Question

Tính giới hạn hàm số :

\(\lim\limits_{x\rightarrow+\infty}\left(\frac{x}{1+x}\right)^x\)

Answer 1

\(L=\lim\limits_{x\rightarrow+\infty}\left(\frac{x}{1+x}\right)^x\)

Ta có : \(L=\lim\limits_{x\rightarrow+\infty}\left(\frac{x}{1+x}\right)^x=\lim\limits_{x\rightarrow+\infty}\left(1-\frac{1}{1+x}\right)^x\)

Đặt \(-\frac{1}{1+x}=\frac{1}{t}\Rightarrow\begin{cases}x=-\left(1+t\right)\\x\rightarrow+\infty;t\rightarrow-\infty\end{cases}\)

\(\Rightarrow L=\lim\limits_{t\rightarrow-\infty}\left(1+\frac{1}{t}\right)^{-\left(1+t\right)}=\lim\limits_{t\rightarrow-\infty}\frac{1}{\left(1+\frac{1}{t}\right)^{1+t}}=\lim\limits_{t\rightarrow-\infty}\frac{1}{\left(1+\frac{1}{t}\right)\left(1+\frac{1}{t}\right)^t}=\frac{1}{1.e}=\frac{1}{e}\)