Compute using L'Hopital's rule:
$$\lim_{x\to 0^+} \frac{\ln(x)}{1/\sin(x)}$$
I kept differentiating, but it's getting too long. How can I tackle this kind of problem?
Also, when I encounter a limit of the form $\infty \cdot 0$ and I want to make it eligible for L'Hopital's rule, should I transform it into $$\frac{\infty}{1/0}=\frac{\infty}{\infty}$$ or
$$\frac{0}{1/\infty}=\frac{0}{0}$$
Best Answer
Or without l'Hopital: $$\frac{\ln x}{\frac1{\sin x}} = (x\ln x)\cdot \frac{\sin x}x,$$ where each factor has a well-known limit as $x\to0^+$.