Continuous (It’s Meaning and Applications)

Continuity is one of the essential concepts in all of the calculus; indeed, all of mathematics. So in this article, I discuss continuous functions and the cases where a discontinuity may occur. I also present one-sided continuity and the removability of a continuity. In other words, how to extend a function so that it has no discontinuities.

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

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 continuous from the left at $c$ if and only if $$ \lim_{x\to c^-}f(x)=f(c). $$

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.

David A. Smith at Dave4Math

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.

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