# How do you use the Squeeze Theorem to find lim x^2 (Sin 1/x)^2  as x approaches zero?

Oct 11, 2015

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

#### Explanation:

Assuming you meant ${x}^{2} {\sin}^{2} \left(\frac{1}{x}\right)$

We know that

$0 \le {\sin}^{2} \left(\theta\right) \le 1$

So for $\theta = \frac{1}{x}$ we have

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

Multiplying both sides by ${x}^{2}$

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

Since

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

The squeeze theorem tells us that

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