怎样用导数求极限？

Question

如题所述

Answer 1

f(x) = 1/(1-x) =>f(0) = 1

f'(x) = 1/(1-x)^2 =>f'(0)/1! = 1
...
f^(n)(x) = n!/(1-x)^(n+1) =>f^(n)(0)/n! = 1
f(x) = f(0) +[f'(0)/1!]x +[f''(0)/2!]x^2+...+[f^(n)(0)/n!]x^n+...
1/(1-x) =1+x+x^2+....+x^n+....
1/(1-x) = ∑(n:0->∞) x^n
两边乘以 x

x/(1-x) = ∑(n:0->∞) x^(n+1)
= ∑(n:1->∞) x^n