limx→1(1-x)^(cosxπ/2)求极限lim(2/π arctanx)^x 其中x趋向于正无穷大

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lim (1-x)^(cosxπ/2)
x→1
=lim (1-x)^[sin(1-x)π/2] (令y=1-x)
x→1
=lim y^[sin(yπ/2)]
y→0
=lim e^{[sin(yπ/2)]lny}
y→0
=lim e^(yπ/2*lny)
y→0
=lim e^[π/2*lny/(1/y)]
y→0
=[e^(π/2)]^lim lny/(1/y) （∞/∞,罗比达法则）
y→0
=[e^(π/2)]^lim 1/y/(-1/y^2)
y→0
=[e^(π/2)]^lim (-y)
y→0
=[e^(π/2)]^0
=1
lim (2/π arctanx)^x
x→+∞
=lim e^[x*ln(2/π arctanx)]
x→+∞
=e^{lim [ln(2/π arctanx)/(1/x)]} （0/0,罗比达法则）
x→+∞
=lim e^[1/(2/π arctanx)*2/π*1/(1+x^2)/(-1/x^2)]
x→+∞
=lim e^[1/(2/π arctanx)*2/π*(-x^2)/(1+x^2)]
x→+∞
=e^[1/(π/2)*(-1)]
=e^(-2/π)

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