Squeeze Theorem Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Squeeze Theorem.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

If g(x) \leq f(x) \leq h(x) near x = a, and \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L, then \lim_{x \to a} f(x) = L.

If f is squeezed between two functions that both approach the same value L, then f has no choice—it must also approach L. Like being caught between two walls closing in to the same point.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: The Squeeze Theorem is a tool for finding limits that can't be computed by direct substitution or algebraic manipulation. Trap the unknown function between two known ones.

Common stuck point: The hard part is finding the bounding functions g and h. For oscillating functions like \sin or \cos, use -1 \leq \sin(\cdot) \leq 1 and multiply by the vanishing factor.

Sense of Study hint: Bound the oscillating part between -1 and 1, then multiply the entire inequality by the non-oscillating factor.

Worked Examples

Example 1

easy
Use the squeeze theorem to find \displaystyle\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right).

Solution

  1. 1
    Since -1 \leq \sin\!\left(\frac{1}{x}\right) \leq 1 for all x \neq 0:
  2. 2
    Multiply by x^2 \geq 0: -x^2 \leq x^2\sin\!\left(\frac{1}{x}\right) \leq x^2.
  3. 3
    \lim_{x\to 0}(-x^2) = 0 and \lim_{x\to 0}(x^2) = 0.
  4. 4
    By the squeeze theorem: \lim_{x\to 0} x^2\sin\!\left(\frac{1}{x}\right) = 0.

Answer

0
The function \sin(1/x) oscillates wildly near 0, so its limit doesn't exist alone. Multiplying by x^2 traps it between -x^2 and x^2, both going to 0, so the product must also go to 0.

Example 2

hard
Use the squeeze theorem to prove \displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

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

Example 2

medium
Show that \displaystyle\lim_{n\to\infty} \frac{\sin n}{n} = 0.
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Background Knowledge

These ideas may be useful before you work through the harder examples.

limit