Find : $\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$ with l'Hôpital's rules I want to calculate :
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$$
with l'Hôpital's rules.
I get
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{x-\sin(x)}{x\sin(x)}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{1-\cos(x)}{\sin(x)+x\cos(x)}\right)$$
but I don't see useful next steps.

As it turns out, you just have to use L'Hospital's rule twice and observe that on the second try the denominator is no longer zero when you use directly substitution:
\begin{align}
\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin x} - \frac{1}{x}\right)
&=\lim\limits_{x \rightarrow 0}\frac{x -\sin x}{x\sin x}\\
&=\lim\limits_{x \rightarrow 0}\frac{(x -\sin x)'}{(x\sin x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{x' -(\sin x)'}{x'\sin x+x(\sin x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{1 - \cos x}{\sin x+x \cos x}\\
&=\lim\limits_{x \rightarrow 0}\frac{(1 - \cos x)'}{(\sin x+x \cos x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{1' - (\cos x)'}{(\sin x)'+x' \cos x +x(\cos x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{\sin x}{\cos x + \cos x - x \sin x}\\
&=\lim\limits_{x \rightarrow 0}\frac{\sin x}{2\cos x - x \sin x}\\
&=\frac{0}{2\cdot 1 - 0 \cdot 0}\\
&=\frac{0}{2}\\
&=0
\end{align}
