当u->0时 ,(1+u)^(1/u) -> e 当x->π/2 时,令 u = sinx-1,u->0 (sinx) ^ (tanx) = (1+ sinx-1) ^ (tanx) = (1+u) ^ {(1/u) * u * tanx }lim(x->π/2) u * tanx 令 t = π/2 -x = lim(t->0) (cost - 1)/ tant = lim(t...