If $a_n\to L$ and $c_n\to L$ as $n\to \infty$, and if $a_n\le b_n\le c_n$, then $b_n\to L$ as $n\to \infty$ as well.

Ex: ${\lim }_{x\to \infty }\frac{\sin (x)}{x}=?$

Notice that $-1\le \sin x\le 1$

$-\frac{1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}$

Since, ${\lim }_{x\to \infty }-\frac{1}{x}\le {\lim }_{x\to \infty }\frac{\sin x}{x}\le {\lim }_{x\to \infty }\frac{1}{x}$

$0\le {\lim }_{x\to \infty }\frac{\sin x}{x}\le 0$ So by squeeze theorem, ${\lim }_{x\to \infty }\frac{\sin (x)}{x}=0$