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

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?

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,

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

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

Then use the squeeze theorem.

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

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 $$0$$.

• Why the downvote? – Dr. Mathva Feb 13 '19 at 21:37
• Because the proof is wrong. Your argument works if I replace $sin(x)/x$ with $sin(x)/x+1/ln(x)$, but the conclusion does not hold. – Mindlack Feb 13 '19 at 21:42
• Careful in general you can´t us $\sin{x} \sim x$ when the limit has addition or subtraction. See $\frac{x - \tan{x}}{x^3}$ – JoseSquare Feb 13 '19 at 21:44
• This is not correct, -1. – MPW Feb 13 '19 at 21:45
• Don't forget the also much "easier" rule $(x+y)^n = x^n + y^n$ – MPW Feb 13 '19 at 21:46