lim （e^x+xe^x）/（e^x-1）-1/x

Question

第三题

Answer 1

简单计算一下即可，答案如图所示

Answer 2

x->0
分母
(e^x-1) = x +o(x)
x(e^x-1) = x^2 +o(x^2)
分子
e^x = 1+x+(1/2)x^2 +o(x^2)
(-1+x+x^2).e^x
=(-1+x+x^2). [ 1+x+(1/2)x^2 +o(x^2)]
=-[ 1+x+(1/2)x^2 +o(x^2)] +x[ 1+x+(1/2)x^2 +o(x^2)] +x^2.[ 1+x+(1/2)x^2 +o(x^2)]
=[-1-x-(1/2)x^2+o(x^2)] + [x+x^2+o(x^2)] + [x^2+o(x^2)] +o(x^2)
=-1 + (3/2)x^2 +o(x^2)
(-1+x+x^2).e^x +1 = (3/2)x^2 +o(x^2)
lim(x->0) [（e^x+xe^x）/（e^x-1）-1/x]
=lim(x->0) [x(e^x+xe^x) - (e^x -1) ]/ [x(e^x-1)]
=lim(x->0) [ (-1+x+x^2).e^x +1 ]/ [x(e^x-1)]
=lim(x->0) (3/2)x^2/ x^2
=3/2本回答被提问者采纳