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求极限:lim<x→π/2>tanx/tan3x

因为当x→π/2，tanx和tan3x均→+∞，所以这个满足应用罗必塔法则的条件，则：
lim<x→π/2>tanx/tan3x
=lim<x→π/2>(tanx)'/(tan3x)'
=lim<x→π/2>sec^x/(3*sec^3x)
=lim<x→π/2>(1/3)*[1/cos^x]/[1/cos^3x]
=lim<x→π/2>(1/3)*[cos^3x/cos^x]
=lim<x→π/2>(1/3)(cos^3x)'/(cos^x)'
=lim<x→π/2>(1/3)[2cos3x*(-sin3x)*3]/[2cosx*(-sinx)]
=lim<x→π/2>(sin6x)/sin2x
=lim<x→π/2>(sin6x)'/(sin2x)'
=lim<x→π/2>(6cos6x)/(2cos2x)
=3