lim(x->0)[cosx-e^(-x^2/2)]/x^2[x+ln(1-x)咋看分母是四阶的?

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)ln(1+2x)(1-cosx)/（(e^x-1)sinx^2) 一道求极限的题 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 利用带有佩亚诺余项的麦克劳林公式求极限lim(x→0)[cosx-e^(-x^2/2)]/{x^2[x+ln(1-x)] 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