<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] You can also use Taylor series:
$$\sin(x)=x-\frac{x^3}{6}+O(x^4)$$
So your function:
$$\frac{1}{\sin(x)}-\frac{1}{x}=\frac{x-\sin(x)}{x\sin(x)}=\frac{\frac{x^3}{6}+O(x^4)}{x^2-\frac{x^4}{6}+O(x^5)}=\frac{\frac{x}{6}+O(x^2)}{1-\frac{x^2}{6}+O(x^3)} \to \frac{0}{1}=0$$
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