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

Question

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 1

$0 \leq x-\sin(x)=\int_0^x{(1-\cos{t})\,dt}=2\int_0^x{\sin^2{\frac{t}{2}}\,dt} \leq 2\int_0^x{\frac{t^2}{4}\,dt}=\frac{x^3}{6}$.

Then use the squeeze theorem.

Answer 2

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,

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

Answer 3

You can use this:$$\sin{x}=x-\frac{x^3}{6}+o(x^3)$$ $$\Longrightarrow \frac{\sin{x}-x}{x^2}=-\frac{x}{6}+o(x)$$

Answer 4

This simplifies into $$\frac{\frac{sin(x)}{x}}{x}-\frac{1}{x}$$ and knowing $$\lim_{x\to 0}\frac{\sin(x)}{x} = 1$$ we get $$\frac{1}{x}-\frac{1}{x} = 0$$

So the limit is $0$.