Mathematics & Statistics
Mahasiswa

有人可以解釋一下極限怎麼算出來的嗎

x an+1 = lim n-> lim n-> An (-1)+1e-(n+1) (-1)"e-n en en+1 || lim n+♡ 1 lim n e 1 || || <1 e
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Answers

因取絕對值所以(-1)可略
e^-(n+1)/e^-n = e^n/e^(n+1)
化簡為1/e

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