Question

rezolvati limita urmatoare:

Answer 1

$l = \lim\limits_{x\to \infty} \Big(x-x^2\ln(1+\frac{1}{x})\Big)\\ \\ 1+\frac{1}{x} = t \Rightarrow \frac{1}{x} = t-1 \Rightarrow x = \dfrac{1}{t-1}\\ x\to +\infty \Rightarrow t\to 1_+\\ \\ l = \lim\limits_{t\to 1_+}\Big(\dfrac{1}{t-1}-\dfrac{\ln t}{(t-1)^2}\Big) = \lim\limits_{x\to 1_+}\dfrac{t-1-\ln t}{(t-1)^2} \overset{\frac{0}{0}}{=} \lim\limits_{t\to 1_+}\dfrac{1-\frac{1}{t}}{2(t-1)} \overset{\frac{0}{0}}{=} \\ \\\overset{\frac{0}{0}}{=} \lim\limits_{t\to 1_+}\dfrac{\frac{1}{t^2}}{2} = \boxed{\dfrac{1}{2}}$