lim(x→0)(1/cosx)=?
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lim(x→0)(1/cosx)=?
lim(x→0)(1/cosx)=?
lim(x→0)(1/cosx)=?
1/cosx在x=0处连续,直接代值即可
lim(x→0)(1/cosx)=1/cos0=1
lim(x→0)(1/cosx)
=1/cos0
=1
原式=lim(sinx^2+cosx^2)/cosx
=lim(sinxtanx+cosx)
=sin0tan0+cos0
=1
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