我们常说曲线是连续的，或说汽车速度的变化是连续的，这都反映了函数的一个 重要特性一连续性。连续在日常生活中有很直观的提现。通常我们说一根曲线是连续的, 因为它没有断开，那么什么叫“断开”呢？你若说，不连续就是断开的，这样从 逻辑的角度看，出现了循环定义的现象，也就是说什么也没有定义，因此有必要 给“连续”一个严格的定义。但连续的定义如何给出，却经历了一个漫长的过程（整个18世纪下半叶！）。 现代数学给出的连续的定义如下:
函数在一点连续的精确定义
定义 设函数 $f$ 在 $x_{0}$ 的某个邻域内有定义。如果 $_{\text {函数 }} f$ 当 $x \rightarrow x_{0}$ 时的极限存
在, 并且等于它在 $x_{0}$ 处的值, 即 $\lim _{x \rightarrow x_{0}} f(x)=f\left(x_{0}\right),$ 则称函数 $f$ 在 $x_{0}$ 点连续。
多数初次学习函数在一点连续定义的同学觉得不可接受，有心理障碍的原因有以下两点.
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。
所以熟练的掌握函数在一点连续的定义对于后续的学习是非常必要的。
下面是一些典型的Math 115A 中期的assignment（连续函数）中可能会出现的典型题目。
$$
\left|\sum_{n=1}^{\infty} x_{n}\right| \leq \sum_{n=1}^{\infty}\left|x_{n}\right|
$$
$$
\left|\sum_{n=1}^{N} x_{n}\right| \leq \sum_{n=1}^{N}\left|x_{n}\right|
$$
Taking $N \rightarrow \infty$ and noticing that
$$
\lim _{N \rightarrow \infty}\left|\sum_{n=1}^{N} x_{n}\right|=\lim _{N \rightarrow \infty} \sum_{n=1}^{N} x_{n} \mid
$$
(since the limit of the right hand side exists) gives the inequality.
Cesàro summable if $C=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} s_{k}$ exists. The value $C$ is called the Cesàro sum of the sequence.
a) Show that if $\left\{a_{n}\right\}_{n=1}^{\infty}$ is a convergent sequence with $\lim _{n \rightarrow \infty} a_{n}=A,$ then the sequence $\left\{b_{n}\right\}_{n=1}^{\infty}$ defined by $b_{n}=\frac{1}{n} \sum_{k=1}^{n} a_{k}$ is also convergent and $\lim _{n \rightarrow \infty} b_{n}=A$.
b) Show that if $S=\sum_{n=1}^{\infty} x_{n}$ converges, then the sequence is Cesàro summable and $S=C .$ Hint: use a).
c) Show that $x_{n}=(-1)^{n}$ is Cesàro summable, even though $\sum_{n=1}^{\infty}(-1)^{n}$ is not convergent.
$$
\left|a_{n}-A\right|<\varepsilon / 2
$$
Since $a_{n}$ is convergent there is $M>0$ such that $\left|a_{n}\right| \leq M$ for all $n .$ Now given $n>N$ we have by triangle inequality
$$
\left|b_{n}-A\right| \leq \sum_{k=1}^{n} \frac{1}{n}\left|a_{n}-A\right| \leq \sum_{k=1}^{N} \frac{1}{n}\left|a_{n}-A\right|+\sum_{k=N}^{n} \frac{1}{n}\left|a_{n}-A\right| \leq \sum_{k=1}^{N} \frac{2 M}{n}+\sum_{k=N}^{n} \frac{\varepsilon}{2 n}<\frac{2 M N}{n}+\frac{\varepsilon}{2}
$$
Now let $N_{1}$ be such that
$$
\frac{2 M N}{N_{1}}<\varepsilon / 2
$$
It follows that for $n>\max \left\{N, N_{1}\right\}$ we have
$$
\left|b_{n}-A\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon
$$
which means $b_{n} \rightarrow A$ as well.
(b) This follows immediately by taking $a_{n}=s_{n}$ in part (a).
