##### 极限的精确定义

1. 大家更加适应和习惯自然语言, 而第一次认真的接触逻辑语言是会不适应的, 就像你第一次学匀 编程一样, 肯定既兴奋又不太适应;
2 . 练习或是接触的频率不够, 接触多了你就会觉得相对直观和舒服多了。其实我们学习中文或是英文这种自然语言也是如此, 母语从小到大一直学习和使用你就觉得亲切, 而第二第三语言要是接触少一些你 自然就会有不少心理障碍, 逻辑语言就更是了。

Problem 1. Problem #1. Let $I$ be a closed interval and $J$ an open interval.
a) Let $\left\{x_{n}\right\}_{n=1}^{\infty}$ be a convergent sequence with $x_{n} \in I$ show that $\lim _{n \rightarrow \infty} x_{n} \in I$.
b) Let $\left\{y_{n}\right\}_{n=1}^{\infty}$ be a convergent sequence with $y_{n} \in J .$ Show by example that it may be possible for $\lim _{n \rightarrow \infty} y_{n} \notin J$
Proof . (a) Let $I=[a, b] .$ Suppose for a contradiction that $x=\lim _{n \rightarrow \infty} x_{n} \notin I .$ Without loss of generality we may assume $x>b .$ Let $\varepsilon=x-b .$ By definition of the limit there is $N \in \mathbb{N}$ such that $n>N$ implies $\left|x_{n}-x\right|<\varepsilon,$ but this implies
$$x_{n} \in(x-\varepsilon, x+\varepsilon)=(b, x+\varepsilon)$$
which is disjoint from $I,$ a contradiction.
(b) Consider $\left\{\frac{1}{n+1}\right\}_{n-1}^{\infty} \subset(0,1),$ the limit is 0 as $n \rightarrow \infty$ which is not in (0,1) .
Problem 2. Let $\left\{x_{n}\right\}_{n=1}^{\infty}$ is a convergent sequence. For $k \in \mathbb{N}$ show that
$$\lim _{n \rightarrow \infty} x_{n}^{k}=\left(\lim _{n \rightarrow \infty} x_{n}\right)^{k}$$
Proof . Let $x=\lim _{n \rightarrow \infty} x_{n},$ we must show that $\lim _{n \rightarrow \infty} x_{n}^{k}=x^{k} .$ Given $\varepsilon>0,$ choose $N_{1} \in \mathbb{N}$ such that
$n>N_{1}$ implies
$$\left|x_{n}-x\right|<\frac{\varepsilon}{k(|x|+1)^{k-1}}$$
Since convergent sequences are bounded, we may choose $N_{2} \in \mathbb{N}$ such that $n>N_{2}$ implies $\left|x_{n}\right| \leq|x|+1$. Let $N=\max \left\{N_{1}, N_{2}\right\} .$ For $n>N$ we have
\begin{aligned} \left|x_{n}^{k}-x^{k}\right| &=\left|x_{n}-x\right|\left|x_{n}^{k-1}+x x_{n}^{k-2}+x^{2} x_{n}^{k-3}+\cdots+x^{k-1}\right| \\ & \leq\left|x_{n}-x\right|\left(\left|x_{n}\right|^{k-1}+|x|\left|x_{n}\right|^{k-2}+\cdots+|x|^{k-1}\right) \\ &\left.\leq \frac{\varepsilon}{k(|x|+1)^{k}}\left((|x|+1)^{k-1}+|x|(|x|+1)^{k-2}+\cdots+|x|^{k-1}\right)\right) \\ &<\frac{\varepsilon}{k(|x|+1)^{k}}\left(k(|x|+1)^{k}\right)<\varepsilon \end{aligned}
which gives the required result by definition.
Problem 3. Let $\left\{a_{n}\right\}_{n=1}^{\infty}$ and $\left\{b_{n}\right\}_{n=1}^{\infty}$ be two sequences.
a) If $\left\{a_{n}\right\}_{n=1}^{\infty}$ is bounded and $\lim _{n \rightarrow \infty} b_{n}=0,$ then show $\left\{a_{n} b_{n}\right\}_{n=1}^{\infty}$ is convergent and $\lim _{n \rightarrow \infty} a_{n} b_{n}=0$.
b) Given an example where $\left\{a_{n}\right\}_{n=1}^{\infty}$ is unbounded and $\lim _{n \rightarrow \infty} b_{n}=0$ and $\left\{a_{n} b_{n}\right\}_{n=1}^{\infty}$ is divergent.
c) Give an example where $\left\{a_{n}\right\}_{n=1}^{\infty}$ is bounded and $\lim _{n \rightarrow \infty} b_{n}$ exists and is $\neq 0$ and $\left\{a_{n} b_{n}\right\}_{n=1}^{\infty}$ is divergent.
Proof . (a) Let $M \in \mathbb{R}$ be such that $\left|a_{n}\right| \leq M .$ Given $\varepsilon>0,$ choose $N \in \mathbb{N}$ such that $n \geq N$ implies
$$\left|b_{n}\right|<\frac{\varepsilon}{M}$$
Then for $n \geq N$ we have
$$\begin{array}{l} \quad\left|a_{n} b_{n}\right| \leq\left|a_{n}\right|\left|b_{n}\right|<M \frac{\varepsilon}{M}<\varepsilon \\ \text { so } \lim _{n \rightarrow \infty} a_{n} b_{n}=0 \end{array}$$
(b) Consider $a_{n}=n^{2}$ and $b_{n}=\frac{1}{n},$ then $a_{n} b_{n}=n \rightarrow \infty$ as $n \rightarrow \infty$
(c) Let $a_{n}=1$ if $n$ is even and -1 if $n$ is odd. Let $b_{n}=1 .$ Then $a_{n} b_{n}=a_{n}$ which is divergent.
Problem 4.

