<s>[INST] To prove the limit of $\frac{x^2+2\cos x-2}{x\sin^3 x}$ at zero is $1/12$ To Prove $$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{1}{12}$$
I tried with L'Hospital rule but in vain.
 [/INST] We can use the basic limits 
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x}{x}=1,\ \ \ \ \ \ \ \ \
\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}}=\dfrac{1%
}{24}
\end{equation*}
as follows
\begin{equation*}
\lim\limits_{x\rightarrow 0}\frac{x^{2}+2\cos x-2}{x\sin ^{3}x}%
=\lim\limits_{x\rightarrow 0}\left( \frac{x}{\sin x}\right)
^{3}\lim\limits_{x\rightarrow 0}\left( \frac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}%
}\right) \left( 2\right) =1^{3}\cdot \frac{1}{24}\cdot 2=\frac{1}{12}.
\end{equation*}
The basic limits can be computed by repeated use of L'Hospital's rule. 
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