<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] Is it compulsory to exploit $\lim_\limits{x\to 0} \dfrac{\sin x}{x}$ and $\lim_\limits{x\to 0} \dfrac{\cos x -1}{x^2}$?
If not, why don't you just expand it, like: $$
\lim_\limits{x\to 0} \frac{ x \cos x -\sin x }{ x^2 }
=\lim_\limits{x\to 0} \frac{ x [1 -x^2/2 +\mathscr{O}(x^4)] -[x -x^3/6 +\mathscr{O}(x^5)] }{x^2}
=\lim_\limits{x\to 0} \frac{x -x^3/2 -x +x^3/6 +\mathscr{O}(x^5)}{x^2}
=\lim_\limits{x\to 0} \frac{-x^3/3}{x^2}
=\lim_\limits{x\to 0} -\frac{x}{3}
=0
$$
</s>