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)]=](/uploads/image/z/17619962-50-2.jpg?t=%E8%AE%BEf%28x%29%E6%98%AF%E5%8F%AF%E5%AF%BC%E5%87%BD%E6%95%B0%2C%E4%B8%94lim+f%27%28x%29%3D5%2C%E5%88%99lim%5Bf%28x%2B2%29-f%28x%29%5D%3D)
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
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
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