Limit without l'Hopital and using another one.

Question

I was trying to find a limit which is trivial using L'Hopital, but I can't seem to find a way to do it without it. For example : $$f(x)= \frac{\ln(1+\sin x)}{e^x-1}$$ as $x$ tends to 0

It seems doable using $\ln(1+x)/x$ tends to 1 as $x$ tends to 0.

Answer 1

Note that we have\begin{align}\lim_{x \to 0}\frac{\ln(1+\sin(x))}{\exp(x)-1} =\lim_{x \to 0}\left(\frac{\ln(1+\sin(x))}{\sin(x)}\right)\left(\frac{x}{\exp(x)-1}\right)\left( \frac{\sin(x)}{x}\right)\end{align}

Answer 2

Hint: For $x$ close to $0$ $\sin(x) \approx x$ so we can consider

$\lim_{x \to 0}\frac{\ln(1+x)}{e^x-1}$. See if you can go from there.