real analysis - Proof that $lim_{x to 0} frac{sin(x) - x}{x^2} = 0$ (Without L'Hospital)

I was studying calculus and I got stuck in proving that

$$\lim_{x \to 0} \frac{\sin(x) - x}{x^2} = 0.$$

Using L'Hospital is easy. However, I want a proof where I don't use L'Hospital.

Help?

Answer

Since $\frac{\sin x-x}{x^2}$ is an odd function, it suffices to show that the right-limit converges to $0$. In this answer, we only use the following fact:

$$\forall x \in (0, \pi/2), \quad \sin x \leq x \leq \tan x.$$

This inequality appears in the standard proof of $\frac{\sin x}{x} \to 1$ as $x \to 0$ in many calculus textbook, so I will skip the proof. And indeed, once we have this inequality, then $\frac{1}{\cos x} \leq \frac{\sin x}{x} \leq 1$ and letting $x \to 0$ together with the squeezing theorem gives $\frac{\sin x}{x} \to 1$.

Next, for $x \in (0, \pi/2)$,

$$\frac{\sin x - \tan x}{x^2} \leq \frac{\sin x - x}{x^2} \leq 0.$$

But since $\sin x - \tan x = \tan x(\cos x - 1) = - \frac{\tan x \sin^2 x}{1+\cos x}$ and $\frac{\sin x}{x} \to 1$ as $x\to 0$, we have

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

So, by the squeezing theorem,

$$\lim_{x\to0^+} \frac{\sin x - x}{x^2} = 0.$$