Topic 1.8 Notes – Determining Limits Using the Squeeze Theorem

What the Squeeze Theorem Says

Here’s the statement in mathematical form:

If

• $f(x) \le g(x) \le h(x)$ for all $x$ near $a$ (not necessarily at $a$), and
• $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$,

then

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

The idea is simple. If a function is trapped between two others, and those outer functions both approach the same number, the middle one has no choice but to approach that number too.

Here’s what that looks like visually.

Visual model of the Squeeze Theorem

In the graph, $g(x)$ stays between $f(x)$ and $h(x)$ as $x$ gets close to $a$. Since the outer curves meet at the same height near $x = a$, the middle curve is forced to head to that same value.

Notice that we only care about behavior near $a$. The function doesn’t even have to be defined at $a$.

The Three Conditions You Must Check

When you use this on a quiz or FRQ, you need all three pieces.

1. An Inequality Near $a$

You must show

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

for values of $x$ sufficiently close to $a$.

It doesn’t have to work at $x = a$. Just around it.

2. The Outer Limits Exist

Both

$$\lim_{x \to a} f(x) \quad \text{and} \quad \lim_{x \to a} h(x)$$

must exist.

3. They Match

They must equal the same number $L$.

If the outer limits are different, the theorem does nothing for you.

On an FRQ, skipping any of these justifications can cost you a point. The AP rubric wants to see that you verified the squeeze setup, not that you guessed correctly.

A Classic Oscillation Example

Consider

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

Direct substitution fails. The sine term oscillates wildly as $x \to 0$.

Start with something you know:

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

Multiply everything by $x$:

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

Now take limits of the outer functions:

• $\lim_{x \to 0} -x = 0$
• $\lim_{x \to 0} x = 0$

Since both go to 0, the middle function must also go to 0.

So,

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

The graph makes the squeeze clear.

Graph of $y = x\sin\left(\frac{2}{x}\right)$ with $y = x$ and $y = -x$

The oscillations get tighter near 0, and the lines $y = x$ and $y = -x$ pinch the graph toward the origin. The $x$ factor forces the height to shrink to 0.

Two Essential Trig Limits

These are foundational results that come from the Squeeze Theorem:

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

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

You are expected to know these. They get used later in derivative definitions and trig derivatives. You won’t reprove them every time, but you should understand they come from a squeeze argument using geometric inequalities.

If you see something that can be rewritten to look like one of these, do it. That’s often faster than building a new squeeze from scratch.

The Absolute Value Strategy

Sometimes it’s cleaner to write:

$$|g(x)| \le h(x).$$

If you can show $h(x) \to 0$, then $g(x) \to 0$.

Example idea:

If

$$|x^2 \cos(5/x)| \le x^2,$$

and $x^2 \to 0$, then the whole expression goes to 0.

This avoids worrying about flipping inequalities when multiplying by negatives.

When You Should Think “Squeeze”

• A trig function like sine or cosine is oscillating.
• The limit involves $x \to 0$.
• Direct substitution fails.
• Algebra doesn’t simplify the expression cleanly.

If it simplifies algebraically, do that instead. Squeeze is a tool, not a requirement.