# Question #8cc23

Nov 30, 2016

#### Explanation:

$- 1 \le \sin \left(\frac{\pi}{x}\right) \le 1$ for all $x \ne 0$.

$\sqrt{{x}^{3} + {x}^{2}} > 0$ so we can multiply without changing the inequalities.

$- \sqrt{{x}^{3} + {x}^{2}} \le \sqrt{{x}^{3} + {x}^{2}} \sin \left(\frac{\pi}{x}\right) \le \sqrt{{x}^{3} + {x}^{2}}$ $\text{ }$ for all $x \ne 0$.

Observe that ${\lim}_{x \rightarrow 0} - \sqrt{{x}^{3} + {x}^{2}} = 0$ and ${\lim}_{x \rightarrow 0} \sqrt{{x}^{3} + {x}^{2}} = 0$.

Therefore, by the squeeze theorem, ${\lim}_{x \rightarrow 0} \sqrt{{x}^{3} + {x}^{2}} \sin \left(\frac{\pi}{x}\right) = 0$