lim(x->0)[cosx-e^(-x^2/2)]/[x^2[x+ln(1-x)]]

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/04/28 03:42:07

lim(x->0)[cosx-e^(-x^2/2)]/[x^2[x+ln(1-x)]]
为什么分母是四阶啊?怎么看的?
为什么ln（1-x）这一项要算到2阶啊

原式=
lim{x->0}[1-x^2/2+x^4/24+o(x^4)-(1-x^2/2+x^4/8+o(x^4))]/[x^2(x-x+x^2/2+o(x^2)]
=lim{x->0}[-x^4/12+o(x^4)]/[x^4/2+o(x^4)]
=lim{x->0}[-/12+o(1)]/[1/2+o(1)]
=-1/6
因为分子是4阶无穷小,分母是高于3阶的无穷小,所以ln(1-x)至少要展开到平方项啦.

lim(x->0)[cosx-e^(-x^2/2)]/[x^2[x+ln(1-x)]] lim(x→0)(cosx)^(1/ln(1+x^2)) lim(x→0)(ln(1+x^2)/(sec-cosx)) lim[ln(1+x^2)]/(secx-cosx) x->0 lim(x->0)ln(1+2x)/e^x-1 lim(x→0)(x^2+cosx-2)/(x^3)*ln(1+x)怎么算 lim x趋于0 (sinx+x^2sin1/x)/[(1+cosx)ln(1+x)] lim x→0[ln(cosx)]/x^2 1、lim ln(1+x+2x^2)+ln(1-x+x^2)/secx-cosx 求 lim ln(1+x+2x^2)+ln(1-x+x^2)/secx-cosx lim e^2x-1/ln(1-X) 0 lim(x→0)ln(1+2x)(1-cosx)/（(e^x-1)sinx^2) 一道求极限的题