325. Without the l'Hospital's rules.

by Virgil Nicula, Oct 27, 2011, 7:51 PM

$1\blacktriangleright$ Find $\lim_{x \to 0}\frac{x-\sin x}{x^2}$ without using the l'Hospital's rules.

Proof. Denote $f(x)=\frac {x-\sin x}{x^2}\ ,\ x\ne 0$ . Observe that $f(-x)=-f(x)$ for any $x\ne 0$ . Suppose w.l.o.g. $0<x<\frac {\pi}{2}$ .

Therefore, $\boxed{0<\sin x<x<\tan x}\implies$ $0<x-\sin x<\tan x-\sin x=\tan x\cdot (1-\cos x)\implies$

$0<f(x)=\frac {x-\sin x}{x^2}<\tan x\cdot\frac {1-\cos x}{x^2}\implies$ $\lim_{x\searrow 0}f(x)=0$ and $\lim_{x\nearrow 0}f(x)=0\implies$ $\lim_{x\to 0}f(x)=0$ .

I used the well-known limits $\lim_{x\to 0}\frac {\sin x}{x}=1$ and $\lim_{x\to 0}\frac {1-\cos x}{x^2}=\frac 12$ . Indeed, $\lim_{x\to 0}\frac {1-\cos x}{x^2}=$ $\lim_{x\to 0}\frac 12\cdot\left( \frac {\sin\frac x2}{\frac x2}\right)^2=\frac 12$ .
This post has been edited 5 times. Last edited by Virgil Nicula, Mar 27, 2019, 8:13 AM

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