= lim(x->；1)e^[cos(πx/2)ln(1-x)]
=e^ { lim(x-& gt；1)[ln(1-x)/sec(πx/2)]]
=e^ { lim(x-& gt；1)[2cos^2(πx/2)/π(x-1)sin(πx/2]
=e^ { 1/π* lim(x->；1)[(1+cos(πx))/(x-1)sin(πx/2)]]
=e^ { lim(x-& gt；1)[-sin(πx)/[sin(πx/2)+π/2 *(x-1)cos(πx/2)]
=e^0
=1