<s>[INST] 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.
 [/INST] 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}
</s>