A function is called ** continuous** whenever sufficiently small changes in the input results in arbitrarily small changes in the output. We discuss continuous functions, one-sided and two-sided continuity, and removable continuity. The infamous Intermediate Value Theorem is considered at the end. First you need a good understanding of limits of functions.

**Definition**. A function is ** continuous** at a point $c$ means $f(c)$ is defined, $\lim_{x\to c}f(x)$ exists, and $\displaystyle \lim_{x\to c}f(x)=f(c).$

## Discontinuity

**Example**. Find three examples of how a discontinuity might arise.

**Solution**. First, the function $$ f(x)=\frac{x^2-2x+1}{x-1} $$ is discontinuous at $x=1$ because $f(1)$ is not defined. So one type of discontinuity is a ** hole** in the function. Secondly, the function \begin{equation} f(x)= \begin{cases} x^2+1 & x\geq 0 \\ -x^2-2 & x<0 \end{cases} \end{equation} is discontinuous at $x=0$ because $$ \lim_{x\to 0^-}f(x)=-2 $$ and $$ \lim_{x\to 0^+}f(x)=1; $$ thus $\lim_{x\to 0}f(x)$ does not exist and so $f$ has a discontinuity at $x=0.$ This type of discontinuity is called a

**. Thirdly, the function $$ f(x)=\frac{x-1}{x-2} $$ is discontinuous at $x=2$ because $f(2)$ is not defined and this type of discontinuity is called a**

*jump***because $f(x)\to+\infty $ as $x\to 2^+.$**

*pole*## Continuous Functions

**Theorem**. If $f$ and $g$ are functions that are continuous at $x=c$ then $f\pm g,$ $f g,$ $f/g,$and $f\circ g$ are continuous at $x=c,$ provided that $c$ is in the domain of the function.

**Example**. Give some examples of continuous functions.

**Solution**. For example, the functions $2x^2-2x+5,$ (polynomial), $(x-1)/x$ (rational), $\csc x$ (trigonometric), and $\sec ^{-1}x$ (inverse trigonometric) are continuous on their domains. Also the functions $$ 2x^2-2x+5+\frac{x-1}{x}(\csc x), $$ $ \sec ^{-1}\left(\frac{x-1}{x}\right),$ and $ \left(2x^2-2x+5\right)\left(\frac{x-1}{x}\right)$ are continuous functions on their domains.

**Theorem**. If $f$ is a polynomial function, rational function, trigonometric function, or inverse trigonometric function then $f$ is continuous where it is defined.

**Theorem**. (** Composition Limit Theorem**) Whenever $$ \lim_{x\to c}g(x)=L $$ and $f$ is a continuous function at $L,$ then $$ \lim_{x\to c}(f\circ g)(x)=f(L). $$

**Example**. Use the Composition Limit Theorem to evaluate the following limits. $$ \lim_{x\to 3} \left(x^2+3\right)^2 \quad\text{and}\quad \lim_{x\to \pi /4} \sin ^4 x. $$

**Solution**. By the composition limit theorem, we have \begin{equation} \lim_{x\to 3}\left(x^2+3\right)^2 =\left(\lim_{x\to 3}\left(x^2+3\right)\right)^2 =12^2 =144. \end{equation} By the composition limit theorem, we have \begin{equation} \lim_{x\to \pi /4}\sin ^4x =\left(\lim_{x\to \pi /4}\sin x\right){}^4 =\left(\frac{\sqrt{2}}{2}\right)^4 =\left(\frac{1}{4}\right) \end{equation} as desired.

## One-Sided Continuity

**Definition**. A function $f$ is ** continuous from the right** at $c$ if and only if $$ \lim_{x\to c^+}f(x)=f(c) $$ and it is

**at $c$ if and only if $$ \lim_{x\to c^-}f(x)=f(c). $$**

*continuous from the left***Example**. Give an example of a function that is continuous from the right (or ** right continuous**) at $x=0.$

**Solution**. The function $f(x)=\sqrt{x}$ is continuous from the right at $x=0$ because $$ \lim_{x\to 0^+}\sqrt{x}=0 $$ as needed.

