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lim【x→a】[e^(x-a)-1](x-a) =lim【x→a】(x-a)/(x-a) =1 或者利用洛必達法則: lim【x→a】[e^(x-a)-1]/(x-a) =lim【x→a】e^(x-a) =e^(a-a) =e^0 =1 答案:1
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