题目太多，1同阶无穷小，2secx-1=(1-cosx)/cosx=2[sin(x/2)]^2/cosx~(sinx)/2高阶无穷小，4=lim[(1-x)sin(πx/2)/cos(πx/2)]=lim[(1-x)/cos(πx/2)]=lim{-1/[-(π/2)sin(πx/2)]} 洛必达法则，=2/π
