Question

Exercitiul e in poza, mai jos.

Rog rezolvare completa sau pasii (corecti) de rezolvare, multumesc!

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$l = \lim\limits_{x\to \infty}\Big(\dfrac{1+x}{2+x}\Big)^{\frac{1-x}{1-\sqrt x}} \\ \\ \ln(l) = \ln\Bigg(\lim\limits_{x\to \infty}\Big(\dfrac{1+x}{2+x}\Big)^{\frac{1-x}{1-\sqrt x}}\Bigg)\\ \\ \ln(l) = \lim\limits_{x\to \infty}\Bigg(\dfrac{1-x}{1-\sqrt x}\cdot \ln\dfrac{1+x}{2+x}\Bigg) \\ \\\\ \sqrt x = t \Rightarrow x = t^2 \Rightarrow t\to\infty\\ \\\\ \ln(l) = \lim\limits_{t\to \infty}\Bigg(\dfrac{1-t^2}{1-t}\cdot \ln \dfrac{1+t^2}{2+t^2}\Bigg)$

$\ln(l) = \lim\limits_{t\to \infty}\Bigg((1+t)\cdot \ln \dfrac{1+t^2}{2+t^2}\Bigg) \\ \\ \ln(l) = \ln 1 + \lim\limits_{t\to \infty}\dfrac{\ln(1+t^2)-\ln(2+t^2)}{\dfrac{1}{t}}\\ \\ \ln(l) = \lim\limits_{t\to \infty}\dfrac{\dfrac{2t}{1+t^2}-\dfrac{2t}{2+t^2}}{-\dfrac{1}{t^2}} = \lim\limits_{t\to \infty}\dfrac{2t(2+t^2)-2t(1+t^2)}{-\dfrac{(1+t^2)(2+t^2)}{t^2}} = \\ \\ = \lim\limits_{t\to \infty}\dfrac{2t\cdot (-t^2)}{(1+t^2)(2+t^2)} = 0\\ \\\\ \Rightarrow l = e^0 \Rightarrow \boxed{l = 1}$