为了这里书写方便,我来计算：lim[x→0] x/(e^x-1)
lim[x→0] x/(e^x-1)
令e^x-1=u,则x=ln(1+u)
=lim[u→0] ln(1+u)/u
=lim[u→0] (1/u)ln(1+u)
=lim[u→0] ln[(1+u)^(1/u)]
=lne
=1
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