设函数f(x)f(x)f(x)在点x=x0x = x_{0}x=x0处可导，且f′(x0)=1f^{\prime}(x_{0}) = 1f′(x0)=1，则limh→0f(x0−2h)−f(x0+h)h=\lim\limits_{h \to 0}\frac{f(x_{0}-2h)-f(x_{0}+h)}{h}=h→0limhf(x0−2h)−f(x0+h)=（ ）.
（A）−2-2−2
（B）222
（C）−3-3−3
（D）333
Copyright ©
