Squeeze Theorem – AP Calculus

Sometimes a limit is difficult to compute directly because the function oscillates or behaves strangely. The Squeeze Theorem (also called the Sandwich Theorem) provides a way out: if we can "squeeze" our function between two simpler functions that have the same limit, then our function must have that same limit.

\boxed{\text{If } g(x) \le f(x) \le h(x) \text{ for all } x \text{ near } a \text{ (except possibly at } a\text{), and}} $$ $$ \boxed{\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \quad \Longrightarrow \quad \lim_{x \to a} f(x) = L}

🎨 The Intuition (Picture This)

Imagine a criminal being squeezed between two police cars. As the police cars get closer and closer together, the criminal has no choice but to be where they are. Similarly, if $g(x)$ and $h(x)$ "squeeze" $f(x)$ from below and above, and they both approach $L$, then $f(x)$ is forced to approach $L$ as well.

📊 Visualizing the Squeeze Theorem

Use the interactive graph below. The blue curve is $f(x) = x^2 \sin(1/x)$ — it oscillates wildly near $x = 0$. The dashed curves $g(x) = -x^2$ and $h(x) = x^2$ squeeze it from below and above, forcing the limit to be $0$:

🔧 How to Use the Squeeze Theorem

Identify the tricky function $f(x)$ whose limit you want to find.
Find bounding functions $g(x)$ and $h(x)$ such that $g(x) \le f(x) \le h(x)$ near the point of interest.
Verify that $\lim g(x) = \lim h(x) = L$.
Conclude that $\lim f(x) = L$.

Example 1The Classic: $\displaystyle \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$

We know that $-1 \le \sin(1/x) \le 1$ for all $x \neq 0$.

Multiply by $x^2$ (which is $\ge 0$): $-x^2 \le x^2 \sin(1/x) \le x^2$
Let $g(x) = -x^2$ and $h(x) = x^2$
$\lim_{x \to 0} g(x) = 0$ and $\lim_{x \to 0} h(x) = 0$
By the Squeeze Theorem, $\displaystyle \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$

Example 2Another Oscillating Function

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

$-1 \le \sin(1/x) \le 1$
Multiply by $x$: but careful! If $x$ is negative, the inequality flips.
Better: $-|x| \le x \sin(1/x) \le |x|$ for all $x \neq 0$
$\lim_{x \to 0} -|x| = 0$ and $\lim_{x \to 0} |x| = 0$
Thus $\displaystyle \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$

Example 3Using the Squeeze Theorem with Cosine

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

$-1 \le \cos(1/x) \le 1$
Multiply by $x$: $-|x| \le x \cos(1/x) \le |x|$
$\lim_{x \to 0} -|x| = 0$, $\lim_{x \to 0} |x| = 0$
Therefore $\displaystyle \lim_{x \to 0} x \cos\left(\frac{1}{x}\right) = 0$

Example 4AP-Style: Limit with Sine

Find $\displaystyle \lim_{x \to \infty} \frac{\sin x}{x}$.

We know $-1 \le \sin x \le 1$
Divide by $x > 0$ (for large $x$): $-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$
$\lim_{x \to \infty} -\frac{1}{x} = 0$ and $\lim_{x \to \infty} \frac{1}{x} = 0$
Thus $\displaystyle \lim_{x \to \infty} \frac{\sin x}{x} = 0$

💡 Note

This limit shows that even though $\sin x$ oscillates forever, the $1/x$ factor "squeezes" it to zero as $x \to \infty$.

⚠️ Common Pitfalls

Inequality direction matters: When multiplying by a negative number, the inequality flips. Use absolute values to stay safe: $|f(x)| \le M$ implies $-M \le f(x) \le M$.
The bounds must squeeze to the same limit: If $g(x)$ and $h(x)$ have different limits, the Squeeze Theorem doesn't apply.
The Squeeze Theorem does NOT find the limit if the bounds don't converge: It only works when you already have a candidate limit from the bounds.

🔍 Key Takeaways

The Squeeze Theorem is useful for limits involving oscillating functions like $\sin(1/x)$ or $\cos(1/x)$.
It requires finding two functions that bound $f(x)$ and have the same limit.
The classic application: $\displaystyle \lim_{x \to 0} x^2 \sin(1/x) = 0$.
Use absolute values to handle sign issues: $|f(x)| \le M$ gives $-M \le f(x) \le M$.
AP exam questions may ask you to apply the Squeeze Theorem to evaluate tricky limits.

← Back to Module Page

📦 Squeeze Theorem (Sandwich Theorem)

📊 Visualizing the Squeeze Theorem

🔧 How to Use the Squeeze Theorem