$\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{1}{60}$ without using L Hospital rule or series expansion Prove that $\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{1}{60}$ without using L Hospital rule or series expansion.

I tried $\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\lim_{x\to 0}\frac{2+\cos x}{x^4\lim_{x\to 0}\frac{\sin x}{x}}-\frac{3}{x^4}$
$=\lim_{x\to 0}\frac{2+\cos x}{x^4}-\frac{3}{x^4}=\lim_{x\to 0}\frac{\cos x-1}{x^4}=\lim_{x\to 0}\frac{-2\sin^2 \frac{x}{2}}{x^4}=\lim_{x\to 0}\frac{-2\sin^2 \frac{x}{2}}{4x^2(\frac{x}{2})^2}$
$=\lim_{x\to 0}\frac{-1}{2x^2}$
Now it has turned into limit does not exist.I dont know where have i made mistake,because as per my knowledge my every step is correct.If i have made mistake please correct me
Please help me.Thanks.

As per the method suggested by Lab bhattacharjee, I got the answer to my problem.Thanks to him.
Let $L=\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}......(1)$
So $L=\lim_{x\to 0}\frac{2+\cos 2x}{8x^3\sin 2x}-\frac{3}{16x^4}$
$L=\lim_{x\to 0}\frac{1+2\cos^2x}{16x^3\sin x\cos x}-\frac{3}{16x^4}$
$16L=\lim_{x\to 0}\frac{1+2\cos^2x}{x^3\sin x\cos x}-\frac{3}{x^4}......(2)$
Subtracting $(1)$ from $(2)$,we get
$15 L=\lim_{x\to 0}\frac{1-2\cos x+\cos^2 x}{x^3\sin x}$
$15 L=\lim_{x\to 0}\frac{(1-\cos x)^2}{x^4\lim_{x\to 0}\frac{\sin x}{x}}$
$15L=\lim_{x\to 0}\frac{4\sin^4\frac{x}{2}}{x^4}$
$15L=\lim_{x\to 0}\frac{4\sin^4\frac{x}{2}}{16(\frac{x}{2})^4}$
$L=\frac{1}{60}\lim_{x\to 0}(\frac{\sin\frac{x}{2}}{\frac{x}{2}})^4$
$L=\frac{1}{60}$
