设f(x)是可导函数,且lim f'(x)=5,则lim[f(x+2)-f(x)]=

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设f(x)是可导函数,且limf''(x)=5,则lim[f(x+2)-f(x)]=设f(x)是可导函数,且limf''(x)=5,则lim[f(x+2)-f(x)]=设f(x)是可导函数,且limf''(

设f(x)是可导函数,且lim f'(x)=5,则lim[f(x+2)-f(x)]=
设f(x)是可导函数,且lim f'(x)=5,则lim[f(x+2)-f(x)]=

设f(x)是可导函数,且lim f'(x)=5,则lim[f(x+2)-f(x)]=
f(x+2) - f(x) = f(x+2) - f(x + 2 - 2/x) + f(x + 2 - 2/x) - f(x + 2 - 4/x) + ...+ f(x + 4/x) - f(x + 2/x) + f(x + 2/x) - f(x) ([x ,x+2]按照2/x的长度分成x份)
而对于每个f(t + 2/x) - f(t),当t趋向于无穷时,有( f(t + 2/x) - f(t) ) / (2/x) = 5 ,这是由题意给的导数知道的(因为2/x趋向于无穷小,满足导数的定义),故每个f(t + 2/x) - f(t) = 5 * (2/x)
前面说过一共分成了x份,故f(x+2) - f(x) = 5 * (2/x) * x = 10

x趋近于多少?

lim(h→0) f(3)-f(3+h)/2h
=0.5lim(h→0) f(3)-f(3+h)/h (导数定义)
=0.5*[-f'(3)]
=5
所以f'(3)=-10

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