Use L'Hôpital's rule to solve
$$\lim_{x\to 0^{+}}\sin(x)\ln(x)$$
My attempt:
$$\lim_{x\to 0^{+}}\sin(x)\ln(x) = \lim_{x\to 0^{+}} \frac{\ln(x)}{\sin(x)}$$
$\frac{\ln(0)}{\sin(0)}$ is in the form $\frac{-\infty}{0}$, it is indeterminate, and as such, using L'Hôpital's rule:
$$\lim_{x\to 0^{+}} \frac{\ln(x)}{\sin(x)} = \lim_{x\to0^{+}}\frac{1}{x\cos(x)}$$
I would have applied L'Hôpital's rule again, but to my horror, I realise that $\frac{1}{0}$ is not an indeterminate form, according to Wikipedia.
I realized that my reasoning, while it could let me get the correct answer, is wrong! How do you solve this question now?
EDIT: Is $\frac{-\infty}{0}$ indeterminate as well? I couldn't find it in Wikipedia. If it is not indeterminate, I couldn't use L'Hôpital's rule too!