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:
Start with a bounded oscillation: $-1 \le \sin(\text{something}) \le 1$ or $-1 \le \cos(\text{something}) \le 1$
Multiply by a function going to zero: If $\vert g(x)\vert \le M$ and $f(x) \to 0$, then $f(x) \cdot g(x) \to 0$
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 $x \to 0$, so 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
Evaluate $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$.
Evaluate $\lim_{x \to 0} x^4 \cos\left(\frac{3}{x}\right)$.
Given that $3x - 1 \le f(x) \le x^2 + 3$ for all $x$ near 2, find $\lim_{x \to 2} f(x)$.
Evaluate $\lim_{x \to 0^+} \sqrt{x} \sin\left(\frac{1}{x^2}\right)$.
Use the Squeeze Theorem to prove that $\lim_{x \to 0}\frac{\sin x}{x} = 1$.
Hint: For $0 < x < \pi/2$, the following inequalities hold (from geometry of the unit circle): $$\cos x < \frac{\sin x}{x} < 1$$
Conceptual Check (CCI-Style)
The limit $\lim_{x \to 0}\sin(1/x)$ does not exist because the function oscillates infinitely.
But $\lim_{x \to 0} x^2 \sin(1/x) = 0$ exists.
How can multiplying by $x^2$ “create” a limit that wasn’t there before?
Common Misconceptions
the Squeeze Theorem requires the bounding functions to equal the target limit at the point. The theorem requires only that $g(x) \le f(x) \le h(x)$ near $a$ and that $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$. Neither $g$ nor $h$ need to equal $L$ at $x = a$, and $f$ need not be defined at $a$ at all. For $\lim_{x \to 0} x^2 \sin(1/x)$, neither the lower bound $-x^2$ nor the upper bound $x^2$ equals zero at $x = 0$ in any special way; they both approach zero, which is sufficient.
any two bounds that trap the function are enough. The two bounding functions must share the same limit at $a$. Choosing $g(x) = -1$ and $h(x) = 1$ to bound $\sin(x)/x$ near $x = 0$ does trap the function, but $\lim g = -1 \neq 1 = \lim h$, so the theorem gives no conclusion. Tight bounds, ones that converge to the same value, are what make the squeeze work.
Mastery Checklist
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.
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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|---|---|---|
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Last updated: 2026-01-22