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+...