题目

当 $x\to 0^+$ 时，$(1+x)^{\frac{1}{x}} - (e+ax+bx^2)$ 是比 $x^2$ 高阶的无穷小，求 $a,b$ 的值

解答

简单推导：

$$
\ln(1 + x) - x = -\dfrac{1}{2}x^2 + \dfrac{1}{3}x^3 - \dfrac{1}{4}x^4 + o(x^4)
$$

$$
\dfrac{\ln(1 + x) - x}{x} = -\dfrac{1}{2}x + \dfrac{1}{3}x^2 - \dfrac{1}{4}x^3 + o(x^3)
$$

$$
e^x - 1 = x + \dfrac{1}{2}x^2 + \dfrac{1}{6}x^3 + o(x^3)
$$

$$
e^{\dfrac{\ln(1 + x) - x}{x}} - 1 = (-\dfrac{1}{2})x + (\dfrac{1}{3} + \dfrac{1}{8})x^2 + o(x^2) = -\dfrac{1}{2}x + \dfrac{11}{24}x^2 + o(x^2)
$$

由于 $e \cdot [e^{\dfrac{\ln(1 + x) - x}{x}} - 1] - ax - bx^2$ 是 $o(x^2)$

故 $a = -\dfrac{1}{2}e$，$b = \dfrac{11}{24}e$