How do I evaluate this limit (without the use of L'Hospital or series expansions)? I have been trying for quite some time to evaluate this limit
$$\lim_{x\to0} \frac{\tan x-\sin x}{x-\sin x}$$
but without using L'Hospital's rule or series expansions. I have tried the common things, I have got this:
$$\lim_{x\to0} \frac{\sin x(1-\cos x)}{\cos x(x-\sin x)}$$
But I have no idea how to proceed further.

Hint: Here the key limit will be derived. The rest should be easy.
Let:
$$
L=\lim_{x\to0}\frac{x-\sin x}{x^3}.
$$ 
We have:
$$
L=\lim_{x\to0}\frac{2x-\sin 2x}{8x^3}\implies
4L=\lim_{x\to0}\frac{x-\frac12\sin 2x}{x^3}\\\implies
3L=\lim_{x\to0}\frac{\sin x-\frac12\sin 2x}{x^3}
=\lim_{x\to0}\frac{\sin x(1-\cos x)}{x^3}
=\lim_{x\to0}\frac{\sin^3 x}{x^3(1+\cos x)}=\frac12\\
\implies L=\frac16.
$$
