题目

(2020年2) 求曲线 $y = \dfrac{x^{1+x}}{(1+x)^x}(x>0)$ 的斜渐近线方程

解答

$$
k = \lim\limits_{x\to+\infty}\bigg(\dfrac{x}{1+x}\bigg)^{x} =
e^{\lim\limits_{x\to+\infty} \frac{-x}{x+1}} = e^{-1}
$$

$$
b = \lim\limits_{x\to+\infty} \Bigg[x\bigg(\dfrac{x}{1+x}\bigg)^{x} - \dfrac{x}{e}\Bigg] =
\lim\limits_{x\to+\infty} \dfrac{x}{e}\Bigg(e^{x\ln\frac{x}{1+x} + 1} - 1\Bigg) =
\lim\limits_{x\to+\infty} \dfrac{x}{e}(x\ln\frac{x}{1+x} + 1)
$$

“$\infty \cdot 0$” 型，直接倒代还即可

$$
\frac{1}{e}\lim\limits_{x\to0^+} \frac{\dfrac{\ln\dfrac{1}{1+x}}{x} + 1}{x} =
\frac{1}{e}\lim\limits_{x\to0^+} \dfrac{\ln 1 - \ln (1+x) + x}{x^2} =
\frac{1}{e}\lim\limits_{x\to0^+} \dfrac{x - \ln (1+x)}{x^2} = \frac{1}{2e}
$$

故该 斜渐近线 为 $y = e^{-1}x + (2e)^{-1}$