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# 1.7: Continuity

Difficulty Level: 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:

$f(x)= (x + 1)/(x^2 -1).$

We know from our study of domains that in order for the function to be defined, we must use $x \neq -1, 1.$ 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 $x = -1$:

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

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

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 $f(x)$ is continuous at $x = a$ if the following conditions all hold:

1. $a$ is in the domain of $f(x)$;
2. $\lim_{x \to a} f(x)$ exists;
3. $\lim_{x \to a} f(x) = f(a)$

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

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

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

One-Sided Limits and Closed Intervals

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

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

Definition:

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

For the example above, we write $\lim_{x \to 0^+}$ $\sqrt{x}=0.$

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

Example 1:

Find $\lim_{x \to 0^+} \frac {x}{|x|}.$

The graph has a discontinuity at $x = 0$ as indicated:

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

Properties of Continuous Functions

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

$f(x)= 1/(x + 1),\ g(x) = -1.$

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

In general, we will require that $f$ be continuous at $x = g(a)$ and $x = g(a)$ must be in the domain of $(f \circ g)$ in order for $\lim_{x \to a} (f \circ g)(x)$ 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 $f(x)$ is continuous in a closed interval $I$, then $f(x)$ has both a maximum value and a minimum value in $I$.

Example 2:

Consider $f(x) = x^3 + 1$ and interval $I = [-2,2].$

The function has a minimum value at value at $x = -2,$ $f(-2) = -7,$ and a maximum value at $x = 2,$ where $f(2) = 9$

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 $[a, b],$ then the function assumes every value between $f(a)$ and $f(b)$.

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.

$f(x) = 3x^4 - 3x^3 - 2x + 1,\ (1,2)$

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 $(1, 2).$ In order to apply the Intermediate Value Theorem, we need to find a pair of $x-$values that have function values with different signs. Let’s try some in the table below.

$x && f(x) \\1.1 && -.80 \\1.2 && -.36 \\1.3 && .37$

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

## Lesson Summary

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

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

1. Generate the graph of $f(x)= (|x + 1|)/(x + 1)$ using your calculator and discuss the continuity of the function.
2. Generate the graph of $f(x)= (3x - 6)/(x^2 - 4)$ using your calculator and discuss the continuity of the function.

Compute the limits in #3 - 6.

1. $\lim_{x \to 0^+}\frac{\sqrt{x}}{\sqrt{1+\sqrt{x}}-1}$
2. $\lim_{x \to 2^-} \frac{x^3 - 8} {|x - 2| (x - 2)}$
3. $\lim_{x \to 1^+} \frac{2x|x - 1|} {x - 1}$
4. $\lim_{x \to -2^-} \frac{|x + 2| + x + 2} {|x + 2| - x - 2}$

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.):

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

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

Feb 23, 2012

Feb 26, 2015

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CK.MAT.ENG.SE.1.Calculus.1.7