推 PttFund:硬算也是好方法 XDDDD140.112.218.142 07/16
※ 引述《PttFund (批踢踢基金)》之銘言：
: Show that
: lim x { (1 + 1/x)^x - e(ln(1 + 1/x)^x) } = 0.
: x→∞
y = (1 + 1/x)^x => y -> e as x -> oo
y' = y{[ln(1 + 1/x)] - 1/(x+1)} => y' -> 0 as x -> oo
y" = y'{[ln(1 + 1/x)] - 1/(x+1)} + y[-1/x(x+1)^2)]
= y {[ln(1 + 1/x)] - 1/(x+1)}^2 + y[-1/x(x+1)^2)]
=> y" -> 0 as x -> oo
(以下極限符號下標皆為 x -> oo)
y' - e[ln(1 + 1/x) - 1/(x+1)]
lim x{y-e[xln(1 + 1/x)]} = lim -------------------------------
-x^-2
y" - e{-1/x(x+1)^2}
= lim --------------------- = lim {y"x^3 + e[x/(x+1)]^2}/2
2x^-3
Claim : y"x^3 -> -e as x -> oo
(we just show (x^3){[ln(1 + 1/x)] - 1/(x+1)}^2 -> 0 as x -> oo )
lim (x^3){[ln(1 + 1/x)] - 1/(x+1)}^2
2{[ln(1 + 1/x)] - 1/(x+1)}[-1/x(x+1)^2]
= lim ------------------------------------------
-3x^-4
= lim (2/3){[x^3/(x+1)^2]ln(1 + 1/x) - [x/(x+1)]^3}
Use the fact that [x^3/(x+1)^2]ln(1 + 1/x) -> 1 as x -> oo
then, the proof is complete obviously.
(好累, 有簡單的方法嗎 =.=)
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