Question

高数求极限这步怎么来的

这一步是怎么来的

Answer 1

lim<x→0>ln[(2+cosx)/3]/x^2 = lim<x→0>ln[(3+2+cosx-3)/3]/x^2
= lim<x→0>ln[1+(2+cosx-3)/3]/x^2 分子等价无穷小代换
= lim<x→0>[(2+cosx-3)/3]/x^2
= lim<x→0>(2+cosx-3)/(3x^2) = lim<x→0>(cosx-1)/(3x^2)
= lim<x→0>(-x^2/2)/(3x^2) = -1/6

Answer 2

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