题目

$$
\lim_{x \to 0} \frac{\ln(1+\sin^2 x) - 6(\sqrt[3]{2-\cos x} - 1)}{x^4}
$$

解答

复合函数处理方法：

  1. 强行泰勒展开（多项式计算量大）
  2. 添项减项（精度随缘）

$$
\ln(1 + \sin^2 x) - \sin ^2x \sim -\dfrac{1}{2}\sin^4x \sim -\dfrac{1}{2}x^4
$$

$$
(1 + x)^{\frac{1}{3}} - 1 - x \sim -\dfrac{1}{9} x^2
$$

$$
[1 + (1 - \cos x)]^{\frac{1}{3}} - 1 - \dfrac{1}{3}(1 - \cos x)\sim
-\dfrac{1}{9} (1 - \cos x)^2 \sim
-\dfrac{1}{36} x^4
$$

$$
\begin{aligned}
&\lim_{x \to 0} \frac{\ln(1+\sin^2 x) - 6(\sqrt[3]{2-\cos x} - 1)}{x^4}
\\
=&
\lim_{x \to 0} \frac{\ln(1+\sin^2 x) - \sin^2x + 2(1-\cos x)- 6(\sqrt[3]{2-\cos x} - 1)+\sin^2x - 2(1 - \cos x)
}{x^4}
\end{aligned}
$$

$$
\dfrac{\sin^2x - 2 + 2\cos x}{x^4} =
\dfrac{\cos x - 1}{2x^2} = -\dfrac{1}{4}
$$

$$
\begin{aligned}
&
\lim_{x \to 0} \frac{\ln(1+\sin^2 x) - \sin^2x}{x^4} -
\lim_{x \to 0} \frac{-2(1-\cos x) + 6(\sqrt[3]{2-\cos x} - 1)}{x^4}
\\
+&
\lim_{x \to 0} \frac{\sin^2x - 2(1 - \cos x)}{x^4}
\\
= & -\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{4} = -\dfrac{7}{12}
\end{aligned}
$$