Math Core

Some limits resist every algebraic trick. What is $\displaystyle \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$ ? The factor $\sin\left(\dfrac{1}{x}\right)$ oscillates wildly and has no limit at $0$ , so the product law is useless. The squeeze theorem handles limits like this by trapping a hard function between two easy ones.

The idea

Imagine two functions $g$ and $h$ that both approach the same number $L$ as $x \to c$ . If a third function $f$ is always caught between them, then $f$ has nowhere to go: it must approach $L$ too.

The squeeze theorem

Suppose that for all $x$ in an open interval containing $c$ (except possibly at $c$ itself),

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

and that

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

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

The theorem is sometimes called the sandwich theorem or pinching theorem. It has three requirements, and on the AP exam you should name each one when you use it:

The inequality $g(x) \le f(x) \le h(x)$ holds near $c$ .
The lower function has limit $L$ at $c$ .
The upper function has the same limit $L$ at $c$ .

A classic example

Return to $f(x) = x^2 \sin\left(\dfrac{1}{x}\right)$ . Whatever the input, sine takes values between $-1$ and $1$ :

-1 \le \sin\left(\frac{1}{x}\right) \le 1 \quad \text{for all } x \ne 0.

Multiplying through by $x^2$ , which is positive for $x \ne 0$ , keeps the inequalities pointing 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 $\displaystyle \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$ .

y = x² sin(1/x) oscillates, but it is trapped between the dashed parabolas y = x² and y = −x², which both approach 0.Open in grapher →

The general pattern: a bounded factor times something that approaches $0$ approaches $0$ . The bounded factor is usually a sine or cosine.

Worked example: Given bounds

Suppose $5 - x^2 \le f(x) \le 5 + 3x^2$ for all $x$ near $0$ . Find $\displaystyle \lim_{x \to 0} f(x)$ .

Solution. The lower bound gives $\displaystyle \lim_{x \to 0} (5 - x^2) = 5$ and the upper bound gives $\displaystyle \lim_{x \to 0} (5 + 3x^2) = 5$ . Since $f$ is trapped between two functions that both approach $5$ , the squeeze theorem gives $\displaystyle \lim_{x \to 0} f(x) = 5$ .

Proving the special trigonometric limit

In the previous lesson you used $\displaystyle \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ without proof. The squeeze theorem proves it.

Take $0 < \theta < \dfrac{\pi}{2}$ and draw an angle $\theta$ in the unit circle. Compare three areas: the triangle inside the sector, the sector itself, and the larger right triangle that reaches up to the tangent line $x = 1$ .

Inside the unit circle: the small triangle (base 1, height sin θ) fits inside the sector, which fits inside the large right triangle (base 1, height tan θ).Open in grapher →

\underbrace{\tfrac{1}{2}\sin \theta}_{\text{small triangle}} \;\le\; \underbrace{\tfrac{1}{2}\theta}_{\text{sector}} \;\le\; \underbrace{\tfrac{1}{2}\tan \theta}_{\text{large triangle}}.

Multiply by $\dfrac{2}{\sin \theta}$ (positive here): $1 \le \dfrac{\theta}{\sin \theta} \le \dfrac{1}{\cos \theta}$ . Taking reciprocals reverses the inequalities:

\cos \theta \le \frac{\sin \theta}{\theta} \le 1.

Both $\cos\theta$ and $\sin\theta/\theta$ are even functions, so the same inequality holds for $-\dfrac{\pi}{2} < \theta < 0$ . As $\theta \to 0$ , $\cos \theta \to 1$ and the constant $1$ stays at $1$ . By the squeeze theorem, $\displaystyle \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ .

The companion limit follows from it:

\frac{1 - \cos \theta}{\theta} = \frac{1 - \cos^2 \theta}{\theta(1 + \cos \theta)} = \frac{\sin \theta}{\theta} \cdot \frac{\sin \theta}{1 + \cos \theta} \;\longrightarrow\; 1 \cdot \frac{0}{2} = 0.

Worked example: Building the bounds yourself

Evaluate $\displaystyle \lim_{x \to 0} \sqrt{x^2 + x^4} \, \cos\left(\frac{3}{x}\right)$ .

Solution. Since $-1 \le \cos\left(\dfrac{3}{x}\right) \le 1$ and $\sqrt{x^2 + x^4} \ge 0$ ,

-\sqrt{x^2 + x^4} \le \sqrt{x^2 + x^4} \, \cos\left(\frac{3}{x}\right) \le \sqrt{x^2 + x^4}.

By direct substitution, $\sqrt{x^2 + x^4} \to \sqrt{0} = 0$ , so both bounds approach $0$ . By the squeeze theorem, the limit is $0$ .

Common mistake

Multiplying an inequality by a negative quantity reverses it. That is why the bounds for $x \sin\left(\dfrac{1}{x}\right)$ are $-\lvert x \rvert$ and $\lvert x \rvert$ , not $-x$ and $x$ : when $x$ is negative, $-x$ is the larger of the two. Using $\lvert x \rvert$ (or an even power such as $x^2$ ) keeps the bounds in the right order on both sides of $0$ .

Tip

If the two bounding functions approach different limits, the squeeze theorem tells you nothing. The trapped function might have a limit anywhere in between, or no limit at all.

Practice

Practice 1

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

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Evaluate $\displaystyle \lim_{x \to 0} x^4 \cos\left(\frac{2}{x}\right)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Suppose $1 - \dfrac{x^2}{4} \le u(x) \le 1 + \dfrac{x^2}{2}$ for all $x \ne 0$ . Find $\displaystyle \lim_{x \to 0} u(x)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Which pair of inequalities, valid for all $x \ne 0$ , can be used with the squeeze theorem to show that $\displaystyle \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$ ?

Practice 5

Evaluate $\displaystyle \lim_{x \to 0} \sqrt{x^3 + x^2} \, \sin\left(\frac{\pi}{x}\right)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

The function $g$ satisfies $3x \le g(x) \le x^3 + 2$ for all $x$ in the interval $(0, 2)$ . What is $\displaystyle \lim_{x \to 1} g(x)$ ?

Practice 7

Functions $f$ , $g$ , and $h$ satisfy $g(x) \le f(x) \le h(x)$ for all $x$ , with $\displaystyle \lim_{x \to 2} g(x) = 2$ and $\displaystyle \lim_{x \to 2} h(x) = 5$ . Which statement is true?