# Question #57637

Apr 12, 2017

${\lim}_{x \to {0}^{+}} \sin x \ln x = 0$

#### Explanation:

This limit is in the indeterminate form $0 \cdot \infty$. To apply L'Hospital's rule we have to transform it in a way that it is in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$:

${\lim}_{x \to {0}^{+}} \sin x \ln x = {\lim}_{x \to {0}^{+}} \ln \frac{x}{\frac{1}{\sin} x}$

we can now apply the rule and have:

${\lim}_{x \to {0}^{+}} \ln \frac{x}{\frac{1}{\sin} x} = {\lim}_{x \to {0}^{+}} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} \frac{1}{\sin} x} = {\lim}_{x \to {0}^{+}} \frac{\frac{1}{x}}{- \cos \frac{x}{\sin} ^ 2 x} = {\lim}_{x \to {0}^{+}} - \sin \frac{x}{x} \tan x = - 1 \cdot 0 = 0$

graph{sinxlnx [-10, 10, -5, 5]}