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Squeeze Theorem

Tags

Calculus

Cegep/1

Word count

86 words

Reading time

1 minute

If $f(x)\le g(x)\le h(x)$ when $x$ is near $a$ (except possibly at $a$),
and

$$\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}h(x)=L$$

, then

$$\underset{x\to a}{lim}g(x)=L$$

[!example]+ $\underset{x\to \mathrm{\infty }}{lim}\frac{\sin x}{x}$
Because $-\frac{1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}$
and $\underset{x\to \mathrm{\infty }}{lim}-\frac{1}{x}=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}=0$,
$\underset{x\to \mathrm{\infty }}{lim}\frac{\sin x}{x}=0$.

Contributors

Ajitani Hifumi

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