∵当x->1时,y->0
∴lim(x->1)[(x-1)tan(πx/2)]
=lim(y->0)[y*tan(π/2+πy/2)]
=lim(y->0)[-y*ctan(πy/2)]
=(-2/π)lim(y->0){[(πy/2)/sin(πy/2)]*[cos(πy/2)]}
=(-2/π){lim(y->0)[(πy/2)/sin(πy/2)]}*{lim(y->0)[cos(πy/2)]}
=(-2/π)*1*1 (应用重要极限lim(x->0)(sinx/x)=1)
=-2/π.