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.

from 
$$
\cos x = 1 -  \frac{x^2}2+\frac{x^4}{24}-\dots
$$
we have
$$
x^2+2\cos x - 2 = \frac{2x^4}{24}-\dots
$$
and
$$
\sin x = x-... \\
x\sin^3 x = x^4-...
$$
