How do I evaluate $\lim _{x\to 0}\left(\frac{x-\sin x}{x\sin x}\right)$ without using L'Hopital or series? 
How do I evaluate $\lim _{x\to 0}\left(\frac{x-\sin x}{x\sin x}\right)$ without using L'Hopital or series?

I've tried expanding the variable such as $x = 2y$ or $x = 3y$, but seemed to still get stuck.

Try the following 
$$L=\lim \dfrac{x-x\cos\frac{x}{2}+x\cos\frac{x}{2}-\sin x}{x\sin x}$$ then use the $\sin(x)=2\sin \frac{x}{2}\cos\frac{x}{2}$ now the first two terms limit is zero and the second two  terms have a relation with the limit $L$
$$\lim \frac{x \cos\frac{x}{2} -2 \sin\frac{x}{2} \cos\frac{x}{2}}{2x \sin\frac{x}{2} \cos\frac{x}{2}}$$
$$\lim \frac{2 \cos\frac{x}{2}\left( \frac{x}{2}- \sin\frac{x}{2}\right)}{2x \sin\frac{x}{2} \cos\frac{x}{2}}$$
$$\lim \frac{1}{2}\frac{\frac{x}{2}-\sin \frac{x}{2}}{\frac{x}{2}\sin\frac{x}{2}}$$ and the last limit is $1/2L$ so $L=1/2L$ hence $L=0$ by this method you can solve something like 
$\frac{\sin x -x}{x^3}$ but you need to add and subtract a certain term can figure it out ?
