Question

Se considera functia f : R -> R, $f(x)=\frac{|x|-1}{|x|+1}ln\frac{x^2+1}{|x|+1}$
Sa se calculeze $\lim_{x \to \ 0} f'(x)\\x\ \textgreater \ 0\\\\ \lim_{x \to \ 0} f'(x)\\x\ \textless \ 0$

Limitele se inmultesc doar ca nu le pot scrie intr-o forma mai inteligibila ....

Answer 1

$f(x) = \dfrac{|x|-1}{|x|+1}\ln \dfrac{x^2+1}{|x|+1} \\ \\ \\\underset{x>0}{\lim\limits_{x\to 0}}\,f'(x)\cdot \underset{x<0}{\lim\limits_{x\to 0}}\,f'(x) = \\ \\ \\=\underset{x>0}{\lim\limits_{x\to 0}}\,\dfrac{f(x)-f(0)}{x-0}\cdot \underset{x<0}{\lim\limits_{x\to 0}}\,\dfrac{f(x)-f(0)}{x-0} = \\ \\ = \underset{x>0}{\lim\limits_{x\to 0}}\,\dfrac{f(x)}{x}\cdot \underset{x<0}{\lim\limits_{x\to 0}}\,\dfrac{f(x)}{x}=$

$=\underset{x>0}{\lim\limits_{x\to 0}}\,\dfrac{\dfrac{|x|-1}{|x|+1}\ln \dfrac{x^2+1}{|x|+1}}{x} \cdot \underset{x<0}{\lim\limits_{x\to 0}}\,\dfrac{\dfrac{|x|-1}{|x|+1}\ln \dfrac{x^2+1}{|x|+1}}{x} = \\ \\ =\underset{x>0}{\lim\limits_{x\to 0}}\,\dfrac{\dfrac{x-1}{x+1}\ln \dfrac{x^2+1}{x+1}}{x} \cdot \underset{x<0}{\lim\limits_{x\to 0}}\,\dfrac{\dfrac{-x-1}{-x+1}\ln \dfrac{x^2+1}{-x+1}}{x} =$

$=\underset{x>0}{\lim\limits_{x\to 0}}\,\dfrac{(x-1)\Big[\ln(x^2+1) - \ln(x+1)\Big]}{x^2+x} \cdot \\ \\ \cdot \underset{x<0}{\lim\limits_{x\to 0}}\,\dfrac{(-x-1)\Big[\ln(x^2+1) -\ln(-x+1)\Big]}{-x^2+x} =\\\\ \\ =\underset{x>0}{\lim\limits_{x\to 0}}\,\dfrac{\Big[\ln(x^2+1) - \ln(x+1)\Big]+(x-1)\cdot \Big[\dfrac{2x}{x^2+1}-\dfrac{1}{x+1}\Big]}{2x+1} \cdot \\ \\ \cdot \underset{x<0}{\lim\limits_{x\to 0}}\,\dfrac{-\Big[\ln(x^2+1) -\ln(-x+1)\Big]+(-x-1)\cdot \Big[\dfrac{2x}{x+1}+\dfrac{1}{-x+1}\Big]}{-2x+1} =$

Trecem la limite.

$= \dfrac{1}{1}\cdot \dfrac{-1}{1} = 1\cdot (-1) = \boxed{-1}$