The Squeeze Theorem

Some limits can’t be found by substituting or by algebra. The squeeze theorem (also called the sandwich theorem) handles them by trapping the function between two simpler functions that head to the same place. It’s also how we prove the most important trig limit in calculus, $\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$. All angles on this page are in radians.

Key ideas

The squeeze theorem

Suppose that for all $x$ near $a$ (except possibly at $a$ itself),

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

and

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

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

Think of $f$ as a person walking between two friends who both arrive at the same door. Wherever they end up, $f$ ends up there too.

Bounded times something going to zero

The most common use: sine and cosine are always between $-1$ and $1$, whatever is inside them:

$$-1 \le \sin(\text{anything}) \le 1$$

So if $f(x) = (\text{something} \to 0) \times (\text{something bounded})$, the squeeze theorem shows $f(x) \to 0$. The classic example is $x^2 \sin\left(\dfrac{1}{x}\right)$ (Example 1).

$x^2 \sin\left(\tfrac{1}{x}\right)$ oscillates wildly near $0$, but it’s trapped between $-x^2$ and $x^2$, so it’s squeezed to $0$. (The vertical scale is stretched to show the wiggles.)

Two special trig limits

With $x$ in radians,

$$\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad\qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0$$

Where the first one comes from. Comparing areas inside the unit circle (a triangle, a sector, and a bigger triangle) shows that for small $x \ne 0$,

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

As $x \to 0$, $\cos x \to 1$, so the squeeze theorem gives $\dfrac{\sin x}{x} \to 1$. This only works in radians: the area of a sector is $\tfrac{1}{2}r^2\theta$ only when $\theta$ is in radians.

Useful consequences. Matching the angle with the denominator:

$$\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1, \quad\text{so}\quad \lim_{x \to 0} \frac{\sin(kx)}{x} = k$$

Justifying a squeeze on the AP exam

State the inequality, state that both outer limits are equal, and name the theorem: “Since $-x^2 \le f(x) \le x^2$ and $\displaystyle\lim_{x \to 0} (-x^2) = \lim_{x \to 0} x^2 = 0$, by the squeeze theorem $\displaystyle\lim_{x \to 0} f(x) = 0$.”

Worked examples

Example 1: The classic squeeze

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

Solution. You can’t substitute, and $\displaystyle\lim_{x \to 0} \sin\left(\frac{1}{x}\right)$ doesn’t exist, so the product law doesn’t apply. Squeeze instead. For all $x \ne 0$,

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

Multiply through by $x^2$, which is positive, so the inequalities stay the same way:

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

Since $\displaystyle\lim_{x \to 0} (-x^2) = 0$ and $\displaystyle\lim_{x \to 0} x^2 = 0$, the squeeze theorem gives

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

Example 2: Matching the angle

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

Solution. To use $\dfrac{\sin u}{u} \to 1$, the denominator must match the angle $5x$. Multiply top and bottom by $5$:

$$\lim_{x \to 0} \frac{\sin(5x)}{x} = \lim_{x \to 0} 5 \cdot \frac{\sin(5x)}{5x} = 5 \cdot 1 = 5$$

(As $x \to 0$, $u = 5x \to 0$ too, so $\dfrac{\sin(5x)}{5x} \to 1$.)

Example 3: Tangent

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

Solution. Write $\tan x = \dfrac{\sin x}{\cos x}$ and split:

$$\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1 \cdot \frac{1}{1} = 1$$

Example 4: Squeezing with given bounds

Suppose $4x - 9 \le f(x) \le x^2 - 4x + 7$ for all $x \ge 0$. Find $\displaystyle\lim_{x \to 4} f(x)$.

Solution. Find the limits of the two bounds:

$$\lim_{x \to 4} (4x - 9) = 16 - 9 = 7, \qquad \lim_{x \to 4} (x^2 - 4x + 7) = 16 - 16 + 7 = 7$$

Both bounds approach $7$, so by the squeeze theorem, $\displaystyle\lim_{x \to 4} f(x) = 7$.

(Check that the bounds make sense: $(x^2 - 4x + 7) - (4x - 9) = x^2 - 8x + 16 = (x - 4)^2 \ge 0$, so the upper bound really is above the lower one, touching at $x = 4$.)

Common mistakes

Using the product law on a limit that doesn’t exist. Writing $\displaystyle\lim x^2 \cdot \lim \sin\left(\frac{1}{x}\right) = 0 \cdot (\text{something})$ is wrong, because the second limit doesn’t exist. Use the squeeze theorem.

Bounds that go to different limits. The squeeze only works if both outer functions approach the same number. If $g \to 0$ and $h \to 2$, you only know $f$ is trapped between $0$ and $2$ (if its limit exists at all).

