The Squeeze Theorem — AP® Calculus AB Lesson

Part 1 · Hook

Why this matters

Some functions are too wild to handle directly. Near 0,

x \sin\left(\frac{1}{x}\right)

wiggles infinitely often, so no table or factoring will tame it. But you can trap it between two simple functions that close in on the same value. When the trap closes, the wild function has nowhere else to go.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. For every real $\theta$ , the values of $\sin\theta$ lie

between $-1$ and $1$
between $0$ and $1$
between $-\pi$ and $\pi$
anywhere on the number line

Show the answer

On the unit circle, $\sin\theta$ is a $y$ -coordinate, so $-1 \le \sin\theta \le 1$ .

Correct: between $-1$ and $1$ :
between $0$ and $1$ :
between $-\pi$ and $\pi$ :
anywhere on the number line:

2. If $\lim_{x \to c} f(x) = 3$ , then $\lim_{x \to c} (2f(x) - 1)$ is

$5$
$6$
$3$
not known

Show the answer

By the limit laws: 2(3) − 1 = 5.

Correct: $5$ :
$6$ :
$3$ :
not known:

3. Which is $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$ ?

$2$
$0$
$1$
It does not exist

Show the answer

Cancel $x - 1$ : the expression is $x + 1$ for $x \ne 1$ , which approaches 2.

Correct: $2$ :
$0$ :
$1$ :
It does not exist:

Part 4 · See it

See it first

The graph of f(x) = x sin(1/x) for x between −0.4 and 0.4, x not 0. It wiggles faster and faster near 0 but stays between the lines y = |x| and y = −|x|. Both bounds reach 0 at x = 0, so f is squeezed to 0: its limit at 0 is 0. An open circle at the origin marks that f(0) is not defined. — $x \sin\left(\frac{1}{x}\right)$ is trapped between $-|x|$ and $|x|$ , which both approach 0. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A function $f$ is hard to evaluate near $c$ , but $g(x) \le f(x) \le h(x)$ thereevery value of $f$ near $c$ lies between the values of $g$ and $h$
$g$ and $h$ have the same limit $L$ at $c$ the gap between them shrinks to nothing near $c$
$f$ is trapped in a gap that closes at $L$ $\lim_{x \to c} f(x) = L$ : the squeeze theorem
Near 0, $\cos x \le \frac{\sin x}{x} \le 1$ and both bounds approach 1 $\lim_{x \to 0} \frac{\sin x}{x}$ $= 1$ (with $x$ in radians)

Part 6 · Key ideas

Key ideas

Squeeze theorem: a function trapped between two functions with the same limit $L$ at $c$ also has limit $L$ at $c$ .
Both bounds must approach the same value. Bounds with different limits give no conclusion.
Build bounds from $-1 \le \sin\theta \le 1$ and $-1 \le \cos\theta \le 1$ , multiplied by a factor that is not negative (use $|x|$ , not $x$ ).
The special limit $\lim_{x \to 0} \frac{\sin x}{x}$ $= 1$ comes from the squeeze. Match the inside: $\frac{\sin(kx)}{x}$ $= k \cdot \frac{\sin(kx)}{kx} \to k$ .
A justification states the inequality, shows the two bounds have the same limit, and names the squeeze theorem.

Part 7 · Misconception

A common mistake

The wrong idea: $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ , so the squeeze theorem shows $\sin\left(\frac{1}{x}\right)$ has a limit at 0.

What actually happens: The bounds $-1$ and $1$ have different limits, so the squeeze theorem says nothing. In fact $\sin\left(\frac{1}{x}\right)$ oscillates and has no limit at 0. Multiply by $x^2$ and the bounds $\pm x^2$ both approach 0; then the squeeze works.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Three functions near $x = 2$

1. The graph shows $g(x) = 3 - (x - 2)^2$ and $h(x) = 3 + (x - 2)^2$ , with $g(x) \le f(x) \le h(x)$ for every $x \ne 2$ . Find $\lim_{x \to 2} f(x)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\lim_{x \to 2} g(x) = 3$ and $\lim_{x \to 2} h(x) = 3$ . Since $g \le f \le h$ near 2, the squeeze theorem gives $\lim_{x \to 2} f(x) = 3$ .

Answer: 3

2. Which pair of facts is exactly what the squeeze theorem needs to conclude $\lim_{x \to 2} f(x) = 3$ ?

$g \le f \le h$ near 2, and both $g$ and $h$ have limit 3 there
$f(2) = 3$ , and $f(2)$ lies between $g(2)$ and $h(2)$ at $x = 2$
$g(2) = h(2)$ , and $f$ wiggles faster and faster as $x$ nears 2
$g(x) \le h(x)$ near 2, and $h$ has limit 3 as $x$ approaches 2

Show the answer

The theorem has two parts: the inequality $g \le f \le h$ for $x$ near 2 (not at 2), and equal limits $L$ for the two bounds. Then $f$ has limit $L$ too.

