\[\underset{x\rightarrow x_0}{\lim}{\frac{f(x)-f(x_0)}{(x-x_0)^2}}
= \underset{x\rightarrow x_0}{\lim}{\frac{\frac{f(x)-f(x_0)}{(x-x_0)}}{(x-x_0)}}
= \frac{f^{'}(x_0)}{\underset{x\rightarrow x_0}{\lim}{(x-x_0)}}=-1\\
由于二阶可导所以一阶导数连续，由\underset{x\rightarrow x_0}{\lim}{(x-x_0)}=0，所以有f^{'}(x_0)=\underset{x\rightarrow x_0}{\lim}f^{'}(x)=0\\
所以\frac{f^{'}(x_0)}{\underset{x\rightarrow x_0}{\lim}{(x-x_0)}}
= \underset{x\rightarrow x_0}{\lim}\frac{f^{'}(x)}{(x-x_0)}
= \underset{x\rightarrow x_0}{\lim}\frac{f^{'}(x)-f^{'}(x_0)}{(x-x_0)}=f^{''}(x_0)=-1
\]
不知道错在哪儿了🤒。