# How do you use the Squeeze Theorem to find lim ((sin n )^2) / 2^n as x approaches zero?

Oct 22, 2015

Assuming that the question should be $\lim \frac{{\left(\sin x\right)}^{2}}{2} ^ x$ as $x \rightarrow 0$, you don't need the squeeze theorem for that limit.

#### Explanation:

${\lim}_{x \rightarrow 0} {\left(\sin x\right)}^{2} = {0}^{2} = 0$

${\lim}_{x \rightarrow 0} {2}^{x} = {2}^{0} = 1$

So,

$\lim \frac{{\left(\sin x\right)}^{2}}{2} ^ x = \frac{0}{1} = 0$