题目

当 $x\to0$ 时，$2\arctan x - \ln\dfrac{1+x}{1-x}$ 是 $x$ 的 $n$ 阶无穷小，求 $n$ 的值

解答

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

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

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

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

故 $n = 3$