急.求极限 lim x趋向于π/2,（sinx）的（1/cosx）^2

设 f(x) = (sinx) ^ (1/cos²x ),ln f(x) = (1/cos²x ) ln(sinx)
lim (x->π/2) ln f(x) = lim (x->π/2) ln(sinx) / cos²x
= lim(t->0) ln(cost) / sin²t 令 t = π/2 - x,t->0
= lim(t->0) ln(1+ cost-1) / t² 等价无穷小代换 ln(1+ cost-1) cost - 1 -t² /2
= lim(t->0) (-1/2) t² / t² = -1/2
原式 = e ^ (-1/2)