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The Squeeze Theorem

Trapping a Limit

Some functions oscillate so wildly near a point that direct methods fail. Consider $\sin(1/x)$ as $x \to 0$: it swings between $-1$ and $1$ infinitely often. How could such a function have a limit?

The answer: trap it between two simpler functions. If you can show that a complicated function is always between two simple ones, and both simple functions approach the same limit, then the complicated function has no choice—it's squeezed into that same limit.

Prerequisite Map

Quick Reference

Property	Value
Concept	Limits
Chapter	1.6
Difficulty	Intermediate
Time	~20 minutes

The Squeeze Theorem (Sandwich Theorem)

Statement

If $f(x) \le g(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$ itself), and

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

then

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

Visualization

    │
h(x)│     ╭──╮
    │   ╱     ╲
  L ├──●────────●───  ← Both bounds approach L
    │   ╲     ╱
f(x)│     ╰──╯
    │
    └───────┬───────
            a

g(x) is "squeezed" between f(x) and h(x)

Why It Works

The function $g(x)$ is trapped in a narrowing corridor. As $x \to a$, the corridor height $h(x) - f(x) \to 0$. With nowhere else to go, $g(x)$ must approach $L$.

The Standard Setup

Most Squeeze Theorem problems follow this pattern:

1. Start with a bounded oscillation: $-1 \le \sin(\text{something}) \le 1$ or $-1 \le \cos(\text{something}) \le 1$

1. Multiply by a function going to zero: If $\vert g(x)\vert \le M$ and $f(x) \to 0$, then $f(x) \cdot g(x) \to 0$

1. The squeeze: $-\vert f(x)\vert \le f(x) \cdot g(x) \le \vert f(x)\vert $

The Classic Example

Evaluate $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$.

The challenge: $\sin(1/x)$ oscillates infinitely as $x \to 0$—no limit exists for this factor alone.

The solution: The oscillations are bounded, and $x^2 \to 0$ crushes them.

Setup the squeeze:

Since $-1 \le \sin(1/x) \le 1$ for all $x \neq 0$:

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

Apply the theorem:

Both $\lim_{x \to 0}(-x^2) = 0$ and $\lim_{x \to 0}(x^2) = 0$.

Therefore: $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$

When to Use the Squeeze Theorem

Situation	Why Squeeze Works
Product: (goes to 0) × (bounded)	Bounded oscillations get crushed
Functions with $\sin(1/x)$ or $\cos(1/x)$	These oscillate but stay in $[-1, 1]$
Proving limits equal zero	Easy to build symmetric bounds
Functions trapped by known limits	When direct computation fails

Practice Problems

Conceptual Check (CCI-Style)

Mastery Checklist

• [ ] I can state the Squeeze Theorem precisely
• [ ] I can set up bounds using $-1 \le \sin(\cdot) \le 1$ and $-1 \le \cos(\cdot) \le 1$
• [ ] I can evaluate limits of the form (goes to 0) × (bounded oscillation)
• [ ] I recognize when the Squeeze Theorem cannot be applied (bounds don't converge to same limit)
• [ ] I can prove $\lim_{x \to 0}\frac{\sin x}{x} = 1$ using the Squeeze Theorem

Mental Model

The narrowing corridor:

Imagine walking through a hallway that gets narrower and narrower. No matter how erratically you move side to side, if the walls close in to the same point, you'll end up exactly there.

The function $g(x)$ is the walker, $f(x)$ and $h(x)$ are the walls, and $L$ is where the walls meet.

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Connections

Looking back:

• Basic Limit Laws compute the limits of the bounding functions

Looking ahead:

• Trigonometric Limits uses Squeeze to prove $\lim_{x \to 0}\frac{\sin x}{x} = 1$
• Limits at Infinity applies Squeeze for oscillating functions as $x \to \infty$

Real-world connections:

• Signal processing: noisy signals bounded by decaying envelopes
• Physics: damped oscillations in mechanical systems

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Last updated: 2026-01-22