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.

First, read this answer. This answer shows that, for $x$ close to $0$, we have the following:
$$\sin x \leq x \leq \tan x$$
Thus, since $\sin x \leq x$ and $\sin x$ has the same sign as $x$ (i.e. either both are positive, both are negative, or both are $0$), we know that $\frac{1}{\sin x} \geq \frac{1}{x}$. Thus, $\frac{1}{\sin x}-\frac{1}{x}\geq 0$.
Also, since $x \leq \tan x$ and $\tan x$ has the same sign as $x$, we know that $\frac{1}{x} \geq \frac{1}{\tan x}$. This implies that $\frac{1}{\sin x}-\frac{1}{x} \leq \frac{1}{\sin x}-\frac{1}{\tan x}$,
Now, we have:
$$0 \leq \frac{1}{\sin x}-\frac{1}{x} \leq \frac{1}{\sin x}-\frac{1}{\tan x}$$
Simplify the trig expressions:
$$0 \leq \frac{x-\sin x}{x\sin x} \leq \frac{1-\cos x}{\sin x}$$
Now, $\lim_{x\to 0} 0=0$ for obvious reasons. Thus, let's consider $\lim_{x\to 0} \frac{1-\cos x}{\sin x}$. Multiply numerator and denominator by $1+\cos x$:
$$\lim_{x\to 0}\frac{1-\cos^2 x}{\sin x(1+\cos x)}=\lim_{x\to 0}\frac{\sin^2 x}{\sin x(1+\cos x)}=\lim_{x\to 0}\frac{\sin x}{1+\cos x}=\frac{\sin 0}{1+\cos 0}=0$$
Thus, $\lim_{x\to 0} 0=\lim_{x\to 0} \frac{1-\cos x}{\sin x}=0$, so by Squeeze Theorem, $\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=0$.
