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$I=\lim _{x \rightarrow 0} \frac{1}{x^2}\left\{\ln \left(1+2 x-x^2\right)-6\left[(1+x)^{\frac{1}{3}}-1\right]\right\}=$

答案：

$-\frac{7}{3} $

解析：

方法 1 (泰勒公式) 由于
$$
\begin{aligned}
\ln \left(1+2 x-x^2\right) & =\left(2 x-x^2\right)-\frac{1}{2}\left(2 x-x^2\right)^2+o\left(\left(2 x-x^2\right)^2\right) \\
& =2 x-x^2-\frac{1}{2}(2 x)^2+o\left(x^2\right)=2 x-3 x^2+o\left(x^2\right), \\
(1+x)^{\frac{1}{3}}-1 & =\frac{1}{3} x+\frac{1}{2} \cdot \frac{1}{3}\left(\frac{1}{3}-1\right) x^2+o\left(x^2\right)=\frac{1}{3} x-\frac{1}{9} x^2+o\left(x^2\right) .
\end{aligned}
$$
于是
$$
\begin{aligned}
I & =\lim _{x \rightarrow 0} \frac{1}{x^2}\left[\left(2 x-3 x^2\right)-6\left(\frac{1}{3} x-\frac{1}{9} x^2\right)+o\left(x^2\right)\right] \\
& =\lim _{x \rightarrow 0} \frac{1}{x^2}\left(-3+\frac{2}{3}\right) x^2=-\frac{7}{3} .
\end{aligned}
$$

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