洛必达法则解决.
lim(n->∞) [tan(π/4 + 2/n)]^n
= e^lim(n->∞) n ln[tan(π/4 + 2/n)]
令t = 1/n,t->0,则
= e^lim(t->0) ln[tan(π/4 + 2t)]/t
= e^lim(t->0) [1/tan(π/4 + 2t)] * sec²(π/4 + 2t) * 2
洛必达法则解决.
lim(n->∞) [tan(π/4 + 2/n)]^n
= e^lim(n->∞) n ln[tan(π/4 + 2/n)]
令t = 1/n,t->0,则
= e^lim(t->0) ln[tan(π/4 + 2t)]/t
= e^lim(t->0) [1/tan(π/4 + 2t)] * sec²(π/4 + 2t) * 2