→ n0204159:感謝啦 140.138.31.205 10/08 13:31
→ n0204159:話說這個ip是中央研究院?! 真神人..! 140.138.31.205 10/08 17:47
推 midarmyman:他一直都很神 140.117.198.78 10/08 17:49
※ 引述《n0204159 (我期待有一天我會回來)》之銘言：
: 1/x
: e-(1+x)
: l i m ---------
: x→0 x
: 這題是碩士班招生考題..哈哈
: 請各位大大指點迷津!!
: 謝謝!!
e = lim (1+x)^(1/x) = lim f(x)
x->0 x->0
e is the limit of f(x) at x=0
or define f(x) = (1+x)^(1/x) for x =/= 0
e for x = 0
1/x
e-(1+x)
l i m ---------
x→0 x
e - f(x)
lim - ---------- = -f'(x) definition of derivative
x->0 0 - x
= lim -{(1/x)(1+x)^[(1/x)-1] - (1/x^2)[ln(1+x)](1+x)^(1/x)}
x->0
x - (1+x)ln(1+x)
= lim - f(x) ──────
x->0 x^2 (1+x)
x - (1+x)ln(1+x)
lim ──────
x->0 x^2 (1+x)
- ln(1+x)
= lim ──────
x->0 2x(1+x) + x^2
-1/(1+x)
= lim ──────
x->0 2(1+x) + 4x
= -1/2
Thus,
1/x
e-(1+x)
l i m ---------
x→0 x
x - (1+x)ln(1+x)
= lim - f(x) ──────
x->0 x^2 (1+x)
= - e * (-1/2)
= e/2
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.109.103.151