Calculus Question: $\lim_{x \rightarrow 0}\sin(x)\ln{\sin{x}}$ I'm having trouble to compute
$$\lim_{x \rightarrow 0}\sin(x)\ln{\sin{x}}$$
I tried to find this limit using Wolfram Alpha and the result is 0, but I don't know how to get this 0.
 A: First we convert it to a quotient:
$$\lim_{x \rightarrow 0}{\sin{x}\ln{\sin{x}}}$$
And now we use the l'Hopital rule:
$$\lim_{x \rightarrow 0}{\frac{\ln{\sin{x}}}{\frac{1}{\sin{x}}}}$$
...and now derive:
$$\lim_{x \rightarrow 0}{\frac{(\ln{\sin{x}})'}{(\frac{1}{\sin{x}})'}}$$
$$\lim_{x \rightarrow 0}{\frac{\frac{1}{\sin{x}}\cos{x}}{-\frac{\cos{x}}{\sin^2{x}}}}$$
$$\lim_{x \rightarrow 0}{\frac{\cot{x}}{-\frac{\cot{x}}{\sin{x}}}}$$
$$\lim_{x \rightarrow 0}{\frac{1}{-\frac{1}{\sin{x}}}}$$
$$\lim_{x \rightarrow 0}{-\sin{x}}$$
...which is trivially $\underline{\underline{0}}$. Q.E.D.
A: If you believe/know $\lim_{x\rightarrow0} x\log(x)=0$ and that $\lim_{x\rightarrow0} \frac{\sin(x)}{x}=1$  you are done:$$\sin(x)\log(\sin(x))=\frac{\sin(x)}{x}x\log(x\frac{\sin(x)}{x})=\frac{\sin(x)}{x}(x\log(x)+x\log(\frac{\sin(x)}{x}))$$
A: Let
$$
I=\lim_{x\to0}\sin (x)\ln\sin (x),
$$
then multiply $I$ by $\dfrac{\sin (x)}{\sin (x)}$ yield
$$
I=\lim_{x\to0}\dfrac{\sin^2 (x)\ln\sin (x)}{\sin (x)}.
$$
Now we use L'Hôpital's rule and we obtain
$$
\begin{align}
I&=\lim_{x\to0}\dfrac{2\sin (x)\cos(x)\ln\sin (x)}{\cos(x)}-\lim_{x\to0}\dfrac{\sin^2 (x)\cdot\dfrac{\cos(x)}{\sin (x)}}{\cos(x)}\\
&=2I-\lim_{x\to0}\sin(x)\\
I&=\lim_{x\to0}\sin(x)\\
&=0.
\end{align}
$$
A: A more elementary argument:
$\sin(x)\ln\sin(x)=\ln(\sin(x)^{\sin x})$ and $\ln t$ is continuous at $t=1$. Here one only uses the fact that $y^y$ tends to $1$ when $y$ tends to $0$.
A: Make a substitution.  Setting $u = \sin x$ (and noting the continuity of $\sin x$), we have
$$
\lim_{x \to 0} \sin(x)\ln(\sin(x)) = \lim_{u \to \sin(0)}u \,\ln u
= \lim_{u \to 0}u \,\ln u
$$
This is a standard problem usually done using L'Hôpital's rule.