(c) The partial sums are $s_{k}=0$ if $k$ is even and -1 if $k$ is odd. So we have
$$
\sum_{k=1}^{n} s_{k}=\left\{\begin{array}{ll}
-\frac{n}{2} & n \text { even } \\
-\frac{n+1}{2} & n \text { odd }
\end{array}\right.
$$
and it follows that
$$
\frac{1}{n} \sum_{k=1}^{n} s_{k}=\left\{\begin{array}{ll}
-\frac{1}{2} & n \text { even } \\
-\frac{1}{2}-\frac{1}{2 n} & n \text { odd }
\end{array}\right.
$$
In either case the limit is $-\frac{1}{2}$ as $n \rightarrow \infty,$ so the series is Cesàro summable with $S=-\frac{1}{2}$
$(c-\varepsilon, c+\varepsilon) \backslash\{c\} \cap S,$ so the latter set is also not empty for any $\varepsilon>0$
Problem #4. Let $c_{1}$ be a cluster point of $A$ and $c_{2} \in B$ be a cluster point of $B$. Suppose $f: A \rightarrow \mathbb{R}$ and $g: B \rightarrow \mathbb{R}$ are functions with $\lim _{x \rightarrow c_{1}} f(x)=c_{2}$ and $\lim _{x \rightarrow c_{2}} g(x)=L=g\left(c_{2}\right) .$ Show $\lim _{x \rightarrow c_{1}} g(f(x))=L$
have
$$
\begin{array}{l}
\qquad|g(f(c))-L|=\left|g(f(c))-g\left(c_{2}\right)\right|<\varepsilon \\
\text { since }\left|f(c)-c_{2}\right|<\delta_{1}
\end{array}
$$
Problem $\# 5 .$ Suppose $S \subset \mathbb{R}$ and $c$ is a cluster point of $S .$ Suppose $f: S \rightarrow \mathbb{R}$ is bounded. Show that there exists a sequence $\left\{x_{n}\right\}_{n=1}^{\infty}$ with $x_{n} \in S \backslash\{c\}$ and $\lim _{n \rightarrow \infty} x_{n}=c$ such that $\left\{f\left(x_{n}\right)\right\}_{n=1}^{\infty}$ converges.
Since $c$ is a cluster point of $S,$ there is $x_{n} \in(c-1 / n, c+1 / n) \backslash\{c\}$ for every $n \in \mathbb{N}$. Clearly $x_{n} \rightarrow c$. Since $f$ is bounded, by Bolzano-Weierstrass we may extract a subsequence $\left\{x_{n_{i}}\right\} \subset\left\{x_{n}\right\}$ such that $\left\{f\left(x_{n_{i}}\right)\right\}$ converges. Since $x_{n} \rightarrow c$ it follows $x_{n_{i}} \rightarrow c$ as well, so the sequnce $\left\{x_{n_{j}}\right\}$ is what we wanted.
Using the $\epsilon-\delta$ -definition of continuity directly prove that $f(x)=\frac{1}{x}$ is continuous on $(0, \infty)$.
$$
\left|\frac{1}{x}-\frac{1}{x_{0}}\right|=\frac{\left|x-x_{0}\right|}{x x_{0}}
$$
Let us first agree that $\left|x-x_{0}\right|<\frac{1}{2}\left|x_{0}\right|,$ so that $x>\frac{x_{0}}{2}$ and
$$
\frac{\left|x-x_{0}\right|}{x x_{0}}<\frac{2\left|x-x_{0}\right|}{x_{0}^{2}}
$$
If further $\left|x-x_{0}\right|<\frac{\varepsilon x_{0}^{2}}{2}$ we will have
$$
\frac{2\left|x-x_{0}\right|}{x_{0}^{2}}<\varepsilon
$$
Thus it suffices to pick $\delta=\min \left\{\frac{1}{2}\left|x_{0}\right|, \frac{\varepsilon\left|x_{0}\right|^{2}}{2}\right\}$ (so that both of the above inequalities hold).
Show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous and $f(c)>0$, then there is a $\delta>0$ so that $f(x)>0$ for all $x \in(c-\delta, c+\delta)$.
$$
|f(x)-f(c)|<\frac{|f(c)|}{2}
$$
which in turn (by triangle inequality) implies $f(x)>f(c)-\frac{f(c)}{2}=\frac{f(c)}{2}>0$ for $|x-c|<\delta$.
A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is said to be Lipschitz if there is a constant $L>0$ so $|f(x)-f(y)| \leq$ $L|x-y|$ for all $x, y \in \mathbb{R} .$ Show that $f$ is continuous.
$$
|f(x)-f(y)| \leq L|x-y|<L \frac{\varepsilon}{L}=\varepsilon
$$
which shows the continuity at $x$. Since $x$ is arbitrary, $f$ is continuous.
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