※ 引述《PttFund (批踢踢基金)》之銘言：
: Using (1+1/n)^n → e as n→∞ to show that
: (1+1/x)^x → e as x→+∞, and
: (1+1/x)^x → e as x→-∞,
: where n in N and x in R.
let f(x) = (1+1/x)^x
f'(x) = (1+1/x)^x [ln(1+1/x) - 1/(x+1)] > 0, for x > 0
==> f(x) is strictly increasing
for any x > 0, there is n_x belongs to |N such that
n_x <= x < n_x + 1
==> (1+1/n_x)^n_x <= (1+1/x)^x < (1+1/(n_x+1))^(n_x+1)
so (1+1/x)^x → e as x→+∞
for x negative, let y = -x
(1+1/x)^x = (1+1/(y-1))^y = (1+1/(y-1))^(y-1)*(1+1/(y-1))
apply the proof above
(1+1/x)^x → e as x→-∞
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