limx→π/2:(cosx)^(π/2-x)

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limx→π/2:(cosx)^(π/2-x)
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limx→π/2:(cosx)^(π/2-x)
limx→π/2:(cosx)^(π/2-x)

limx→π/2:(cosx)^(π/2-x)
y=(cosx)^(π/2-x)
取对数:lny=(π/2-x)lncosx
lny=lncosx/[1/(π/2-x)]
应用罗必达法则求极限:
lny=lim (-sinx/cosx)/[1/(π/2-x)^2]
=lim -tanx (π/2-x)^2
=lim -(π/2-x)^2/cosx
=lim -(π/2-x)/sin(π/2-x) * (π/2-x)
=lim -(π/2-x)
=0
因此y=1
即原式极限=1