# How do you use the Squeeze Theorem to find lim sqrt(x^3+x^2) * sin (pi/x)  as x approaches zero?

Oct 7, 2015

See the explanation below.

#### Explanation:

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

For all $x \ne 0$ for which the square root is real, $\sqrt{{x}^{3} + {x}^{2}} > 0$, so we can multiply the inequality without changing the direction.

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

We observe that ${\lim}_{x \rightarrow 0} - \sqrt{{x}^{3} + {x}^{2}} = - \sqrt{0 + 0} = 0$,

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

So, by the Squeeze Theorem,

${\lim}_{x \rightarrow 0} \sqrt{{x}^{3} + {x}^{2}} \sin \left(\frac{\pi}{x}\right) = 0$.