Đáp án:
$\lim\limits_{x\to +\infty}\left(\dfrac{x+2}{x+1}\right)^{\displaystyle{{x+3}}} = e$
Giải thích các bước giải:
$\begin{array}{l}\quad L = \lim\limits_{x\to +\infty}\left(\dfrac{x+2}{x+1}\right)^{\displaystyle{{x+3}}}\\ \to L = \lim\limits_{x\to +\infty}e^{\displaystyle{\ln\left(\dfrac{x+2}{x+1}\right)^{\displaystyle{{x+3}}}}}\\ \to L = e^{\displaystyle{\lim\limits_{x\to +\infty}\left[(x+3)\ln\left(\dfrac{x+2}{x+1}\right) \right]}}\\ Xét\,\,L' =\lim\limits_{x\to +\infty}\left[(x+3)\ln\left(\dfrac{x+2}{x+1}\right) \right]\\ \to L' =\lim\limits_{x\to +\infty}\dfrac{\ln\left(\dfrac{x+2}{x+1}\right)}{\dfrac{1}{x+3}}\\ \to L' =\lim\limits_{x\to +\infty}\dfrac{-\dfrac{1}{(x+1)^2}:\left(\dfrac{x+2}{x+1}\right)}{-\dfrac{1}{(x+3)^2}}\\ \to L' = \lim\limits_{x\to +\infty}\dfrac{(x+3)^2}{(x+1)(x+2)}\\ \to L' = \lim\limits_{x\to +\infty}\dfrac{2(x+3)}{2x+3}\\ \to L' = 2\lim\limits_{x\to +\infty}\dfrac{1 + \dfrac3x}{2 + \dfrac3x}\\ \to L' = 2\cdot\dfrac12 = 1\\ \to L = e^{\displaystyle{L'}}\\ \to L = e^1 = e \end{array}$
