Question

17题不用洛必达，用泰勒该怎么计算？

答案用的洛必达，我用泰勒该怎么计算

Answer 1

lim{x->0} [1/(2sinx)]ln[(2-x-1+x^2/2)/(1+x)]
= lim{x->0} [1/(2sinx)]ln[1 - 2x/(1+x)]
= lim{x->0} [1/(2sinx)][- 2x/(1+x)]
= lim{x->0} [1/(2sinx)][- 2x]
= -1
所以，原极限 = 1/e

Answer 2

x->0
sinx = x+o(x)

cosx = 1+o(x)

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