Question

Problema nr.10
Admitere UPT timisoara
Va multumesc!

Answer 1

$f(x) = x-\tan x \\ \\\lim\limits_{x\to 0}\dfrac{f(\tan x)}{3x^3}=\lim\limits_{x\to 0}\dfrac{\tan x-\tan(\tan x))}{3x^3} =\\ \\\tan x = t\Rightarrow x = \arctan t\\x\to 0\Rightarrow t\to 0 \\ \\=\lim\limits_{x\to 0}\dfrac{t - \tan(t)}{3\arctan^3 t} =\lim\limits_{x\to 0}\left(\dfrac{t - \tan(t)}{3\arctan^3 t} \cdot \dfrac{t^3}{t^3}\right)= \lim\limits_{t\to 0}\dfrac{t - \tan(t)}{3t^3}\overset{\frac{0}{0}}{=}\\ \\\overset{\frac{0}{0}}{=}\lim\limits_{t\to 0}\dfrac{1-(1-\tan^2 t)}{9t^2}=$

$=\lim\limits_{t\to 0}\dfrac{-\tan^2 t}{9t^2}= -\dfrac{1}{9}\cdot\lim\limits_{t\to 0}\dfrac{\tan^2 t}{t^2}=\boxed{-\dfrac{1}{9}}$

$\lim\limits_{x\to 0}\dfrac{\arctan x}{x} = 1,\quad \lim\limits_{x\to 0}\dfrac{\tan x}{x} = 1$