1. 计算极限 $\lim\limits_{n\rightarrow\infty}\prod_{k=2}^{n}\cos(\frac{\pi}{2^k})$.
Posted by haifeng on 2025-06-19 10:04:10 last update 2025-06-19 10:04:10 | Answers (1) | 收藏
计算极限
\[\lim_{n\rightarrow\infty}\prod_{k=2}^{n}\cos(\frac{\pi}{2^k}).\]
Posted by haifeng on 2025-06-19 10:04:10 last update 2025-06-19 10:04:10 | Answers (1) | 收藏
计算极限
\[\lim_{n\rightarrow\infty}\prod_{k=2}^{n}\cos(\frac{\pi}{2^k}).\]
Posted by haifeng on 2025-03-23 18:20:52 last update 2025-03-24 12:29:40 | Answers (2) | 收藏
如何证明 $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1$.
问题的关键是证明:
\[
\sin x < x < \tan x,\quad\forall x\in (0,\frac{\pi}{2}).
\]
Posted by haifeng on 2025-01-18 10:34:34 last update 2025-01-18 10:34:34 | Answers (1) | 收藏
设 $x_n=\sum\limits_{k=1}^{n}\frac{1}{\sqrt{k}}-2\sqrt{n}$, 证明 $\lim\limits_{n\rightarrow\infty}x_n$ 存在.
Posted by haifeng on 2025-01-17 23:17:27 last update 2025-01-17 23:17:27 | Answers (2) | 收藏
求极限 \[\lim\limits_{x\rightarrow+\infty}\frac{e^x}{(1+\frac{1}{x})^{x^2}}.\]
Posted by haifeng on 2024-10-28 10:28:47 last update 2024-10-28 10:28:47 | Answers (3) | 收藏
设 $a$ 是正实数. 递归定义序列 $\{x_n\}$ 为 $x_{n+1}=\sqrt{a(a+1)+x_n}$, $x_1=\sqrt{a(a+1)}$. 证明 $\{x_n\}$ 收敛, 并求其极限.
Posted by haifeng on 2024-10-14 12:43:07 last update 2024-10-14 12:44:07 | Answers (2) | 收藏
若 $\lim\limits_{x\rightarrow 0}\frac{f(2x)}{x}=2$, 则 $\lim\limits_{x\rightarrow\infty}xf(\frac{1}{2x})$ 的值是多少?
Posted by haifeng on 2024-09-25 16:47:34 last update 2024-09-25 16:50:27 | Answers (0) | 收藏
证明
\[\lim\limits_{n\rightarrow\infty}\dfrac{n}{2^n}=0.\]
Hint. 当 $n\geqslant 2$ 时,
\[
2^n=(1+1)^n >\frac{n(n-1)}{2}
\]
Posted by haifeng on 2024-09-25 10:15:09 last update 2024-09-25 10:15:09 | Answers (1) | 收藏
证明:
\[\lim\limits_{x\rightarrow+\infty}\frac{x}{e^x}=0.\]
Posted by haifeng on 2024-09-07 10:35:21 last update 2024-09-07 10:51:03 | Answers (1) | 收藏
设 $\lim\limits_{n\rightarrow\infty}(a_{n+1}-a_n)=A$, 则 $\lim\limits_{n\rightarrow\infty}\frac{a_n}{n}=A$.
[Hint] 看到 $a_{n+1}-a_n$ 就要联想到 $\frac{a_{n+1}-a_n}{(n+1)-n}$, 进而 Stolz 公式. 当然这里我们并不需要应用 Stolz 公式, 只需引理3344即可.
[Remark] 下面的证法不行.
由条件得 $\forall\ \varepsilon > 0$, $\exists N$, 当 $n > N$ 时, 总有
\[A-\varepsilon < a_{n+1}-a_n < A+\varepsilon .\]
于是, 有
\[
\begin{cases}
A-\varepsilon < &a_{n+1}-a_n < A+\varepsilon,\\
A-\varepsilon < &a_{n+2}-a_{n+1} < A+\varepsilon,\\
&\vdots\\
A-\varepsilon < &a_{2n}-a_{2n-1} < A+\varepsilon .
\end{cases}
\]
相加得
\[
n(A-\varepsilon) < a_{2n}-a_n < n(A+\varepsilon),
\]
从而推出
\[
A-\varepsilon < \frac{a_{2n}-a_n}{n} < A+\varepsilon .
\]
但这无法证明 $\lim\limits_{n\rightarrow\infty}\frac{a_n}{n}=A$, 除非知道数列 $\{\frac{a_n}{n}\}$ 是收敛的.
Posted by haifeng on 2024-04-26 22:06:23 last update 2024-04-26 22:06:23 | Answers (1) | 收藏
证明 $\lim\limits_{n\rightarrow\infty}\sqrt[n]{n}=1$.