lim(x→π/2) secx-tanx
=lim(x→π/2) 1/cosx-sinx/cosx
=lim(x→π/2) (1-sinx)/cosx
=lim(x→π/2) [(1-sinx)(1+sinx)]/[cosx(1+sinx)]
=lim(x→π/2) (1-sin²x)/[cosx(1+sinx)]
=lim(x→π/2) cosx/(1+sinx)
=cos(π/2)/[1+sin(π/2)]
=0/(1+1)
=0
求极限.lim(x→pi/2)(secx-tanx)
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