“导数定义应用”练习题

2019-03-17 本文已影响78人 7300T

已知 $f^{\prime}\left(x_{0}\right)=a$ 求下列极限：
1. $\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+\Delta x\right)-f\left(x_{0}-\Delta x\right)}{\Delta x}$

$\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h^{2}\right)-f\left(x_{0}\right)}{h}$

解：

$\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+\Delta x\right)-f\left(x_{0}-\Delta x\right)}{\Delta x}$

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

$=\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+\Delta x\right)-f\left(x_{0}\right)}{\Delta x}+\lim _{-\Delta r \rightarrow 0} \frac{f\left(x_{0}-\Delta x\right)-f\left(x_{0}\right)}{-\Delta x}$

$=2 f^{\prime}\left(x_{0}\right)$
$=2 a$

$\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h^{2}\right)-f\left(x_{0}\right)}{h}$

$=\lim _{h \rightarrow 0}\left[\frac{f\left(x_{0}+h^{2}\right)-f\left(x_{0}\right)}{h^{2}} \cdot h\right]$

$=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h^{2}\right)-f\left(x_{0}\right)}{h^{2}} \cdot \lim _{h \rightarrow 0} h$

$=a \cdot 0$

“导数定义应用”练习题

猜你喜欢

热点阅读