## Determining Parameters for Continuity

**Example**. Find constants $a$ and $b$ so that \begin{equation} \begin{cases} a x^2+b & x>2 \\ 4 & x=2 \\ x^2-a x+b & x<2 \end{cases} \end{equation} is continuous on $\mathbb{R}.$

**Solution**. Since $f$ is defined on $\mathbb{R},$ and $f$ is continuous for all $x\neq 2$ for any $a$ and $b$ that we choose, itis left to find an $a$ and $b$ such that $$ \lim_{x\to 2^+}a x^2+b=4 $$ and $$ \lim_{x\to 2^-}\left(x^2-a x+b\right)=4. $$ Thus we have the system $4 a+b=4$ and $4-2 a+b=4.$ Solving this system we have, $a=2/3$ and $b=4/3.$

**Example**. Find constants $a$ and $b$ such that $f$ is continuous at $x=1$ where \begin{equation} \begin{cases} a x+b & x>1 \\ 3 & x=1 \\ x^2-4x+b+3 & x<1 \end{cases} \end{equation}

**Solution**. To have continuity at $x=1$ we must have $$ \lim_{x\to 1^-}f(x)=3 $$ and $$ \lim_{x\to 1^+}f(x)=3, $$ thus $a (1)+b=3$ and $(1)^2-4(1)+b+3=3.$ Therefore, $a+b=3$ and $b=3.$ So $a+3=3$ and $a=0.$

## Removable Continuity

**Example**. Determine the value for which $f(2)$ should be assigned, if any, to have $$ f(x)=\sqrt{\frac{x^2-4}{x-2}} $$ continuous at $x=2.$

**Solution**. Since $$ \lim_{x\to 2^-}f(x)=2 $$ and $$ \lim_{x\to 2^+}f(x)=2 $$ we have $\lim_{x\to 2}f(x)=2.$ Therefore, if we define $f(2)=2$ the function $f$ will be continuous at $x=2.$

## Intermediate Value Theorem

The Intermediate Value Theorem is extremely useful.

**Theorem**. (** Intermediate Value Theorem**) If $f$ is a continuous function on the closed interval $[a,b]$ and $L$ is some number strictly between$f(a)$ and $f(b)$, then there exists at least one number $c$ on the open interval $(a,b)$ such that $f(c)=L.$

**Example**. The population (in thousands) of a colony of bacteria $t$ minutes after the application of a toxin is given by the function \begin{equation} P(t)= \begin{cases} t^2+1 & \text{if } 0\leq t<5 \\ -8t+66 & \text{if } t\geq 5 \end{cases} \end{equation} When does the colony die out? Show that at some time between $t=2$ and $t=7,$ the population is $9,000.$

**Solution**. Since $$ \lim_{t\to 5^+}P(t)=\lim_{t\to 5^+}(-8t+66)=26 $$ and $$ \lim_{t\to 5^-}P(t)=\lim_{t\to 5^-}\left(t^2+1\right)=26$$ we know that $P$ is continuous at $t=5$ and thus, also for all $t\geq 0.$ (a) The colony dies out when $-8t+66=0$ which means $t=33/4\approx 8.25.$ Therefore, the colony dies out in 8 minutes and 15 seconds. (b) Since $P(2)=5,$ $P(7)=10,$ and $P$ is continuous on $(2,7),$ the intermediate value theorem yields at least one number $c$ between 2 and 7 such $P(c)=9.$ Therefore, there is some time $t=c$ between $t=2$ and $t=7$ such that the population is 9,000.

