题目

若极限lim _(harrow 0)dfrac (f({x)_(0)+2h)-f((x)_(0))}(h)=dfrac (1)(2)，则导数值lim _(harrow 0)dfrac (f({x)_(0)+2h)-f((x)_(0))}(h)=dfrac (1)(2)lim _(harrow 0)dfrac (f({x)_(0)+2h)-f((x)_(0))}(h)=dfrac (1)(2)

若极限 $\lim_{h\rightarrow0}\frac{f(x_0+2h)-f(x_0)}{h}=\frac{1}{2}$ ，则导数值 $f'(x_0)=\left(\right)$

$A.-\frac{1}{4}\ \ \ B.\frac{1}{4}\ \ \ C.\frac{1}{2}\ \ \ D.\frac{1}{3}$

题目解答

答案

由题意得，已知极限 $\lim_{h\rightarrow0}\frac{f(x_0+2h)-f(x_0)}{h}=\frac{1}{2}$ ，则有 $\lim_{h\rightarrow0}\frac{f(x_0+2h)-f(x_0)}{h}$ $=2\lim_{h\rightarrow0}\frac{f(x_0+2h)-f(x_0)}{2h}=2f'(x_0)=\frac{1}{2}$ ，所以 $f'(x_0)=\frac{1}{4}$ ，因此本题答案选 $B$ 。