Correct: $g \le f \le h$ near 2, and both $g$ and $h$ have limit 3 there: These are the two conditions: $f$ is trapped, and the trap closes at the single value 3.
$f(2) = 3$ , and $f(2)$ lies between $g(2)$ and $h(2)$ at $x = 2$ : $f(2)$ does not exist here, and the theorem never uses the value at the point.
$g(2) = h(2)$ , and $f$ wiggles faster and faster as $x$ nears 2: Wiggling is not a condition, and equal values at 2 are not the same as equal limits.
$g(x) \le h(x)$ near 2, and $h$ has limit 3 as $x$ approaches 2: $f$ must be trapped between the bounds, and both bounds need limit 3; one bound is not enough.

3. Suppose the upper bound were $k(x) = 4 + (x - 2)^2$ instead of $h$ , still with $g(x) \le f(x) \le k(x)$ near 2. What does the squeeze theorem say about $\lim_{x \to 2} f(x)$ ?

The limit is 3.5, halfway between the limits of the bounds
Nothing: the bounds approach different values, 3 and 4
The limit is 4, the larger of the two bounds
The limit does not exist, since the bounds differ

Show the answer

The squeeze theorem applies only when both bounds have the same limit. With limits 3 and 4 it gives no conclusion; $f$ might have a limit, or might not.

The limit is 3.5, halfway between the limits of the bounds: Nothing in the theorem averages the bounds.
Correct: Nothing: the bounds approach different values, 3 and 4: Unequal bound limits leave a gap from 3 to 4, so $f$ is not pinned to one value.
The limit is 4, the larger of the two bounds: The upper bound does not decide the limit by itself.
The limit does not exist, since the bounds differ: Unequal bounds do not prove the limit fails to exist; the theorem simply gives no answer.

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

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{\sin(5x)}{x}$ $= 5 \cdot \frac{\sin(5x)}{5x}$ . As $x \to 0$ , $5x \to 0$ , so $\frac{\sin(5x)}{5x} \to 1$ . The limit is 5.

Answer: 5

5. To squeeze $x \sin\left(\frac{1}{x}\right)$ as $x \to 0$ , which pair of bounds is valid on both sides of 0?

$-x$ and $x$
$-|x|$ and $|x|$
$-1$ and $1$
$0$ and $x^2 + |x|$

Show the answer

$\left|x \sin\left(\frac{1}{x}\right)\right| \le |x|$ , so $-|x| \le x \sin\left(\frac{1}{x}\right) \le |x|$ for all $x \ne 0$ . Both bounds approach 0, so the limit is 0.

$-x$ and $x$ : For $x < 0$ , multiplying by $x$ flips the inequalities, so $-x \le x\sin\left(\frac{1}{x}\right) \le x$ fails there.
Correct: $-|x|$ and $|x|$ : Multiplying $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ by $|x| \ge 0$ keeps the direction, and both bounds go to 0.
$-1$ and $1$ : These are true bounds, but their limits are $-1$ and $1$ , which are not equal.
$0$ and $x^2 + |x|$ : The function is negative for many $x$ , so 0 is not a lower bound.

6. A student writes: " $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ , so by the squeeze theorem $\lim_{x \to 0} \sin\left(\frac{1}{x}\right)$ exists and is between $-1$ and $1$ ." What is wrong?

Nothing; the limit exists and is $0$ , the midpoint of the two bounds $-1$ and $1$
The inequality is false near 0, since $\sin\left(\frac{1}{x}\right)$ can be larger than 1 there
The theorem requires the function to be defined at $x = 0$ , and $\frac{1}{x}$ is not defined at 0
The bounds $-1$ and $1$ have different limits, so the theorem says nothing; this limit does not exist

Show the answer

The squeeze theorem needs the two bounds to share a limit. Here they are the constants $-1$ and $1$ . In fact $\sin\left(\frac{1}{x}\right)$ swings between $-1$ and $1$ infinitely often near 0, so the limit does not exist.

Nothing; the limit exists and is $0$ , the midpoint of the two bounds $-1$ and $1$ : Being bounded does not make a limit exist, and the theorem never averages bounds.
The inequality is false near 0, since $\sin\left(\frac{1}{x}\right)$ can be larger than 1 there: Sine never exceeds 1, so the inequality is true; it just does not help.
The theorem requires the function to be defined at $x = 0$ , and $\frac{1}{x}$ is not defined at 0: The theorem never needs the value at the point.
Correct: The bounds $-1$ and $1$ have different limits, so the theorem says nothing; this limit does not exist: Unequal bounds give no conclusion, and the oscillation shows the limit does not exist.

Part 9 · Summary

Summary

If

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

near

c

and

g

and

h

have the same limit

L

at

c

, then

f

has limit

L

too. Bounds usually come from

-1 \le \sin \le 1

. The squeeze gives

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

= 1

in radians, and

\frac{\sin(kx)}{x} \to k

.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Three functions near x=2x = 2

Summary

What comes next

Connections

Three functions near $x = 2$