# How do you find the limit of sinxlnx as x approaches 0 using l'hospital's rule?

Jul 7, 2016

0

#### Explanation:

to use Lhopital we need to get it into an indeterminate form. as $\sin 0 = 0$ and $\ln 0 = - \infty$, we can do that as follows

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

which by LHopital

$= {\lim}_{x \to 0} \frac{\frac{1}{x}}{- \csc x \cot x}$

$= {\lim}_{x \to 0} - {\sin}^{2} \frac{x}{x \cos x}$

$= - \frac{1}{\cos} x {\lim}_{x \to 0} \sin \frac{x}{x} \sin x \text{ as} {\lim}_{x \to 0} \cos x = 1$

$= - 1 {\lim}_{x \to 0} \sin \frac{x}{x} \sin x q \quad \triangle$

we know that ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$ and ${\lim}_{x \to 0} \sin x = \sin 0 = 0$

so

$= - 1 \cdot 1 \cdot 0 = 0$

or you could do L'Hopital again on $\triangle$ to get

$= - 1 {\lim}_{x \to 0} {\left({\sin}^{2} x\right)}^{p} r i m \frac{e}{{x}^{p} r i m e}$

$= - 1 {\lim}_{x \to 0} \frac{2 \sin x \cos x}{1}$

$= - 1 {\lim}_{x \to 0} \sin 2 x = 0$