# How do you use the Squeeze Theorem to find lim sqrtx[1+ sin^2 (2π /x)] as x approaches zero?

Sep 28, 2015

See explanation, below.

#### Explanation:

For all $x \ne 0$,

$- 1 \le \sin \left(\frac{2 \pi}{x}\right) \le 1$, so

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

$1 \le 1 + {\sin}^{2} \left(\frac{2 \pi}{x}\right) \le 2$.

For all $x > 0$, we have $\sqrt{x} > 0$, so we can multiply the inequality by $\sqrt{x}$ without changing the inequalities.

$\sqrt{x} \le \sqrt{x} \left(1 + {\sin}^{2} \left(\frac{2 \pi}{x}\right)\right) \le 2 \sqrt{x}$.

Observe that ${\lim}_{x \rightarrow {0}^{+}} \sqrt{x} = 0$ and ${\lim}_{x \rightarrow {0}^{+}} 2 \sqrt{x} = 0$.

Therefore, ${\lim}_{x \rightarrow {0}^{+}} \sqrt{x} \left(1 + {\sin}^{2} \left(\frac{2 \pi}{x}\right)\right) = 0$