Re: [考古] 清大˙微分
※ 引述《afulet (阿弗雷特)》之銘言：
: Find
: 1/100
: ╭ 100 99 ╮
: lim │ (n ＋ n ) │ － n = ?
: n→∞ ╰ ╯
: 求解法 （答案為１／１００)
(1 + 1/n)^(1/100) - 1
(n^100 + n^99)^(1/100) - n = -----------------------
1/n
因此 lim { (n^100 + n^99)^(1/100) - n }
n→∞
(1 + 1/n)^(1/100) - 1
= lim -----------------------
n→∞ 1/n
(1+h)^(1/100) - 1^(1/100)
= lim --------------------------- = f'(1) = 1/100,
h→0 h
其中 f(x) = x^(1/100). (f'(x) = 1/100 * x^(-99/100)
f'(1) = 1/100)
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