<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] Hint. You may use Taylor expansions, as $x \to 0$,
$$
\cos x =1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)
$$
$$
\sin x =x+O(x^3)
$$ giving
$$
x^2+2\cos x-2=\frac{x^4}{12}+O(x^6)
$$ $$
x\sin^3 x =x^4+O(x^6)
$$ and

$$
\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{\frac{x^4}{12}+O(x^6)}{x^4+O(x^6)}=\frac{1}{12}+O(x^2)
$$

</s>