Limits with Trig

Is there any way to evaluate

$$\lim_{x\to0} \frac{x \cos x - \sin x}{x^2}$$
without using L'Hopital's Rule? I was trying to use some of the standard trig limits (e.g. $\lim_{x\to 0} \frac{\sin x}{x} = 1)$, etc. but couldn't figure it out.

Thank you.

Answer

Adding and subtracting $x$ in nummerator we can split the expression under limit as

$$\frac{\cos x-1} {x} +\frac{x-\sin x} {x^2}$$ The first expression tends to $0$ because $(1-\cos x) /x^2\to 1/2$ and the second expression also tends to $0$ as shown below.

Since the second expression is an odd function it is sufficient to prove that the expression tends to $0$ as $x\to 0^{+}$. Next note the famous inequality $$\sin x