微积分数学分析求解[ln(1+2x+x∧2)+ln(1-2x+x∧2 )]/(sec x-cosx)
lim(x→0)(ln(1+x^2)/(sec-cosx))
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
微积分 ∫ 1/(x ln^2 x )dx
lim(x→0)(cosx)^(1/ln(1+x^2))
lim[ln(1+x^2)]/(secx-cosx) x->0
lim(x->0)[cosx-e^(-x^2/2)]/[x^2[x+ln(1-x)]]
【急】大一微积分lim[x-x^2ln(1+1/x)] x趋近于无穷
求导y=ln ln ln(x^2+1)
lim(x→0) (ln cosx)/[ln(1+x^2)] 等于多少?
limx->∞ ln(1+3x∧2)/ln(3+x∧4)极限求解
lim(x→0)(x^2+cosx-2)/(x^3)*ln(1+x)怎么算