61. 求极限 $\lim\limits_{x\rightarrow 0^+}\frac{x^x-(\sin x)^x}{x-\sin x}$.
Posted by haifeng on 2017-04-10 20:42:31 last update 2017-04-10 20:42:46 | Answers (2) | 收藏
求极限
\[
\lim_{x\rightarrow 0^+}\frac{x^x-(\sin x)^x}{x-\sin x}
\]
Posted by haifeng on 2017-04-10 20:42:31 last update 2017-04-10 20:42:46 | Answers (2) | 收藏
求极限
\[
\lim_{x\rightarrow 0^+}\frac{x^x-(\sin x)^x}{x-\sin x}
\]
Posted by haifeng on 2017-04-09 23:26:32 last update 2017-04-10 19:35:12 | Answers (0) | 收藏
设 $f(x)$ 是定义在 $a$ 点某开邻域 $U$ 上的实值函数, 且 $f$ 在点 $a$ 处可微. 证明: 如果 $U$ 中的两个点列, $x_n$ 递增趋于 $a$, $y_n$ 递减趋于 $a$, 则有
\[
\lim_{n\rightarrow\infty}\frac{f(y_n)-f(x_n)}{y_n-x_n}=f'(a).
\]
应用此结论, 求解极限
\[
\lim_{x\rightarrow\infty}x^4\Bigl(\arctan(2+\frac{3}{x^2+1})-\arctan(2+\frac{3}{x^2+2})\Bigr)
\]
References:
Paulo Ney de Souza, Jorge-Nuno Silva, Berkeley Problems in Mathematics, Third Edition. Problem 1.4.21 (Sp91).
Posted by haifeng on 2017-04-09 08:10:09 last update 2017-04-09 08:10:09 | Answers (1) | 收藏
设 $x_1 > 0$, $x_{n+1}=\ln(x_n+1)$. 求
(1) $\lim\limits_{n\rightarrow\infty}x_n$;
(2) $\lim\limits_{n\rightarrow\infty}\frac{x_n\cdot x_{n+1}}{x_n-x_{n+1}}$.
Posted by haifeng on 2017-04-09 07:40:00 last update 2017-04-09 07:40:17 | Answers (2) | 收藏
求极限
\[
\lim_{n\rightarrow\infty}\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}{\ln n}
\]
Posted by haifeng on 2017-04-08 20:11:02 last update 2022-09-29 21:09:37 | Answers (1) | 收藏
求极限
\[
\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}{n}
\]
[Hint] 用积分
或者求
\[
\lim_{n\rightarrow\infty}\frac{n}{\sqrt[n]{n!}}
\]
如果不使用积分, 怎么算?
利用此极限, 求
\[
\lim_{n\rightarrow\infty}\frac{\sqrt[n]{(a+b)(2a+b)\cdots(na+b)}}{n},
\]
这里 $a,b$ 均为正数.
注: 对于 $\alpha > 0$,
\[
\lim_{n\rightarrow\infty}\frac{\sqrt[n]{(1+\alpha)(2+\alpha)\cdots(n+\alpha)}}{n}
\]
与
\[
\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}{n}
\]
的结果是一样的.
Posted by haifeng on 2017-04-08 20:04:38 last update 2017-04-08 20:28:22 | Answers (1) | 收藏
求极限
\[
\lim_{n\rightarrow\infty}\biggl[\frac{1}{n+1}+\frac{1}{(n^2+1)^{1/2}}+\cdots+\frac{1}{(n^n+1)^{1/n}}\biggr]
\]
[Hint] 使用夹逼准则.
Posted by haifeng on 2017-04-08 18:57:17 last update 2017-04-08 20:28:42 | Answers (1) | 收藏
求
\[
\lim_{n\rightarrow\infty}\frac{1}{n^2}\biggl[\sqrt{n^2-1}+\sqrt{n^2-4}+\cdots+\sqrt{n^2-(n-1)^2}\biggr]
\]
[Hint] 用积分
Posted by haifeng on 2017-04-08 18:47:51 last update 2017-04-08 20:28:54 | Answers (0) | 收藏
求极限
\[\lim_{n\rightarrow\infty}n^2(x^{\frac{1}{n}}-x^{\frac{1}{n+1}}),\]
这里 $x > 0$.
Posted by haifeng on 2016-12-18 13:32:40 last update 2016-12-18 13:33:04 | Answers (1) | 收藏
求极限 $\lim\limits_{x\rightarrow 0^+}(x^x -1)\ln x$.
Posted by haifeng on 2016-09-03 21:11:52 last update 2022-10-27 12:52:19 | Answers (1) | 收藏
求极限
\[\lim\limits_{n\rightarrow+\infty}\sin(\sin(\cdots(\sin x)))\]
($n$ 次复合函数).
若记 $a_1=\sin x$, $a_n=\sin(a_{n-1})$. 即 $a_n=\sin(\sin(\cdots(\sin x)))$, ($n$ 次复合函数). 则证明
\[
\prod_{n=1}^{\infty}\cos a_n
\]
发散.
若记 $d_n=a_n-a_{n+1}$, 则可以证明:
Claim 1. $\{d_n\}$ 严格递减趋于 0.
Claim 2. $\frac{1}{2}d_{n+1} < \sin\frac{d_n}{2}$.
Claim 3. $d_n < \sin d_{n-1}$.
Claim 4. $\frac{d_{n+1}}{d_n} < \cos\frac{d_n}{2}$