Calculus I · Limits and Continuity · lesson

The Squeeze Theorem

Concept

Learning objectives

Use upper and lower bounds to determine a limit even when the middle function oscillates or is difficult to evaluate directly.

The Squeeze Theorem and Trigonometric Limits

When Direct Algebra Is Not Enough

Some functions refuse to simplify into a friendly formula. The Squeeze Theorem handles them by trapping their outputs between two simpler functions.

Concept

Suppose three people walk through a doorway side by side. The person in the middle cannot end up above the person on the top or below the person on the bottom. If the top and bottom people are forced toward the same height, the middle person is forced there too. That is the Squeeze Theorem.

Theorem

Squeeze Theorem

Suppose that for all $x$ sufficiently close to $a$ ,

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

and suppose

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

Then

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

Guided walkthrough

A number trapped between shrinking walls

Suppose $f(x)$ satisfies

-x^2\le f(x)\le x^2

near $x=0$ . Find $\lim_{x\to0}f(x)$ .

Show worked solution

The lower wall approaches

\lim_{x\to0}(-x^2)=0.

The upper wall approaches

\lim_{x\to0}x^2=0.

Because $f(x)$ is trapped between two functions that both approach $0$ ,

\boxed{\lim_{x\to0}f(x)=0}.

We do not need a formula for $f$ . The trap is enough.

Interactive checksqueeze-flow-01

Evaluate $\lim_{x\to0}x^2\sin(1/x)$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Use $|\sin(1/x)|\le1$ .

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Bounded oscillation times a shrinking factor

The inequalities

-1\le\sin u\le1

and

-1\le\cos u\le1

hold for every real $u$ . If a bounded trigonometric factor is multiplied by something that approaches zero, the product is often squeezed to zero.

Worked example

An oscillating function forced to zero

Evaluate

\lim_{x\to0}x^2\sin\left(\frac1x\right).

Show worked solution

Start with the universal sine bound:

-1\le\sin\left(\frac1x\right)\le1.

Since $x^2\ge0$ , multiply every part by $x^2$ without reversing the inequalities:

-x^2\le x^2\sin\left(\frac1x\right)\le x^2.

Now take limits of the outer functions:

\lim_{x\to0}(-x^2)=0, \qquad \lim_{x\to0}x^2=0.

Therefore,

\boxed{\lim_{x\to0}x^2\sin\left(\frac1x\right)=0}.

The sine factor continues oscillating. The shrinking factor $x^2$ reduces the size of every oscillation until the graph is trapped near zero.

Oscillating function squeezed between x squared and negative x squared. — **Read the graph**
The oscillating function is trapped between $-x^2$ and $x^2$ , both of which approach zero.

Interactive graph exploration

Functions: $y=x^2\sin(1/x)$ , $y=x^2$ , and $y=-x^2$ . Window: $-0.35\le x\le0.35$ , $-0.13\le y\le0.13$ . Use at least 1500 samples on each side of zero for the oscillating curve. Do not connect across $x=0$ , where the displayed formula is undefined.

Worked example

Absolute-value squeeze

Evaluate

\lim_{x\to0}x\cos\left(\frac{5}{x}\right).

Show worked solution

Because $|\cos u|\le1$ ,

\left|x\cos\left(\frac5x\right)\right| \le |x|.

This means

-|x|\le x\cos\left(\frac5x\right)\le|x|.

Both outer functions approach zero, so

\boxed{\lim_{x\to0}x\cos\left(\frac5x\right)=0}.

Problem 1

When a function contains $\sin(1/x)$ , $\cos(1/x)$ , or another bounded oscillating factor, ask whether the rest of the expression approaches zero. The inequality

|\text{bounded factor}\times\text{shrinking factor}| \le C|\text{shrinking factor}|

is often the entire argument.

After the explanation

Use the section idea

Reading lens

Can the expression be rewritten around a known small-angle limit, with every scaling factor accounted for?

Mental model

The fundamental sine limit is a reusable local shape: other trigonometric limits work when you expose that shape through identities and scaling.

Decision

Look for a bounded oscillation times a shrinking factor, or rewrite the expression into sine-over-angle factors whose arguments match their denominators.

Common trap

The sine function is not equal to its angle; their ratio merely approaches one near zero, and that statement requires radian measure.

Check yourself

Mastery means you can mark every scaling factor before simplifying and can explain where the Squeeze Theorem enters the argument.

Source & rights

Original instruction with traceable references.

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