Question

求解答极限

Answer 1

追问

分母1+sinx怎么消掉的

追答

1+sinx的极限是1，直接求出来了，它是因子，所以它的极限可以求出来

追问

兄弟啊，有没有QQ微信什么的，以后有问题可以直接找你，谢谢

追答

有问题可以私信

Answer 2

x->0
tanx = x+(1/3)x^3 +o(x^3)
ln(1+tanx)

=ln[1+x+(1/3)x^3 +o(x^3) ]
=[x+(1/3)x^3] - (1/2)[x+(1/3)x^3]^2 + (1/3)[x+(1/3)x^3]^3 +o(x^3)
=[x+(1/3)x^3] - (1/2)[x^2+o(x^3)] + (1/3)[x^3+o(x^3)] +o(x^3)
=x - (1/2)x^2 + [ 1/3 +1/3 ]x^3 +o(x^3)
=x - (1/2)x^2 + (2/3 )x^3 +o(x^3)
sinx = x-(1/6)x^3 +o(x^3)
ln(1+sinx)
=ln[1 +x-(1/6)x^3 +o(x^3) ]
=[x-(1/6)x^3] - (1/2)[x-(1/6)x^3]^2 + (1/3)[x-(1/6)x^3]^3 +o(x^3)
=[x-(1/6)x^3] - (1/2)[x^2+o(x^3)] + (1/3)[x^3+o(x^3)] +o(x^3)
=x -(1/2)x^2 +( -1/6 +1/3) x^3 +o(x^3)
=x -(1/2)x^2 +(1/6) x^3 +o(x^3)
ln(1+tanx) -ln(1+sinx)

=[x - (1/2)x^2 + (2/3 )x^3 +o(x^3) ] -[x -(1/2)x^2 +(1/6) x^3 +o(x^3)]
= (1/2)x^3 +o(x^3)
lim(x->0) [(1+tanx)/(1+sinx) ]^(1/3)

=lim(x->0) e^{ ln[(1+tanx)/(1+sinx) ] /x^3 }
=lim(x->0) e^{ [ ln(1+tanx)-ln(1+sinx) ] /x^3 }
=lim(x->0) e^[ (1/2)x^3/x^3 ]
=e^(1/2)