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

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

x＾f(x)化为e^f(x)linx 利用符合函数的性质limx＾f(x)=e^limf(x)linx (x趋向于0正）那么原问题转化为证明lim(f(x)*ln(x))=0这时候将原式看成f(x)/1/linx就是0/0型未定式，用罗比大，条件都有了

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