lim(x→+∞)(π/2-arctanx)/(sin1/x)

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0/0型,洛必达法则得
lim(x→+∞)(π/2-arctanx)/(sin1/x)
=lim(x→+∞)(-1/(1+x^2))/(cos1/x)(-1/x^2)
lim(x→+∞)x^2/[(1+x^2)cos(1/x)]
=1

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