求limx趋向0ln(1+x^2)⼀secx-cosx

2025-06-26 20:56:47
推荐回答(1个)
回答1:

lim(x->0) ln(1+x^2)/(secx-cosx) (0/0)
=lim(x->0) [2x/(1+x^2)]/(secx.tanx +sinx)
=lim(x->0) 2/[(secx. +1)(1+x^2)]
=1