calculus - A limit without invoking L'Hopital: $lim_{x to 0} frac{x cos x - sin x}{x^2}$

The following limit

$$\ell=\lim_{x \rightarrow 0} \frac{x \cos x - \sin x}{x^2}$$

is a nice candidate for L'Hopital's Rule. This was given at a school before L'Hopital's Rule was covered. I wonder how we can skip the rule and use basic limits such as:

$$\lim_{x \rightarrow 0} \frac{\sin x}{x} \quad , \quad \lim_{x \rightarrow 0} \frac{\cos x -1}{x^2}$$

Answer

We have,

$$\lim_{x \to 0} \dfrac{x\cos x - \sin x}{x^2} = \lim_{x \to 0} \dfrac{\cos x -1}{x} + \lim_{x \to 0}\dfrac{x - \sin x}{x^2} $$

$$ = -2\lim_{x \to 0} \dfrac{\sin^2 \left(\frac{x}{2}\right)}{x} + \lim_{x \to 0}\dfrac{x - \sin x}{x^2} $$

The first limit is zero since $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1$, and,

$$ 0 \leq \lim_{x \to 0}\dfrac{x - \sin x}{x^2} \leq \lim_{x \to 0}\dfrac{\tan x - \sin x}{x^2}$$

But,

$$\lim_{x \to 0}\dfrac{\tan x - \sin x}{x^2} = \lim_{x \to 0} \ \left( \sin x \times \dfrac{1-\cos x}{x^2 \cos x} \right) = \lim_{x \to 0} \dfrac{1 - \cos x}{x} = 0$$

Thus, by the Squeeze Theorem,

$$\lim_{x \to 0} \dfrac{x\cos x - \sin x}{x^2} =0$$