Question

Vreau si eu ajutor la aceasta limita multumesc anticipat
Dau coronita

Answer 1

Ai solutia in atasament.

Answer 2

$\it Fie\ f(x)=\dfrac{x-lnx}{x+lnx}\\ \\ \\ f'(x)=\dfrac{(1-\dfrac{1}{x})(x+lnx)-(1+\dfrac{1}{x})(x-lnx)}{(x+lnx)^2}=\\ \\ \\ =\dfrac{x+lnx-1-\dfrac{1}{x}lnx-x+lnx-1+\dfrac{1}{x}lnx}{(x+lnx)^2}=\dfrac{2(-1+lnx)}{(x+lnx)^2}$

Limita dată este valoarea derivatei în e, adică:

$\it f'(e) =\dfrac{2(-1-lne)}{(e+lne)^2}=\dfrac{2(-1+1)}{(e+1)^2}=0$