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

From the geometric proof of $\frac{\sin x}{x} \to 1$ as $x\to0$ we know $\cos x<\frac{\sin x}{x} <1$ near $0$. Since $\frac{x\cos x -\sin x}{x^2} = \frac{\cos x -\frac{\sin x}{x}}{x}$, we see that $$ 0=\lim_{x\to0}\frac{\cos x -1}{x} \le \ell \le \lim_{x\to0} \frac{\cos x-\cos x}{x}=0,$$so $\ell=0$.

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,

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

The function $\frac{x\cos(x)-\sin(x)}{x^2}$ is odd, so we only need to look at $0\lt x\lt\frac\pi2$. As shown in this answer, $0\le\sin(x)\le x\le\tan(x)$. Furthermore, $x\cos(x)-\sin(x)=(x-\tan(x))\cos(x)\le0$. So we have $$ \begin{align} 0 &\ge\color{#C00000}{\frac{x\cos(x)-\sin(x)}{x^2}}\\ &\ge\frac{\sin(x)(\cos(x)-1)}{x^2}\\ &=-\frac{\sin(x)}x\frac{1-\cos^2(x)}{x(1+\cos(x))}\\ &=-\frac{\sin^3(x)}{x^2(1+\cos(x))} \end{align} $$ By the Squeeze Theorem, we have $$ \lim_{x\to0}\frac{x\cos(x)-\sin(x)}{x^2}=0 $$