# 1.7: Continuity

**At Grade**Created by: CK-12

## Learning Objectives

A student will be able to:

- Learn to examine continuity of functions.
- Find one-sided limits.
- Understand properties of continuous functions.
- Solve problems using the Min-Max theorem.
- Solve problems using the Intermediate Value Theorem.

## Introduction

In this lesson we will discuss the property of continuity of functions and examine some very important implications. Let’s start with an example of a rational function and observe its graph. Consider the following function:

\begin{align*}f(x)= (x + 1)/(x^2 -1).\end{align*}

We know from our study of domains that in order for the function to be defined, we must use \begin{align*}x \neq -1, 1.\end{align*} Yet when we generate the graph of the function (using the standard viewing window), we get the following picture that appears to be defined at \begin{align*}x = -1\end{align*}:

The seeming contradiction is due to the fact that our original function had a common factor in the numerator and denominator, \begin{align*}x + 1,\end{align*} that cancelled out and gave us a picture that appears to be the graph of \begin{align*}f(x)= 1/(x -1).\end{align*}

But what we actually have is the original function, \begin{align*}f(x)= (x +1)/(x^2 -1),\end{align*} that we know is not defined at \begin{align*}x = -1.\end{align*} At \begin{align*}x = -1,\end{align*} we have a hole in the graph, or a discontinuity of the function at \begin{align*}x =-1.\end{align*} That is, the function is defined for all other \begin{align*}x-\end{align*}values close to \begin{align*}x = -1.\end{align*}

Loosely speaking, if we were to hand-draw the graph, we would need to take our pencil off the page when we got to this hole, leaving a gap in the graph as indicated:

Now we will formalize the property of continuity of a function and provide a test for determining when we have continuous functions.

*Continuity of a Function*

**Definition:**

The function \begin{align*}f(x)\end{align*} is ** continuous at** \begin{align*}x = a\end{align*} if the following conditions all hold:

- \begin{align*}a\end{align*} is in the domain of \begin{align*}f(x)\end{align*};
- \begin{align*}\lim_{x \to a} f(x)\end{align*} exists;
- \begin{align*}\lim_{x \to a} f(x) = f(a)\end{align*}

Note that it is possible to have functions where two of these conditions are satisfied but the third is not. Consider the piecewise function

\begin{align*}f(x) = \begin{cases} x, {\rm if}\ x \neq 1\\ 3, {\rm if}\ x = 1\\ \end{cases}\end{align*}

In this example we have \begin{align*}\lim_{x \to 1} f(x)\end{align*} exists, \begin{align*}x = 1\end{align*} is in the domain of \begin{align*}f(x)\end{align*}, but \begin{align*}\lim_{x \to 1} f(x) \neq f(1)\end{align*}.

*One-Sided Limits and Closed Intervals*

Let’s recall our basic square root function, \begin{align*}f(x)=\sqrt{x}\end{align*}.

Since the domain of \begin{align*}f(x)=\sqrt{x}\end{align*} is \begin{align*}x \ge 0\end{align*}, we see that that \begin{align*}\lim_{x \to 0}\sqrt{x}\end{align*} does not exist. Specifically, we cannot find open intervals around \begin{align*}x = 0\end{align*} that satisfy the limit definition. However we do note that as we approach \begin{align*}x = 0\end{align*} from the right-hand side, we see the successive values tending towards \begin{align*}x = 0\end{align*}. This example provides some rationale for how we can define ** one-sided limits**.

**Definition:**

We say that the *right-hand**limit* of a function \begin{align*}f(x)\end{align*} at \begin{align*}a\end{align*} is \begin{align*}b\end{align*}, written as \begin{align*}\lim_{x \to a^+} f(x)=b\end{align*}, if for every open interval \begin{align*}N\end{align*} of \begin{align*}b\end{align*}, there exists an open interval \begin{align*}(a, a + \delta)\end{align*} contained in the domain of \begin{align*}f(x),\end{align*} such that \begin{align*}f(x)\end{align*} is in \begin{align*}N\end{align*} for every \begin{align*}x\end{align*} in \begin{align*}(a, a + \delta).\end{align*}

For the example above, we write \begin{align*} \lim_{x \to 0^+}\end{align*} \begin{align*} \sqrt{x}=0.\end{align*}

