TZQIQI
幼苗
共回答了17个问题采纳率:100% 举报
根据 自然对数底数e的性质
e = (x→∞) lim[(1 + 1/x)^x]
则有,
等式左边 = (x→∞) lim[(1 + 1/x)^x]^a
= e^a
等式右边 = ∫td(e^t) ←(符号难打,这里 积分区域省略)
= te^t - ∫e^t*dt
= (t-1)e^t | -∞→a
= (a-1)*e^a - 0
= (a-1)*e^a
∴e^a = (a-1)*e^a
即,(a-2)*e^a = 0
∵e^a >0,∴a-2 = 0,即 a=2
1年前
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