Forgetting radians. $\dfrac{\sin x}{x} \to 1$ is true only in radians. In degrees the limit is $\dfrac{\pi}{180}$.

Not matching the angle. $\dfrac{\sin(5x)}{x}$ does not approach $1$. You need $5x$ in the denominator too, which gives a factor of $5$.

Leaving out the justification. On free-response questions, write the inequality, the two limits, and the words “squeeze theorem”. Just writing the answer earns little credit.

Practice

1. (Warm-up) Find $\displaystyle\lim_{x \to 0} \frac{\sin(7x)}{x}$.

Solution

$$\lim_{x \to 0} 7 \cdot \frac{\sin(7x)}{7x} = 7 \cdot 1 = 7$$

2. (Warm-up) Find $\displaystyle\lim_{x \to 0} \frac{x}{\sin x}$.

Solution

This is the reciprocal of $\dfrac{\sin x}{x}$, which approaches $1$:

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

3. (Warm-up) Suppose $1 - x^2 \le f(x) \le 1 + x^2$ for all $x$. Find $\displaystyle\lim_{x \to 0} f(x)$.

Solution

Both bounds approach $1$ as $x \to 0$, so by the squeeze theorem $\displaystyle\lim_{x \to 0} f(x) = 1$.

4. (Core) Find $\displaystyle\lim_{x \to 0} x\cos\left(\frac{1}{x}\right)$. Justify your answer.

Solution

For $x \ne 0$, $-1 \le \cos\left(\dfrac{1}{x}\right) \le 1$. Multiplying by $x$ could flip the inequalities when $x \lt 0$, so use $|x|$ instead:

$$\left| x\cos\left(\frac{1}{x}\right) \right| \le |x| \quad\Rightarrow\quad -|x| \le x\cos\left(\frac{1}{x}\right) \le |x|$$

Since $\displaystyle\lim_{x \to 0} (-|x|) = \lim_{x \to 0} |x| = 0$, by the squeeze theorem the limit is $0$.

5. (Core) Find $\displaystyle\lim_{x \to 0} \frac{\sin(2x)}{5x}$.

Solution

$$\lim_{x \to 0} \frac{2}{5} \cdot \frac{\sin(2x)}{2x} = \frac{2}{5} \cdot 1 = \frac{2}{5}$$

6. (Core) Find $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{\sin x}$.

Solution

Divide top and bottom by $x$ to use both special limits:

$$\lim_{x \to 0} \frac{\dfrac{1 - \cos x}{x}}{\dfrac{\sin x}{x}} = \frac{0}{1} = 0$$

7. (Core) Find $\displaystyle\lim_{x \to \infty} \frac{\sin x}{x}$. (This is a limit as $x$ grows without bound; see limits at infinity.)

Solution

For $x \gt 0$, dividing $-1 \le \sin x \le 1$ by the positive number $x$ gives

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

As $x \to \infty$, both $-\dfrac{1}{x}$ and $\dfrac{1}{x}$ approach $0$. By the squeeze theorem, the limit is $0$.

8. (Challenge) Find $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.

Solution

Multiply top and bottom by $1 + \cos x$, and use $1 - \cos^2 x = \sin^2 x$:

$$\begin{aligned} \lim_{x \to 0} \frac{1 - \cos x}{x^2} \cdot \frac{1 + \cos x}{1 + \cos x} &= \lim_{x \to 0} \frac{\sin^2 x}{x^2 (1 + \cos x)} \\ &= \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1 + \cos x} \\ &= 1^2 \cdot \frac{1}{2} = \frac{1}{2} \end{aligned}$$

9. (Challenge) Suppose $2x \le g(x) \le x^4 - x^2 + 2$ for all real $x$.

• (a) Find $\displaystyle\lim_{x \to 1} g(x)$.
• (b) Show that the inequality is possible, that is, $2x \le x^4 - x^2 + 2$ for every $x$. (Hint: $x^4 - x^2 - 2x + 2 = (x - 1)^2(x^2 + 2x + 2)$.)

Solution

(a) $\displaystyle\lim_{x \to 1} 2x = 2$ and $\displaystyle\lim_{x \to 1} (x^4 - x^2 + 2) = 1 - 1 + 2 = 2$. Both bounds approach $2$, so by the squeeze theorem $\displaystyle\lim_{x \to 1} g(x) = 2$.

(b) $(x^4 - x^2 + 2) - 2x = (x - 1)^2(x^2 + 2x + 2)$. The first factor is a square, so it’s never negative. The second is $(x + 1)^2 + 1$, which is always positive. So the difference is never negative, and $2x \le x^4 - x^2 + 2$ for all $x$.