Question

这道极限怎么做呢

具体步骤，谢谢

Answer 1

Answer 2

x->0

e^x -1 = x +(1/2)x^2 +(1/6)x^3 +(1/24)x^4 +o(x^4)

sin(e^x -1)

=sin[x +(1/2)x^2 +(1/6)x^3 +(1/24)x^4 +o(x^4) ]

=[x +(1/2)x^2 +(1/6)x^3 +(1/24)x^4 +o(x^4) ] 

-(1/6)[x +(1/2)x^2 +(1/6)x^3 +(1/24)x^4 +o(x^4) ]^3 + o(x^4)

=[x +(1/2)x^2 +(1/6)x^3 +(1/24)x^4 +o(x^4) ] 

-(1/6)[x^3 +(3/2)x^4 +o(x^4) ] + o(x^4)

=x +(1/2)x^2 - (5/24)x^4 +o(x^4)

sinx = x-(1/6)x^3 + o(x^4)

e^(sinx) -1

=e^[x-(1/6)x^3 + o(x^4)] -1

=[x-(1/6)x^3] +(1/2)[x-(1/6)x^3]^2 +(1/6)[x-(1/6)x^3]^3 

+(1/24)[x-(1/6)x^3]^4+o(x^4)

= [x-(1/6)x^3] +(1/2)[x^2-(1/3)x^4+o(x^4)] +(1/6)[x^3 +o(x^4)]

+ (1/24)[x^4+o(x^4)]+o(x^4)

=x +(1/2)x^2  -(1/8)x^4 +o(x^4)

 sin(e^x -1) -[ e^(sinx) -1 ]

=[ x +(1/2)x^2 - (5/24)x^4 +o(x^4) ] -[x +(1/2)x^2  -(1/8)x^4 +o(x^4)]

=-(1/12)x^4 +o(x^4)

lim(x->0) { sin(e^x -1) -[ e^(sinx) -1 ] }/x^4

=lim(x->0) -(1/12)x^4 /x^4

=-1/12

Answer 3

这个题目方法有几个，我是用凑第二个重要极限

追问

能不能写一下具体步骤，谢谢啦