Question

求助一道 极限题，各路高人帮帮忙啊～谢谢了

Answer 1

解：原式=lim(x->π/4){[[1+(tanx-1)]^[1/(tanx-1)]]^[(tanx-1)tan(2x)]}
={lim(x->π/4)[[1+(tanx-1)]^[1/(tanx-1)]]}^{lim(x->π/4)[(tanx-1)tan(2x)]}
(应用初等函数的连续性)
=e^{lim(x->π/4)[(tanx-1)tan(2x)]} (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=e^{lim(x->π/4)[2tanx*(tanx-1)/(1-(tanx)^2)]} (应用正切倍角公式)
=e^{lim(x->π/4)[-2tanx/(1+tanx)]}
(应用两数平方差公式，再分子分母同除(1-tanx))
=e^[-2tan(π/4)/(1+tan(π/4))]
=e^[-2*1/(1+1)]
=e^(-1)
=1/e。