Find limits of $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor Expansion. Find the limit $\displaystyle \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor expansion.
My Try
$\displaystyle =\lim _{x \to 0} \frac {\cos x - \frac{\sin x}{x}} {x \sin x}$
$=\frac {\displaystyle\lim _{x \to 0}\cos x - \lim _{x \to 0}\frac{\sin x}{x}} {\displaystyle\lim _{x \to 0}x \sin x}$
$=\frac{1-1}{0}$
But still I end up with $\frac00$
Any hint for me to proceed would be highly appreciated.
P.S: I did some background check on this question on mathstack and found they have solved this with l'Hopital's rule and the answer seems to be $\frac{-1}{3}$.
What is $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$?

While searching more easy way, let me suggest one possible way to solve main difficult part of suggested limit. I change denumerator to $x^3$, for simplicity, as it's equivalent $x^2\sin x$
Suppose we know existence of limit. Then
$$L=\lim_{x\to0}\frac{x-\sin x}{x^3} = \lim_{x\to0}\frac{x-3\sin \frac{x}{3}+4 \sin^3 \frac{x}{3}}{x^3}=\\ =\lim_{x\to0}\left(3\frac{\frac{x}{3} - \sin \frac{x}{3}}{x^3} + \frac{4 \sin^3 \frac{x}{3}}{x^3}\right) =\lim_{x\to0}\left(\frac{\frac{x}{3} - \sin \frac{x}{3}}{9\left(\frac{x}{3}\right)^3} + \frac{4 \sin^3 \frac{x}{3}}{x^3}\right)=\frac{L}{9}+\frac{4}{27}$$
From obtained equation $L=\frac{1}{6}$