Similarly, we say that the ** left-hand limit** of \begin{align*}f(x)\end{align*} at \begin{align*}a\end{align*} is \begin{align*}b\end{align*}, written as \begin{align*}\lim_{x \to a^-} f(x)=b\end{align*}, if for every open interval \begin{align*}N\end{align*} of \begin{align*}b\end{align*} there exists an open interval \begin{align*}(a - \delta, a)\end{align*} contained in the domain of \begin{align*}f(x),\end{align*} such that \begin{align*}f(x)\end{align*} is in \begin{align*}N\end{align*} for every \begin{align*}x\end{align*} in \begin{align*}(a - \delta, a).\end{align*}

**Example 1:**

Find \begin{align*}\lim_{x \to 0^+} \frac {x}{|x|}.\end{align*}

The graph has a discontinuity at \begin{align*}x = 0\end{align*} as indicated:

We see that \begin{align*}\lim_{x \to 0^+} \frac {x}{|x|} = 1\end{align*} and also that \begin{align*}\lim_{x \to 0^-} \frac {x}{|x|}= -1\end{align*}.

*Properties of Continuous Functions*

Let’s recall our example of the limit of composite functions:

\begin{align*}f(x)= 1/(x + 1),\ g(x) = -1.\end{align*}

We saw that \begin{align*}f(g(x))\end{align*} is undefined and has the indeterminate form of \begin{align*}1/0\end{align*}. Hence \begin{align*}\lim_{x \to -1} (f \circ g)(x)\end{align*} does not exist.

In general, we will require that \begin{align*}f\end{align*} be continuous at \begin{align*}x = g(a)\end{align*} and \begin{align*}x = g(a)\end{align*} must be in the domain of \begin{align*}(f \circ g)\end{align*} in order for \begin{align*}\lim_{x \to a} (f \circ g)(x)\end{align*} to exist.

We will state the following theorem and delay its proof until Chapter 3 when we have learned more about real numbers.

** Min-Max Theorem**: If a function \begin{align*}f(x)\end{align*} is continuous in a closed interval \begin{align*}I\end{align*}, then \begin{align*}f(x)\end{align*} has both a maximum value and a minimum value in \begin{align*}I\end{align*}.

**Example 2:**

Consider \begin{align*}f(x) = x^3 + 1\end{align*} and interval \begin{align*}I = [-2,2].\end{align*}

The function has a minimum value at value at \begin{align*}x = -2,\end{align*} \begin{align*}f(-2) = -7,\end{align*} and a maximum value at \begin{align*}x = 2,\end{align*} where \begin{align*}f(2) = 9\end{align*}

We will conclude this lesson with a theorem that will enable us to solve many practical problems such as finding zeros of functions and roots of equations.

*Intermediate Value Theorem*

If a function is continuous on a closed interval \begin{align*}[a, b],\end{align*} then the function assumes every value between \begin{align*}f(a)\end{align*} and \begin{align*}f(b)\end{align*}.

The proof is left as an exercise with some hints provided. (Homework #10).

We can use the Intermediate Value Theorem to analyze and approximate zeros of functions.

**Example 3:**

Use the Intermediate Value Function to show that there is at least one zero of the function in the indicated interval.

\begin{align*}f(x) = 3x^4 - 3x^3 - 2x + 1,\ (1,2)\end{align*}

We recall that the graph of this function is shaped somewhat like a parabola; viewing the graph in the standard window, we get the following graph:

Of course we could zoom in on the graph to see that the lowest point on the graph lies within the fourth quadrant, but let’s use the **[CALC VALUE]** function of the calculator to verify that there is a zero in the interval \begin{align*}(1, 2).\end{align*} In order to apply the Intermediate Value Theorem, we need to find a pair of \begin{align*}x-\end{align*}values that have function values with different signs. Let’s try some in the table below.

\begin{align*}x && f(x) \\ 1.1 && -.80 \\ 1.2 && -.36 \\ 1.3 && .37\end{align*}

We see that the sign of the function values changes from negative to positive somewhere between \begin{align*}1.2\end{align*} and \begin{align*}1.3\end{align*}. Hence, by the Intermediate Value theorem, there is some value \begin{align*}c\end{align*} in the interval \begin{align*}(1.2,1.3)\end{align*} such that \begin{align*}f(c) = 0\end{align*}.

## Lesson Summary

- We learned to examine continuity of functions.
- We learned to find one-sided limits.
- We observed properties of continuous functions.
- We solved problems using the Min-Max theorem.
- We solved problems using the Intermediate Value Theorem.

## Multimedia Links

For a presentation of continuity using limits **(2.0)**, see Math Video Tutorials by James Sousa, Continuity Using Limits (5:44).

For a video presentation of the Intermediate Value Theorem **(3.0)**, see Just Math Tutoring, Intermediate Value Theorem (7:53).

