设f(0)=0,f'(x)在x=0的领域内连续,又f'(x)≠0证明:lim(x趋向0)x^f(x)=1
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设f(0)=0,f'(x)在x=0的领域内连续,又f'(x)≠0证明:lim(x趋向0)x^f(x)=1
f'(0)=lim[f(x)-f(0)]/x
lim(x趋向0)x^f(x)
=e^[lim(x趋向0)f(x)lnx]
=e^[lim(x趋向0)lnx/(1/f(x))]
=e^[lim(x趋向0)1/x/(-f'(x)/f^2(x))]
=e^[-lim(x趋向0)f^2(x)/(xf'(x))]
=e^[-1/f'(0)lim(x趋向0)f^2(x)/(x)]
=e^[-1/f'(0)lim(x趋向0)2f(x)*f'(x)/(1)]
=e^(-f(0)*f'(0)/f'(0))
=e^0
=1
lim(x趋向0)x^f(x)
=e^[lim(x趋向0)f(x)lnx]
=e^[lim(x趋向0)lnx/(1/f(x))]
=e^[lim(x趋向0)1/x/(-f'(x)/f^2(x))]
=e^[-lim(x趋向0)f^2(x)/(xf'(x))]
=e^[-1/f'(0)lim(x趋向0)f^2(x)/(x)]
=e^[-1/f'(0)lim(x趋向0)2f(x)*f'(x)/(1)]
=e^(-f(0)*f'(0)/f'(0))
=e^0
=1
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