Calculate $\lim\limits_{x\to0} \frac{1}{\sin x} \cdot \ln \left(\frac{e^x -1}{x}\right)$ I was trying to calculate the limit of the following function:
$$ \lim_{x\to0} \frac{1}{\sin x} \cdot \ln \left(\frac{e^x -1}{x}\right) $$
My first thought was using L'Hopital's rule since $\Large \frac{e^x -1}{x}$ goes to 1 so the whole $\ln$ goes to 0.
But then I get another complicated expression, and finally I end up using L'Hopital's rule at least 5 times before getting an actual result.
Is there a wiser way for dealing this limit? (I mean, without using this rule?)
Thanks.

Use Taylor expansion: $e^x-1=x+x^2/2+o(x^2)$. Thus
\begin{align*}
\frac{1}{\sin x}\log\left(\frac{e^x-1}{x}\right)
&=\frac{x}{\sin x}\frac1x\log\left(1+\frac{x^2/2+o(x^2)}{x}\right)\\
&=\frac{x}{\sin x}\frac1x\frac{\log\left(1+\frac{x^2/2+o(x^2)}{x}\right)}{(x^2/2+o(x^2))/x}\frac{x^2/2+o(x^2)}{x}\\
&=\underbrace{\frac{x}{\sin x}\frac{\log\left(1+\frac{x^2/2+o(x^2)}{x}\right)}{(x^2/2+o(x^2))/x}}_{\to1}\underbrace{\frac{x^2/2+o(x^2)}{x^2}}_{\to1/2}\\
\end{align*}
Thus your limit goes to $1/2$.
