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.

L'Hopital's rule may be used repeatedly until you find an answer:
$$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\lim_{x \to 0}\frac{2x-2\sin x}{\sin^2 x(\sin x+3x\cos x)}$$
$$= \lim_{x \to 0}\frac{2-2\cos x}{ 6\sin ^2(x)\cos (x) - 3x(\sin^3(x)-2\sin(x)\cos^2(x))}$$
$$= \lim_{x \to 0}\frac{2\sin (x)}{ 3(-3\sin^3 (x) + 6\cos^2(x)\sin(x) + x(2\cos^3(x)-7\cos(x)\sin^2(x)))}$$
$$= \lim_{x \to 0}\frac{-2\cos(x)}{ -24\cos^3(x)+84\sin^2(x)\cos(x)+x(60\sin(x)\cos^2(x)-21\sin^3(x))}$$
$$= \frac{-2\cos(0)}{ -24\cos^3(0)+84\sin^2(0)\cos(0)+0(60\sin(0)\cos^2(0)-21\sin^3(0))}$$
$$=\frac{1}{12}$$
