Question

求这道题极限解法

第22题，e^x-1不是可以等于x本身吗，这题可以多少种解法？

Answer 1

e^x-1在一级近似时，是约等于x，本题中有必要取到更高x的更高幂次的项。
e^x≈1+x+x^2/2
所以，原极限≈lim(x->0)[1/x - 1/(x+x^2/2)]
=lim(x->0)[ x^2/(x(x+x^2/2)) ]
=lim(x->0)[(1/2)/(1+x/2)]
=1/2

Answer 2

x->0

e^x = 1+ x+ (1/2)x^2 +o(x^2)

e^x -1 +x =(1/2)x^2 +o(x^2)

lim(x->0) [ 1/x -1/(e^x-1) ]

=lim(x->0) [e^x -1 +x]/[x(e^x-1) ]

=lim(x->0) [e^x -1 +x]/x^2

=lim(x->0) (1/2)x^2/x^2

=1/2