Find $\lim_{x\to 0^+}\sin(x)\ln(x)$ Find $\lim_{x\to 0^+}\sin(x)\ln(x)$
By using l'Hôpital rule: because we will get $0\times\infty$ when we substitute, I rewrote it as:
$$\lim_{x\to0^+}\dfrac{\sin(x)}{\dfrac1{\ln(x)}}$$
to get the form $\dfrac 00$
Then I differentiated the numerator and denominator and I got:
$$\dfrac{\cos x}{\dfrac{-1}{x(\ln x)^2}}$$ 
when substitute in this form I get: $\dfrac{1}{0\times\infty^2}$ 
Can we have the result $0\times\infty^2=0$? Then the limit will be  $\dfrac10=\infty$?
 A: Hint: do you know how to compute
$$
\lim_{x\to0}\frac{\sin(x)}{x}
$$
and
$$
\lim_{x\to0}x\log(x)=\lim_{x\to0}\frac{\log(x)}{1/x}
$$
If so, then you can use
$$
\lim_{x\to0}f(x)g(x)=\lim_{x\to0}f(x)\lim_{x\to0}g(x)
$$
provided the limits on the right hand side exist.
A: As a rule of thumb, you should keep the logarithm at the numerator:
$$
\lim_{x\to0^+}\frac{\log x}{1/\sin x}
$$
This is of the form $\infty/\infty$, so we can apply l'Hôpital's theorem:
$$
\lim_{x\to0^+}\frac{1/x}{-\cos x/\sin^2x}=
\lim_{x\to0^+}-\frac{\sin^2x}{x\cos x}
$$
that you should be able to manage.
A: Your computation can continue. 
After you have applied L'H rule once, it suffices to compute 
\begin{align*}
 \lim x \ln ^2 x &=\lim \frac{\ln ^2 x}{\frac{1}{x}} \\
&=\lim \frac{2 \ln x \cdot \frac{1}{x}}{-\frac{1}{x^2}} \\
&=\lim \frac{2\ln x}{ - \frac{1}{x}} \\
&=0
\end{align*}
A: We can use approximation arguments : when $x$ is small $\sin(x) \approx x$ and any polynomial grows faster than logarithm. Hence $\lim_{x \to 0^+} \sin(x) \ln(x) = \lim_{x \to 0^+} x = 0$
