Question

极限limx→0(sec^2x-sin^2x)^1/ln(1+x^4)？

Answer 1

简单分析一下，详情如图所示

Answer 2

x->0
sinx= x-(1/6)x^3 +o(x^3)
(sinx)^2= [x-(1/6)x^3 +o(x^3)]^2 = x- (1/3)x^4 +o(x^4)
sin2x= 2x-(3/2)x^3 +o(x^3)
(sin2x)^2= [2x-(3/2)x^3 +o(x^3)]^2 = 4x^2- 6x^4 +o(x^4)
(1/4)(sin2x)^2= x^2- (3/2)x^4 +o(x^4)
(sinx)^2-(1/4)(sin2x)^2= x^4 +o(x^4)
//
(secx)^2 - (sinx)^2
=[1 - (sinx)^2.(cosx)^2]/(cosx)^2
=[1 - (1/4)(sin2x)^2]/(cosx)^2
= 1+ [(sinx)^2 - (1/4)(sin2x)^2]/(cosx)^2
~1 +[(sinx)^2 - (1/4)(sin2x)^2]
~ 1+x^4
//
lim(x->0) [ (secx)^2-(sinx)^2]^[1/ln(1+x^4)]
=lim(x->0) [ (secx)^2-(sinx)^2]^(1/x^4)
=lim(x->0) [ 1+x^4]^(1/x^4)
=e