Squeeze Theorem Math Example 1

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Example 1

easy

Use the squeeze theorem to find

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

.

Solution

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

About Squeeze Theorem

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 $\lim_{x \to a} f(x) = L$ .

Learn more about Squeeze Theorem →

More Squeeze Theorem Examples

Use the squeeze theorem to prove [formula].

Find [formula].

Example 4 medium

Show that [formula].

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Related Concepts

Limit Types of Continuity and Discontinuity