{x} \to 1$ , you can also show $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ . Both limits are used to find the derivatives of the trig functions.

Practice problems

If $2x - 1 \le f(x) \le x^2$ for x near 1, find $\lim_{x \to 1} f(x)$ .
Find $\lim_{x \to 0} x^2 \cos\frac{3}{x}$ .
Find $\lim_{x \to 0} x \sin\frac{1}{x}$ .
Find $\lim_{x \to \infty} \frac{\cos x}{x^2}$ .
Find $\lim_{x \to \infty} \frac{2 + \sin x}{x}$ .
Find $\lim_{x \to 0} x^4 \sin\frac{1}{x^2}$ .
If $1 - \frac{x^2}{4} \le u(x) \le 1 + \frac{x^2}{2}$ for all x, find $\lim_{x \to 0} u(x)$ .
Find $\lim_{x \to 0^+} \sqrt{x}\, e^{\sin(\pi/x)}$ .
Explain why the squeeze theorem with $-1 \le \sin\frac{1}{x} \le 1$ does not find $\lim_{x \to 0} \sin\frac{1}{x}$ .
A buoy bobs so its height above the average water level is $y(t) = \frac{4\cos(2t)}{t + 1}$ meters after t minutes. Find $\lim_{t \to \infty} y(t)$ .

Show answers

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

Lower bound.

\lim_{x \to 1} x^2 = 1

Upper bound.

Answer: both bounds go to 1, so $\lim_{x \to 1} f(x) = 1$ .

-x^2 \le x^2 \cos\frac{3}{x} \le x^2

Multiply $-1 \le \cos\frac{3}{x} \le 1$ by $x^2 \ge 0$ .

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

Both bounds go to 0.

Answer: 0

-|x| \le x \sin\frac{1}{x} \le |x|

Since $\left|\sin\frac{1}{x}\right| \le 1$ , we get $\left|x \sin\frac{1}{x}\right| \le |x|$ .

\lim_{x \to 0} |x| = 0

Both bounds go to 0.

Answer: 0

-\frac{1}{x^2} \le \frac{\cos x}{x^2} \le \frac{1}{x^2}

Divide the cosine bounds by $x^2 \gt 0$ .

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

Both bounds go to 0.

Answer: 0

1 \le 2 + \sin x \le 3

Add 2 to all parts of $-1 \le \sin x \le 1$ .

\frac{1}{x} \le \frac{2 + \sin x}{x} \le \frac{3}{x}

Divide by x, which is positive for large x.

\lim_{x \to \infty} \frac{1}{x} = \lim_{x \to \infty} \frac{3}{x} = 0

Both bounds go to 0.

Answer: 0

-x^4 \le x^4 \sin\frac{1}{x^2} \le x^4

Multiply the sine bounds by $x^4 \ge 0$ .

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

Both bounds go to 0.

Answer: 0

\lim_{x \to 0} \left(1 - \frac{x^2}{4}\right) = 1

Lower bound.

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

Upper bound.

Answer: 1

e^{-1} \le e^{\sin(\pi/x)} \le e^{1}

Since $-1 \le \sin(\pi/x) \le 1$ and $e^u$ increases.

\frac{\sqrt{x}}{e} \le \sqrt{x}\, e^{\sin(\pi/x)} \le e\sqrt{x}

Multiply by $\sqrt{x} \ge 0$ .

\lim_{x \to 0^+} \frac{\sqrt{x}}{e} = \lim_{x \to 0^+} e\sqrt{x} = 0

Both bounds go to 0.

Answer: 0

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

The lower bound’s limit.

\lim_{x \to 0} 1 = 1

The upper bound’s limit.

Answer: the bounds approach different values (−1 and 1), so the theorem gives no conclusion. In fact $\sin\frac{1}{x}$ swings between −1 and 1 forever, so its limit does not exist.

10.

-\frac{4}{t + 1} \le \frac{4\cos(2t)}{t + 1} \le \frac{4}{t + 1}

Multiply the cosine bounds by $\frac{4}{t + 1} \gt 0$ .

\lim_{t \to \infty} \frac{4}{t + 1} = 0

Both bounds go to 0.

Answer: 0. The bobbing dies down and the buoy settles at the average water level.

Squeeze theorem FAQ

What is the squeeze theorem?

If a function stays between two others near a point, and those two have the same limit there, the function in the middle has that limit too.

When should you use the squeeze theorem?

Use it when a function contains an oscillating part, like $\sin\frac{1}{x}$ or $\cos x$ , multiplied by something that goes to 0, or when the limit laws can’t be used.

Why is it called the sandwich theorem?

The function you want is the filling between two slices of bread, the bounding functions. Squeeze the slices together and the filling has nowhere else to go.

What is the limit of sin x / x as x approaches 0?

It is 1, with x in radians. The squeeze theorem proves it using $\cos x \le \frac{\sin x}{x} \le 1$ .

Summary

If $g(x) \le f(x) \le h(x)$ near a and g and h both approach L, then f approaches L.
Build the bounds from $-1 \le \sin(\cdot) \le 1$ or $-1 \le \cos(\cdot) \le 1$ , multiplying by a positive quantity.
The two bounds must have the same limit, or the theorem says nothing.
It works for one-sided limits and for limits at infinity.
Key results: $\lim_{x \to 0} x^2 \sin\frac{1}{x} = 0$ , $\lim_{x \to \infty} \frac{\sin x}{x} = 0$ and $\lim_{x \to 0} \frac{\sin x}{x} = 1$ .

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