Lesson PreviewLesson 8 of 16 in this unit

1.8 Determining Limits Using the Squeeze Theorem

Part of Limits and Continuity in AP® Calculus BC. Read the opening section, then create a free account for the full interactive lesson, practice, and progress tracking.

Start free

The Squeeze Theorem is used when a function is trapped between two other functions that approach the same value. If g(x)≤ f(x)≤ h(x) near a point and both outer functions have the same limit, then the middle function must have that same limit too. This theorem is especially useful when a function is difficult to evaluate directly but can be bounded by simpler expressions.

Many classic squeeze problems involve trigonometric functions because sin x and cos x are always bounded between -1 and 1. If an oscillating trig expression is multiplied by something that becomes very small, the whole product may be forced toward zero. The theorem is powerful because it allows us to prove a limit exists even when the middle function does not simplify nicely on its own.

Example 1: Find

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

Since

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

multiplying by

x^2 \ge 0

gives

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

Now

\lim_{x \to 0}(-x^2)=0

and

\lim_{x \to 0}x^2=0

So by the Squeeze Theorem,

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

In this unit

16 lessons

Continue the full lesson in Stellar

Create a free account to unlock the remaining sections, checkpoints, course practice, and progress tracking for AP® Calculus BC.

Start free