Question

Salut, am si eu nevoie de ajutor la ex 38 .

Answer 1

Răspuns:

Explicație pas cu pas:

Rezolvarea în imginea de mai jos.

Answer 2

$l = \lim\limits_{x\to \infty} \Big(x-x^2\ln\frac{x+1}{x}\Big) \\ \\ \ln\dfrac{x+1}{x} = t \Rightarrow \dfrac{x+1}{x} = e^t \Rightarrow x+1 = e^tx \Rightarrow xe^t-x = 1 \Rightarrow \\ \Rightarrow x(e^t-1) = 1 \Rightarrow x = \dfrac{1}{e^t-1} \\ \\ x \to \infty \Rightarrow t \to 0 \\ \\ l = \lim\limits_{t \to 0} \Big(\dfrac{1}{e^t-1}-\dfrac{1}{(e^t-1)^2}\cdot t\Big) = \lim\limits_{t \to 0} \dfrac{e^t-1-t}{(e^t-1)^2} \overset{L'H}{=}$

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