设 $f(x)$ 在 $x_0$ 处可导, 且 $\boldsymbol{a b} \neq 0$, 证明: $\lim _{h \rightarrow 0} \frac{f\left(x_0+\boldsymbol{a}\right)-f\left(x_0-b \boldsymbol{h}\right)}{\boldsymbol{h}}=(\boldsymbol{a}+\boldsymbol{b}) \boldsymbol{f}^{\prime}\left(\boldsymbol{x}_0\right)$.
【答案】 证明 因为 $a b \neq 0$, 因此我们有
$$
\begin{aligned}
& \lim _{h \rightarrow 0} \frac{f\left(x_0+a h\right)-f\left(x_0-b h\right)}{h} \\
= & \lim _{h \rightarrow 0} \frac{\left[f\left(x_0+a h\right)-f\left(x_0\right)\right]+\left[f\left(x_0\right)-f\left(x_0-b h\right)\right]}{h} \\
= & a \lim _{h \rightarrow 0} \frac{f\left(x_0+a h\right)-f\left(x_0\right)}{a h}+b \lim _{h \rightarrow 0} \frac{f\left(x_0-b h\right)-f\left(x_0\right)}{-b h} \\
= & a f^{\prime}\left(x_0\right)+b f^{\prime}\left(x_0\right) \\
= & (a+b) f^{\prime}\left(x_0\right)
\end{aligned}
$$


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