## Review Questions

- Generate the graph of \begin{align*}f(x)= (|x + 1|)/(x + 1)\end{align*} using your calculator and discuss the continuity of the function.
- Generate the graph of \begin{align*}f(x)= (3x - 6)/(x^2 - 4)\end{align*} using your calculator and discuss the continuity of the function.

Compute the limits in #3 - 6.

- \begin{align*}\lim_{x \to 0^+}\frac{\sqrt{x}}{\sqrt{1+\sqrt{x}}-1}\end{align*}
- \begin{align*}\lim_{x \to 2^-} \frac{x^3 - 8} {|x - 2| (x - 2)}\end{align*}
- \begin{align*}\lim_{x \to 1^+} \frac{2x|x - 1|} {x - 1}\end{align*}
- \begin{align*}\lim_{x \to -2^-} \frac{|x + 2| + x + 2} {|x + 2| - x - 2}\end{align*}

In problems 7 and 8, explain how you know that the function has a root in the given interval. (Hint: Use the Intermediate Value Function to show that there is at least one zero of the function in the indicated interval.):

- \begin{align*} f(x) = x^3 + 2x^2 - x + 1\end{align*}, in the interval \begin{align*}(-2, -3)\end{align*}
- \begin{align*}f(x) = \sqrt{x} - \sqrt[3]{x} - 1\end{align*}, in the interval \begin{align*}(9, 10)\end{align*}
- State whether the indicated \begin{align*}x-\end{align*}values correspond to maximum or minimum values of the function depicted below.
- Prove the Intermediate Value Theorem: If a function is continuous on a closed interval \begin{align*}[a, b]\end{align*}, then the function assumes every value between \begin{align*}f(a)\end{align*} and \begin{align*}f(b)\end{align*}.

## Review Answers

- The graph of \begin{align*}f(x)\end{align*} indicates the function is not continuous at \begin{align*}x = -1\end{align*}.
- While the graph of the function appears to be continuous everywhere but \begin{align*}x=-2\end{align*}, a check of the table values indicates that the function is also not continuous at \begin{align*}x=2\end{align*}.
- \begin{align*}\lim_{x \to 0^+}\frac{\sqrt{x}}{\sqrt{1+\sqrt{x}}-1}=2\end{align*}
- \begin{align*}\lim_{x \to 2^-} \frac{x^3 - 8} {|x - 2| (x - 2)}\end{align*} does not exist
- \begin{align*}\lim_{x \to 1^+} \frac{2x|x - 1|} {(x - 1)} = 2\end{align*}
- \begin{align*}\lim_{x \to -2^-} \frac{|x + 2| + x + 2} {|x + 2| - x - 2} = 0\end{align*}
- \begin{align*}f(-2.5) = .375,\end{align*} \begin{align*}f(-2.9) = -3.669.\end{align*} By the Intermediate Value Theorem, there is an \begin{align*}x-\end{align*}value \begin{align*}c\end{align*} with \begin{align*}f(c) = 0\end{align*}.
- \begin{align*}f(9.1) = -.071,\end{align*} \begin{align*}f(9.99) = .006.\end{align*} By the Intermediate Value Theorem, there is an \begin{align*}x-\end{align*}value \begin{align*}c\end{align*} with \begin{align*}f(c) = 0\end{align*}.
- \begin{align*}x = a\end{align*} is a relative maximum, \begin{align*}x = b\end{align*} is an absolute minimum, \begin{align*}x = c\end{align*} is an absolute maximum and \begin{align*}x = d\end{align*} is not a maximum nor a minimum.
- Here is an outline of the proof: we need to show that for every number d between \begin{align*}f(a)\end{align*} and \begin{align*}f(b)\end{align*}, there exists a number such that \begin{align*}f(c) = d\end{align*}.
- Assume that \begin{align*}f(a) < f(c) < f(b)\end{align*}.
- Let \begin{align*}S\end{align*} be the set of \begin{align*}x \ \epsilon \ [a, b]>\end{align*} for which \begin{align*}f(x) < d\end{align*}. Note that \begin{align*}a \ \epsilon \ S, b \ \epsilon \ S\end{align*}. so \begin{align*}b\end{align*} is an upper bound for set \begin{align*}S\end{align*}. Hence by the completeness property of the reals, \begin{align*}S\end{align*} has an upper bound, \begin{align*}c\end{align*}.
- There are then three possibilities to explore: \begin{align*}f(c) < d, f(c) = d\end{align*}, or \begin{align*}f(c) > d\end{align*}. Explore these and show why \begin{align*}f(c) = d\end{align*}.

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