Question

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

Answer 1

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