Question

求函数的极限 （x→π/4）lim(tanx)^tan2x

Answer 1

x→π/4
lim (tanx)^tan2x
换元t=π/4-x,x=π/4-t
=lim (tan(π/4-t))^tan(π/2-2t)
=lim (tan(π/4-t))^cot2t
=lim (tan(π/4-t))^(1/tan2t)
=lim e^ln(tan(π/4-t))^(1/tan2t)
=e^lim ln(tan(π/4-t))^(1/tan2t)
现在考虑
lim ln(tan(π/4-t))^(1/tan2t)
=lim ln(tan(π/4-t)) / tan2t
=lim ln((1-tant)/(1+tanx)) / tan2t
=lim ln(1-tant)/tan2t - lim ln(1+tanx)/tan2t
因为ln(1-tant)~-tant~-t,ln(1+tant)~tant~t,tan2t~2t
=lim -t/2t - lim t/2t
=-1/2-1/2
=-1
因此,原极限
=e^(-1)
有不懂欢迎追问