如题所述
lim(x→0+)x^f(x)
=e^[lim(x→0+)f(x)lnx]
=e^[lim(x→0+)lnx/(1/f(x))]
=e^[lim(x→0+)1/x/(-f'(x)/f²(x))] (洛必达法则)
=e^[-lim(x→0+)f²(x)/(xf'(x))]
=e^[-1/f'(0)lim(x→0+)f²(x)/(x)]
=e^[-1/f'(0)lim(x→0+)2f(x)·f'(x)/(1)](再次洛必达法则)
=e^(-f(0)·f'(0)/f'(0)) (f'(0)>0)
=e^0
=1
=e^[lim(x→0+)f(x)lnx]
=e^[lim(x→0+)lnx/(1/f(x))]
=e^[lim(x→0+)1/x/(-f'(x)/f²(x))] (洛必达法则)
=e^[-lim(x→0+)f²(x)/(xf'(x))]
=e^[-1/f'(0)lim(x→0+)f²(x)/(x)]
=e^[-1/f'(0)lim(x→0+)2f(x)·f'(x)/(1)](再次洛必达法则)
=e^(-f(0)·f'(0)/f'(0)) (f'(0)>0)
=e^0
=1
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