Question

Doar pct c va rog mult ​

Answer 1

$\lim\limits_{x\to \infty}\Big(2x\cdot \ln \dfrac{x-1}{x}\Big) = \lim\limits_{x\to \infty}\dfrac{\ln \dfrac{x-1}{x}}{\dfrac{1}{2x}} \overset{\frac{\infty}{\infty}}{=} \lim\limits_{x\to \infty}\dfrac{\Big[\ln(x-1)-\ln x\Big]'}{\Big(\dfrac{1}{2x}\Big)'} =\\ \\ \\ =\lim\limits_{x\to \infty}\dfrac{\dfrac{1}{x-1}-\dfrac{1}{x}}{-\dfrac{1}{2x^2}} = \lim\limits_{x\to \infty}\dfrac{-2x^2\Big[x-(x-1)\Big]}{x(x-1)} =\lim\limits_{x\to \infty}\dfrac{-2x^2\cdot 1}{x^2-1} = \boxed{-2}$