Question

求极限:lim┬(x→1)⁡〖(tan π/4 x)^(tan π/2 x) 〗

Answer 1

方法如下，请作参考：

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Answer 2

L =lim(x->1)⁡ [ (tan (πx/4) ]^[tan(πx/2)]
两边取 ln
lnL
=lim(x->1)⁡ tan(πx/2).ln[tan(πx/4)]
=lim(x->1)⁡ ln[tan(πx/4)]/ cot(πx/2)
洛必达
=lim(x->1)⁡ { (π/4)[sec(πx/4)]^2/[tan(πx/4)] }/ { -(π/2).[csc(πx/2)]^2 }
化简
=-2.lim(x->1)⁡ [sin(πx/2)]^2 /[sin(πx/4).cos(πx/4)]
=-2. [1/(1/2)]
=-4
L = e^(-4)
得出结果
lim(x->1)⁡ [ (tan (πx/4) ]^[tan(πx/2)] =e^(-4)

Answer 3

令 y = [tan(πx/4)]^[tan(πx/2)],
则 lny = [tan(πx/2)ln[tan(πx/4)] = ln[tan(πx/4)]/[cot(πx/2)]
lim<x→1> lny = lim<x→1> ln[tan(πx/4)]/[cot(πx/2)] (0/0)
= lim<x→1> {(π/4)[sec(πx/4)]^2/tan(πx/4)}/{(-π/2)[csc(πx/2)]^2}
= lim<x→1> (-1/2)[sin(πx/2)]^2/{tan(πx/4)[cos(πx/4)]^2}
= (-1/2)/(1/2) = -1
lim<x→1> [tan(πx/4)]^[tan(πx/2)] = e^(-1) = 1/e