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

Question

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 1

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

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

Answer 2

Use the answer here and

$$\lim_{x\to 0} \frac{\sin x - x}{x^2} = -\lim_{x \to 0}x\left(\frac{x - \sin(x)}{x^3}\right) = -\lim_{x \to 0}x\lim_{x \to 0}\left(\frac{x - \sin(x)}{x^3}\right) = \frac{-1}{6}\cdot 0 = 0$$

Answer 3

If you know the function $h(x)=\frac{\sin{x}}{x}$ is analytic, then $$h^{\prime}(0)=\lim_{x\to0}\frac{h(x)-h(0)}{x}=0$$ since it is an even function.

Then we obtain the desired limit.

Answer 4

For $x>0,$

$$\tag 1 0\le x-\sin x = \int_0^x(1-\cos t)\, dt.$$

Now $1-\cos t \le t^2$ for all $t.$ Why? Because both sides are even, the inequality is true for $t=0,$ and it's true for the derivatives on $[0,\infty).$ Thus the right side of $(1)$ is bounded above by

$$\int_0^xt^2\, dt = x^3/3.$$ That is enough to show $\lim_{x\to 0^+}(x-\sin x)/x^2$ is $0,$ and since this function is odd, $\lim_{x\to 0^-}(x-\sin x)/x^2$ is also $0.$