Let $\left\{x_{n}\right\}_{n=1}^{\infty}$ be sequence with $\lim _{n \rightarrow \infty} x_{2 n}=\lim _{n \rightarrow \infty} x_{2 n-1}=x .$ Show $\lim _{n \rightarrow \infty} x_{n}=x$.

Proof . Let $\varepsilon>0 .$ By definition, there is $N_{1} \in \mathbb{N}$ such that $n>N_{1}$ implies $\left|x_{2 n}-x\right|<\varepsilon .$ Similarly there is $N_{2}$ such that $n>N_{2}$ implies $\left|x_{2 n-1}-x\right|<\varepsilon .$ Let $N=\max \left\{2 N_{1}, 2 N_{2}\right\}$ and clearly $n>N$ implies $\left|x_{n}-n\right|<\varepsilon$ (since $n$ is either even or odd). So $\lim _{n \rightarrow \infty} x_{n}=x$ as well.
Problem 5.

Recall, the Fibonacci sequence is defined by $f_{1}=f_{2}=1$ and $f_{n}=f_{n-1}+f_{n-2}$ for $n \geq 3$.
a) Show that the sequence $x_{n}=\frac{f_{n+1}}{f_{n}}$ is well-defined and satisfies $x_{1}=1$ and $x_{n+1}=1+x_{n-1}^{-1}$.
b) Show that for $l \in \mathbb{N},$ the $x_{n}$ satisfy $1 \leq x_{1} \leq x_{3} \leq \ldots \leq x_{2 l+1} \leq x_{2 l} \leq x_{2 l-2} \leq \ldots \leq x_{2} \leq 2$.
c) Show that $\left\{x_{n}\right\}_{n=1}^{\infty}$ is convergent and determine $\lim _{n \rightarrow \infty} x_{n} .$ (Hint: Problem 4 is helpful).

Proof .

which is true for $\frac{1-\sqrt{5}}{2} \leq x_{2 i-1} \leq \frac{1+\sqrt{5}}{2}$.
Note that $x_{1}=1$ is between 0 and $\frac{1+\sqrt{5}}{2},$ so $x_{1} \leq x_{3} .$ More generally one can show that this relation is preserved under the recursive relation. Indeed
\begin{aligned} 0 \leq x_{2 i-1} \leq \frac{1+\sqrt{5}}{2} & \Longrightarrow x_{2 i}=1+x_{2 i-1}^{-1} \geq 1+\frac{2}{1+\sqrt{5}}=\frac{1+\sqrt{5}}{2} \\ & \Longrightarrow x_{2 i+1}=1+\left(1+x_{2 i-1}^{-1}\right)^{-1} \leq 1+\frac{2}{1+\sqrt{5}}=\frac{1+\sqrt{5}}{2} \end{aligned}
This means $0 \leq x_{2 i-1} \leq \frac{1+\sqrt{5}}{2}$ for all $i$ and it follows from the previous paragraph that $x_{2 i-1} \leq$ $x_{2 i+1}$
The fourth item follows in a similar fashion: one shows that $x_{2 i} \geq \frac{1+\sqrt{5}}{2}$ is preserved under the recursive relation and that
$$x_{2 i-2} \geq x_{2 i} \Longleftrightarrow x_{2 i-2} \geq 1+x_{2 i-2}^{-1} \Longleftrightarrow x \geq \frac{1+\sqrt{5}}{2} \text { or } x \leq \frac{1-\sqrt{5}}{2}$$
Finally the fifth item follows from the above since
$$x_{2 i-1} \leq \frac{1+\sqrt{5}}{2} \leq x_{2 i}$$
Combining all these gives the desired inequalities.
(c) We use Problem 4 . By the recurrence relation we have that the odd-numbered terms satisfy
$$x_{2 i+1}=1+x_{2 i}^{-1}=1+\left(1+x_{2 i-1}^{-1}\right)^{-1}$$
Now by part b) the odd-numbered term is a monotonically increasing sequence bounded above by
2. So by the monotone convergence theorem the limit exists, and one can directly solve the above equation to get
$$\lim _{n \rightarrow \infty} x_{2 n-1}=\frac{1+\sqrt{5}}{2}$$
since by the previous part $x_{2 n+1} \geq 1$ for all $n .$ Similarly the even-numbered terms also have a limit, and a similar computation shows that this limit is also $\frac{1+\sqrt{5}}{2},$ so Problem 4 implies that
$$\lim _{n \rightarrow \infty} x_{n}=\frac{1+\sqrt{5}}{2}$$

Problem 6.

Let $\left\{x_{n}\right\}_{n=1}^{\infty}$ be a bounded sequence. Show that
$$\limsup _{n \rightarrow \infty} x_{n}=-\liminf _{n \rightarrow \infty}\left(-x_{n}\right)$$

Proof . Recall that
$$\limsup _{n \rightarrow \infty}=\lim _{n \rightarrow \infty} \sup _{k>n} x_{k}$$
Moreover we have
$$\sup _{k>n} x_{k}=-\inf _{k>n}\left(-x_{k}\right)$$
Combining these and noticing that the negative sign can be pulled out of the limit since the limit exists gives the result.

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