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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优质解答：

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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