
高数!求详细过程!
2020-01-02 · 知道合伙人教育行家
关注

展开全部
(x→+∞) lim (π/2-arctanx)^(1/lnx)
令t=arctanx,则x=tant
(t→π/2) lim (π/2-t)^(1/lntant)
= e^(t→π/2) lim ln(π/2-t) / lntant ....[0/0]
= e^(t→π/2) lim [-1/(π/2-t)] / [1/tant * 1/cos²t]
= e^(t→π/2) lim [-1/(π/2-t)] / [1/(sintcost)]
= e^(t→π/2) lim (1/2)[sin2t/(t-π/2)] ....[0/0]
= e^(t→π/2) lim (1/2) [2cos2t / 1]
= e^(t→π/2) lim cos2t
= e^(-1)
= 1/e
令t=arctanx,则x=tant
(t→π/2) lim (π/2-t)^(1/lntant)
= e^(t→π/2) lim ln(π/2-t) / lntant ....[0/0]
= e^(t→π/2) lim [-1/(π/2-t)] / [1/tant * 1/cos²t]
= e^(t→π/2) lim [-1/(π/2-t)] / [1/(sintcost)]
= e^(t→π/2) lim (1/2)[sin2t/(t-π/2)] ....[0/0]
= e^(t→π/2) lim (1/2) [2cos2t / 1]
= e^(t→π/2) lim cos2t
= e^(-1)
= 1/e
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询