calculus - $lim_{xto 0} frac{sin x - x}{x^2}$ without L'Hospital or Taylor

It is easy to see that $$\lim_{x\to 0} \frac{\sin x - x}{x^2} =0,$$ but I can't figure out for the life of me how to argue without using L'Hospital or Taylor. Any ideas?

Answer

In THIS ANSWER, I used the integral definition of the arcsine function to show that for $0 \le x\le \pi/2$, we have the inequalities

$$x\cos(x)\le \sin(x)\le x \tag 1$$

Using the trigonometric identity $1-\cos(x)=2\sin^2(x/2)$, we see from $(1)$ that

$$-2x\,\,\underbrace{\left(\frac{\sin^2(x/2)}{x^2}\right)}_{\to \frac14}\le \frac{\sin(x)-x}{x^2}\le 0 \tag2$$

Applying the squeeze theorem to $(2)$ yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{\sin(x)-x}{x^2}=0}$$