Question

Salut,mă puteți ajuta La problema 335 ?

Answer 1

$l=\lim\limits_{x\to \infty}\Bigg(x\Big(1+\dfrac{1}{x}\Big)^x-ex\Bigg) =\\ \\ = \lim\limits_{x\to \infty}\dfrac{\Big(1+\dfrac{1}{x}\Big)^x-e}{\dfrac{1}{x}} \\ \\ \dfrac{1}{x} = t \Rightarrow t\to 0 \\ \\ l=\lim\limits_{t\to 0}\dfrac{(1+t)^{\frac{1}{t}}-e}{t} \quad(L'H) \\ \\ \\y = (1+t)^{\frac{1}{t}} \\\\ \ln y = \dfrac{\ln(1+t)}{t}\Big|' \\ \\ \dfrac{y'}{y} =\dfrac{\dfrac{t}{1+t}-\ln(1+t)}{t^2} \\ \\ y' =(1+t)^{\frac{1}{t}}\cdot \dfrac{\dfrac{t}{1+t}-\ln(1+t)}{t^2}$

$\Rightarrow l = \lim\limits_{t\to 0}\dfrac{y'-0}{1} = \lim\limits_{t\to 0}\Bigg[(1+t)^{\frac{1}{t}}\cdot \dfrac{\dfrac{t}{1+t}-\ln(1+t)}{t^2}\Bigg]\\ \\\\ (1+t)^{\frac{1}{t}} \to e\\ \\ \Rightarrow l = e\cdot\lim\limits_{t\to 0}\dfrac{\dfrac{t}{1+t}-\ln(1+t)}{t^2}$

$\lim\limits_{t\to 0}\dfrac{\dfrac{t}{1+t}-\ln(1+t)}{t^2} =\lim\limits_{t\to 0}\dfrac{1-\dfrac{1}{1+t}-\ln(1+t)}{t^2}=\\ \\ = \lim\limits_{t\to 0} \dfrac{\dfrac{1}{(1+t)^2}-\dfrac{1}{1+t}}{2t} =\lim\limits_{t\to 0} \dfrac{-\dfrac{2}{(1+t)^3}+\dfrac{1}{(1+t)^2}}{2} = -\dfrac{1}{2} \\ \\ \Rightarrow l = e\cdot \Big(-\dfrac{1}{2}\Big) \Rightarrow \boxed{l = -\dfrac{e}{2} }$