# Question #962f6

Nov 15, 2016

Use the squeeze (pinch, sandwich) theorem (right limit version).

#### Explanation:

In order to stay in the real numbers, I assume that we want ${\lim}_{x \rightarrow {0}^{+}} \sqrt{x} \sin \left(\frac{1}{x}\right)$.

Note the for all $x \ne 0$, we have

$- 1 \le \sin \left(\frac{1}{x}\right) \le 1$

Since $\sqrt{x} > 0$, we can multiply through by $\sqrt{x}$ without changing the inequalities.

$- \sqrt{x} \le \sqrt{x} \sin \left(\frac{1}{x}\right) \le \sqrt{x}$.

Of, course, ${\lim}_{x \rightarrow {0}^{+}} - \sqrt{x} = 0 = {\lim}_{x \rightarrow {0}^{+}} \sqrt{x}$.

By the squeeze theorem, ${\lim}_{x \rightarrow {0}^{+}} \sqrt{x} \sin \left(\frac{1}{x}\right) = 0$