若 $f^{\prime}\left(x_{0}\right)=-3$, 则 $\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}-3 h\right)}{h}=(\quad)$
解析: 解答: 因为 $f^{\prime}\left(x_{0}\right)=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}=-3$, 所以

$\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}-3 h\right)}{h}=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)-\left[f\left(x_{0}-3 h\right)-f\left(x_{0}\right)\right]}{h}$

$=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}-\lim _{h \rightarrow 0} \frac{f\left(x_{0}-3 h\right)-f\left(x_{0}\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}+3 \lim _{-3 h \rightarrow 0} \frac{f\left(x_{0}-3 h\right)-f\left(x_{0}\right)}{-3 h}$

$=f^{\prime}\left(x_{0}\right)+3 f^{\prime}\left(x_{0}\right)=4 f^{\prime}\left(x_{0}\right)=-12$, 选 B;
分析: 本题主要考查了导数的概念, 解决问题的关键是根据导数的概念进行分析计算即可.