lim(x→∞)(cos1/x)^x
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lim(x→∞)(cos1/x)^xlim(x→∞)(cos1/x)^xlim(x→∞)(cos1/x)^x令y=(cos1/x)^xlny=xlncos(1/x)则limlny=limlncos(1
lim(x→∞)(cos1/x)^x
lim(x→∞)(cos1/x)^x
lim(x→∞)(cos1/x)^x
令y=(cos1/x)^x
lny=xlncos(1/x)
则limlny
=limlncos(1/x)/(1/x)
是∞/∞型,用洛必达法则
=lim[1/cos(1/x)*(-sin1/x)*(1/x)']/(1/x)'
=lim[-tan(1/x)]
=0
所以原式=e^0=1
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