Definition: Let $S \subset \R$, $c \in S$, and let $f \colon S \to \R$ be a function. We say that $f$ is continuous at $c$ if for every $\epsilon \gt 0$ there is a $\delta \gt 0$ such that whenever $x \in S$ and $\abs{x-c} \lt \delta$, then $\abs{\,f(x)-f(c)} \lt \epsilon$.

When $f \colon S \to \R$ is continuous at all $c \in S$, then we simply say $f$ is a continuous function.

Theorem: Consider a function $f \colon S \to \R$ defined on a set $S \subset \R$ and let $c \in S$. Then:

1. If $c$ is not a cluster point of $S$, then $f$ is continuous at $c$.
2. If $c$ is a cluster point of $S$, then $f$ is continuous at $c$ if and only if the limit of $f(x)$ as $x \to c$ exists and \begin{equation*} \lim_{x\to c} f(x) = f(c) . \end{equation*}
3. The function $f$ is continuous at $c$ if and only if for every sequence $\{ x_n \}$ where $x_n \in S$ and $\lim\, x_n = c$, the sequence $\{ \,f(x_n) \}$ converges to $f(c)$.

Theorem: Let $S\subset \R$, let $f \colon S \to \R$ and let $c\in S$.

there exists a sequence $\{x_n\}$ in $S$ such that $\lim x_n = c$, but the sequence $\left\{\,f(x_n)\right\}$ does not converge to $f(c)$.

Theorem: Let $f \colon S \to \R$ and $g \colon S \to \R$ be functions continuous at $c \in S$.

1. The function $h \colon S \to \R$ defined by $h(x) := f(x)+g(x)$ is continuous at $c$.
2. The function $h \colon S \to \R$ defined by $h(x) := f(x)-g(x)$ is continuous at $c$.
3. The function $h \colon S \to \R$ defined by $h(x) := f(x)g(x)$ is continuous at $c$.
4. If $g(x)\not=0$ for all $x \in S$, the function $h \colon S \to \R$ defined by $h(x) := \frac{f(x)}{g(x)}$ is continuous at $c$.

Theorem: Let $A, B \subset \R$ and $f \colon B \to \R$ and $g \colon A \to B$ be functions. If $g$ is continuous at $c \in A$ and $f$ is continuous at $g(c)$, then $f \circ g \colon A \to \R$ is continuous at $c$.

Theorem: Let $f \colon \R \to \R$ be a polynomial. That is \begin{equation*} f(x) = a_d x^d + a_{d-1} x^{d-1} + \cdots + a_1 x + a_0 , \end{equation*} for some constants $a_0, a_1, \ldots, a_d$. Then $f$ is continuous.

• Rational functions where they are defined.
• $\sin x$, $\cos x$ are continuous.
• The functions $\tan x$, $\cot x$, $\sec x$, $\csc x$ where they are defined.

Theorem: A continuous function $f \colon [a,b] \to \R$ is bounded.

Theorem (Extreme value theorem): A continuous function $f \colon [a,b] \to \R$ on a closed and bounded interval $[a,b]$ achieves both an absolute minimum and an absolute maximum on $[a,b]$.

Definition: Let $S \subset \R$, and let $f \colon S \to \R$ be a function. Suppose for every $\epsilon \gt 0$ there exists a $\delta > 0$ such that whenever $x, c \in S$ and $\abs{x-c} \lt \delta$, then $\abs{\,f(x)-f(c)} \lt \epsilon$. Then we say $f$ is uniformly continuous.

Theorem: Let $f \colon [a,b] \to \R$ be a continuous function. Then $f$ is uniformly continuous.

Theorem: Let $S\subseteq \R$ and let $f \colon S \to \R$. Then the following statements are equivalent:

1. $f$ is not uniformly continuous on $S$.
2. There exists an $\epsilon_0>0$ such that for every $\delta>0$ there are points $x_{\delta}$, $y_{\delta}$ in $S$ such that $|x_{\delta} - y_{\delta}|\lt \delta$ and $|\,f(x_{\delta}) - f(y_{\delta})|\geq\epsilon_0$.
3. There exists an $\epsilon_0\gt 0$ and two sequences $\{x_n\}$ and $\{y_n\}$ in $S$ such that $\lim (x_n-y_n)=0$ and $|\,f(x_{n}) - f(y_{n})|\geq\epsilon_0$ for all $n\in \N$.

Example: We can apply the previous result to show that $f(x)=\dfrac{1}{x}$ is not uniformly continuous on $(0,\infty)$.

Consider $x_n=\dfrac{1}{n}$ and $y_n=\dfrac{1}{n+1}$ in $(0,\infty)$. Then we have \[ \lim_{n\ra \infty} (x_n-y_n)=\lim_{n\ra \infty} \left(\frac{1}{n}-\frac{1}{n+1}\right)=\lim_{n\ra \infty}\frac{1}{n(n+1)} =0, \] but $\abs{\,f(x_n)- f(y_n)}=1$ for all $n\in \N$.

Definition: A function $f \colon S\subseteq \R \to \R$ is Lipschitz continuous, if there exists a $K > 0$, such that \begin{equation*} \abs{\,f(x)-f(y)} \leq K \abs{x-y} \qquad \text{for all } x \text{ and } y \text{ in } S. \end{equation*}

Theorem: A Lipschitz continuous function is uniformly continuous.

Remark: Not every uniformly continuous function is Lipschitz continuous. For example $f(x)=\sqrt{x}$ defined on $I=[0,2]$. $f$ is uniformly continuous on $I$ but there is no number $K>0$ such that $|\,f(x)-f(0)|\lt K|x-0|$ for all $x\in I.$

Definition: Let $S \subset \R$. We say $f \colon S \to \R$ is increasing (resp. strictly increasing) if $x,y \in S$ with $x \lt y$ implies $f(x) \leq f(y)$ (resp. $f(x) \lt f(y)$). We define decreasing and strictly decreasing in the same way by switching the inequalities for $f$.

If a function is either increasing or decreasing, we say it is monotone. If it is strictly increasing or strictly decreasing, we say it is strictly monotone.

Theorem: If $I \subset \R$ is an interval and $f \colon I \to \R$ is strictly monotone, then the inverse $f^{-1} \colon f(I) \to I$ is continuous.