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  $\dfrac10=\infty$?

We can use approximation arguments : when $x$ is small $\sin(x) \approx x$ and any polynomial grows faster than logarithm. Hence $\lim_{x \to 0^+} \sin(x) \ln(x) = \lim_{x \to 0^+} x = 0$
