极限

(1)\[ \lim_{n \to \infty} \frac{n (\sqrt[n] {n}-1)}{\ln{n}}=1 \]
证明：
令$ h_n=\sqrt[n] {n}-1 $ 则有 $n=(1+h_n)^n$ 且 $h_n \to 0$  $(n \to \infty)$

$\displaystyle \lim_{n \to \infty} \frac{n (\sqrt[n] {n}-1)}{\ln{n}} = \lim_{n \to \infty} \frac{n h_n}{n \ln (1+h_n)}=1$

(2)\[ \lim_{n \to \infty} \frac{n^2 (\sqrt[n] {n+1}-\sqrt[n+1]{n})}{\ln(n+1)} =1 \]
证明：
先证明\[ \lim_{n \to \infty} \frac{n^2 (\sqrt[n+1] {n+1}-\sqrt[n+1]{n})}{\ln(n+1)} =0 \]
$\displaystyle \lim_{n \to \infty} \frac{n^2 (\sqrt[n+1] {n+1}-\sqrt[n+1]{n})}{\ln(n+1)} $

$\displaystyle =\lim_{n \to \infty} \frac{ n^2 \sqrt[n+1]{n+1} \left(1-\left(1-\frac{1}{n+1}\right)^{ \frac{1}{n+1}}\right)}{\ln(n+1)} $

$\displaystyle = \lim_{n \to \infty} \frac{n^2 \left( \left(\frac{1}{(n+1)^2}\right)+o\left( \frac{1}{n^2} \right)\right)}{\ln(n+1)}=0$

于是，只需证明\[ \lim_{n \to \infty} \frac{n^2 (\sqrt[n] {n+1}-\sqrt[n+1]{n+1})}{\ln(n+1)} =1 \]
$\displaystyle \lim_{n \to \infty} \frac{n^2 (\sqrt[n] {n+1}-\sqrt[n+1]{n+1})}{\ln(n+1)} $

$\displaystyle = \lim_{n \to \infty} \frac{ n^2 \sqrt[n+1] {n+1}\left((n+1)^{ \frac{ 1 }{ n(n+1) }}-1\right) } { \ln(n+1) }  $

类似于(1)，令$ h_n = (n+1)^{\frac{1}{n(n+1)}}-1$即得结论。

http://tieba.baidu.com/p/1358122765

(3)
数列$\{x_n\}$满足$x_1+x_2+ \cdots +x_n=\dfrac{1}{\sqrt{x_{n+1}}}$
求极限：\[ \lim_{n \to \infty} \frac{n}{\ln{n}}\left(n^2 x_n^3-\frac{1}{9}\right)\]

http://tieba.baidu.com/f?kz=1354206812

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2012-1-12 16:30