求了limit(x→2/π) (sinx-1)tanx
问题描述：
求了limit(x→2/π) (sinx-1)tanx
答
(sinx-1)*tanx [x-->π/2]
==>[sin(x+π/2)-1]*tan(x+π/2) [x-->0]
即,[cosx-1]*(-cotx)[x-->0]
=-1/2x^2(-1/x)
=1/2x
=0.