L =lim(x->1-) (1-x)^tan(πx/2)lnL =lim(x->1-) ln(1-x) / tan(πx/2) ( ∞/ ∞)=lim(x->1-) [-1/(1-x)] /[ (π/2)[sec(πx/2)]^2 ] =(-2/π) lim(x->1-) [cos(πx/2)]^2/(1-x) (0/0)=(-2/π) lim(x->1-) 2(π/2)[c...可是這道題答案是0啊不好意思， 該是這樣
lim(x->1-) (1-x)^tan(πx/2)


x->1- 0<(1-x)<1
lim(x->1-) tan(πx/2) ->∞


lim(x->1-) (1-x)^tan(πx/2)=0