<s>[INST] 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:
$$\lim_{x \rightarrow 0} \frac{\sin x}{x} \quad , \quad \lim_{x \rightarrow 0} \frac{\cos x -1}{x^2}$$
 [/INST] 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
$$
\lim_{x\to0}\frac{x\cos(x)-\sin(x)}{x^2}=0
$$
</s>