## Exercises on Continuous

**Exercise**. Sketch the graph of the function \begin{equation} f(x)= \begin{cases} x^2-1 & -1\leq x<0 \\ 2x & 0<x<1 \\ 1 & x=1 \\ -2x+4 & 1<x<2 \\ 0 & 2<x<3. \end{cases} \end{equation} Use the graph of the function $f$ to answer the following.

$(a) \quad$ Does $f(-1)$ exist?

$(b) \quad$ Does the limit, $\lim_{x\to -1^+}f(x)$ exist?

$(c) \quad$ Does the limit, $\lim_{x\to -1^+}f(x)=f(-1)$ ?

$(d) \quad$ Does $f(1)$ exist?

$(d) \quad$ Does the limit, $\lim_{x\to 1}f(x)$ exist?

$(e) \quad$ Does the limit, $\lim_{x\to 1}f(x)=f(1)$ exist?

$(f) \quad$ Is $f$ defined at $x=2?$

$(g) \quad$ Is $f$ continuous at $x=2?$

$(h) \quad$ At what values is $f$ continuous?

$(i) \quad$ What value should be assigned to $f(2)$ , to make the extended function continuous at $x=2?$

$(j) \quad$ To what new value should $f(1)$ be assigned to remove the discontinuity?

**Exercise**. For each of the following functions determine the largest set on which the function will be continuous.

$(1) \quad$ $\displaystyle g(x)=\frac{x+1}{x^2-4x+3}$

$(2) \quad$ $\displaystyle g(x)=\frac{1}{|x|+1}-\frac{x^2}{2}$

$(3) \quad$ $\displaystyle g(x)=\frac{\sqrt{x^4+1}}{1+\sin ^2x}$

$(4) \quad$ $\displaystyle g(x)=\sqrt[4]{3x-1}$

**Exercise**. For each of the following functions find constants $a$ and $b$ so that the function will be continuous for all $x$ in the domain.

$(1) \quad$ $f(x)=\begin{cases} \displaystyle \frac{\tan a x}{\tan b x} & x<0 \\ 4 & x=0 \\ a x+b & x>0 \end{cases}$

$(2) \quad$ $f(x)=\begin{cases} \displaystyle \frac{\sqrt{a x+b}-1}{x} & x\neq 0 \\ 1 & x=0 \end{cases}$

$(3) \quad$ $f(x)=\begin{cases} a x+3 & x<1 \\ 5 & x=1 \\ x^2+b & x>1\end{cases}$

**Exercise**. Let $u(x)=x$ and $$ f(x)= \begin{cases} 0 & x\neq 0 \\ 1 & x=0. \end{cases} $$ Show, for this given $u$ and $f$ , that \begin{equation} \lim_{x\to 0}f[u(x)]\neq f\left(\lim_{x\to 0}u(x)\right). \end{equation}

**Exercise**. If a function $f$ is not continuous at $x=c$ , but can be made continuous at $x=c$ by being assigned a new value of that point, it is said to have a ** removable discontinuity** at $x=c.$ Which of the following functions have a removable discontinuity at $x=c$ ?

$(1) \quad$ $\displaystyle f(x)=\frac{2x^2+x-15}{x+3}$ at $c=-3$

$(2) \quad$ $\displaystyle f(x)=\frac{x-2}{|x-2|}$ at $c=2$

$(3) \quad$ $\displaystyle f(x)=\frac{2-\sqrt{x}}{4-x}$ at $c=4$

$(4) \quad$ $\displaystyle f(x)=\frac{2-x}{4-\sqrt{x}}$ at $c=16$

**Exercise**. Prove that the function $f(x)=x^3-x^2+x+1$ must have at least one real root.

**Exercise**. Prove that the function $(x)=\sqrt{x+3}-e^x$ must have at least one real root.

**Exercise**. Find a function(s) with the following properties. (a) Find functions $f$ and $g$ such that $f$ is discontinuous at $x=1$ but $f g$ is continuous there. (b) Give an example of a function defined for all real numbers that is continuous at only one point.