[tex]L =\lim\limits_{x\to \infty}x\Big(\ln(x+1)-\ln(x-1)\Big) =\lim\limits_{x\to \infty}\dfrac{\ln\Big(\dfrac{x+1}{x-1}\Big)}{\dfrac{1}{x}} =[/tex]
Aplic L'Hôpital deoarece sunt în cazul de nedeterminare 0/0.
[tex]=\lim\limits_{x\to \infty}\dfrac{\dfrac{1}{x+1}-\dfrac{1}{x-1}}{-\dfrac{1}{x^2}}= \lim\limits_{x\to \infty}\dfrac{\dfrac{x-1-(x+1)}{(x+1)(x-1)}}{-\dfrac{1}{x^2}} = \\ \\ \\= \lim\limits_{x\to \infty}\dfrac{-2\cdot (-x^2)}{(x+1)(x-1)} = \lim\limits_{x\to \infty}\dfrac{2x^2}{x^2-1^2} = \boxed{2}[/tex]
