Continuity: Learn It 1

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Continuity: Learn It 1

Outline the three criteria a function must meet to be continuous at a specific point
Explain the different types of breaks a function can have that make it not continuous
Explain what it means for a function to be continuous over a range of values
Explain the rule for calculating limits of functions that are combined
Show how a continuous function reaches every value between its start and end points using the Intermediate Value Theorem

Continuity

Many functions can be traced without lifting your pencil, indicating they are continuous. Functions that cannot be traced this way have points of discontinuity. To understand continuity at a point, consider the following conditions:

The function $f(a)$ is defined:
- The function must have a value at $a$ .
The limit $\underset{x\to a}{\lim}f(x)$ exists:
- The value that $f(x)$ approaches as $x$ gets closer to $a$ must exist.
The limit equals the function value $\underset{x\to a}{\lim}f(x)=f(a)$ :
- The value that $f(x)$ approaches must be the same as the value of the function at $a$ .

Let’s illustrate these conditions:

Undefined Function:
- If $f(a)$ is not defined, the function is not continuous at $a$ .
  
  Figure 1. The function $f(x)$ is not continuous at a because $f(a)$ is undefined.
Non-existent Limit:
- Even if $f(a)$ is defined, if $\underset{x\to a}{\lim}f(x)$ does not exist, the function is not continuous at $a$ .
  
  Figure 2. The function $f(x)$ is not continuous at a because $\underset{x\to a}{\lim}f(x)$ does not exist.
Limit Not Equal to Function Value:
- If $f(a)$ is defined and $\underset{x\to a}{\lim}f(x)$ exists, but they are not equal, the function is not continuous at $a$ .
  
  Figure 3. The function $f(x)$ is not continuous at a because $\underset{x\to a}{\lim}f(x)\ne f(a)$ .

continuity

A function $f(x)$ is continuous at a point $a$ if and only if the following three conditions are satisfied:

$f(a)$ is defined
$\underset{x\to a}{\lim}f(x)$ exists
$\underset{x\to a}{\lim}f(x)=f(a)$

A function is discontinuous at a point $a$ if it fails to be continuous at $a$ .

The following procedure can be used to analyze the continuity of a function at a point using this definition.

How to: Determine Continuity at a Point

Check to see if f(a) $f (a)$ is defined.
- If $f(a)$ is undefined, we need go no further. The function is not continuous at $a$ .
- If $f(a)$ is defined, continue to step 2.
Compute limx→af(x) $lim x \to a f (x)$ . In some cases, we may need to do this by first computing limx→a−f(x) $lim x \to a^{-} f (x)$ and limx→a+f(x) $lim x \to a^{+} f (x)$ .
- If $\underset{x\to a}{\lim}f(x)$ does not exist (that is, it is not a real number), then the function is not continuous at $a$ .
- If $\underset{x\to a}{\lim}f(x)$ exists, then continue to step 3.
Compare f(a) $f (a)$ and limx→af(x) $lim x \to a f (x)$ .
- If $\underset{x\to a}{\lim}f(x)\ne f(a)$ , then the function is not continuous at $a$ .
- If $\underset{x\to a}{\lim}f(x)=f(a)$ , then the function is continuous at $a$ .

The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.

Using the definition, determine whether the function $f(x)=\dfrac{(x^2-4)}{(x-2)}$ is continuous at $x=2$ . Justify the conclusion.

Let’s begin by trying to calculate $f(2)$ . We can see that $f(2)=\frac{0}{0}$ , which is undefined. Therefore, $f(x)=\frac{x^2-4}{x-2}$ is discontinuous at 2 because $f(2)$ is undefined. The graph of $f(x)$ is shown in Figure 4.

A graph of the given function. There is a line which crosses the x axis from quadrant three to quadrant two and which crosses the y axis from quadrant two to quadrant one. At a point in quadrant one, there is an open circle where the function is not defined. — Figure 4. The function $f(x)$ is discontinuous at 2 because $f(2)$ is undefined.

Using the definition, determine whether the function $f(x)=\begin{cases} -x^2+4 & \text{ if } \, x \le 3 \\ 4x-8 & \text{ if } \, x > 3 \end{cases}$ is continuous at $x=3$ . Justify the conclusion.

Let’s begin by trying to calculate $f(3)$ .

f (3) = - (3^{2}) + 4 = - 5

$f(3)=-(3^2)+4=-5$

Thus, $f(3)$ is defined. Next, we calculate $\underset{x\to 3}{\lim}f(x)$ . To do this, we must compute $\underset{x\to 3^-}{\lim}f(x)$ and $\underset{x\to 3^+}{\lim}f(x)$ :

lim x \to 3^{-} f (x) = - (3^{2}) + 4 = - 5

$\underset{x\to 3^-}{\lim}f(x)=-(3^2)+4=-5$

and

lim x \to 3^{+} f (x) = 4 (3) - 8 = 4

$\underset{x\to 3^+}{\lim}f(x)=4(3)-8=4$

Therefore, $\underset{x\to 3}{\lim}f(x)$ does not exist. Thus, $f(x)$ is not continuous at $3$ . The graph of $f(x)$ is shown in Figure 5.

Graph showing the discontinuous function f(x)

Using the definition, determine whether the function $f(x)=\begin{cases} \frac{\sin x}{x} & \text{ if } \, x \ne 0 \\ 1 & \text{ if } \, x = 0 \end{cases}$ is continuous at $x=0$ .

First, observe that

f (0) = 1

$f(0)=1$

Next,

lim x \to 0 f (x) = lim x \to 0 \frac{sin x}{x} = 1

$\underset{x\to 0}{\lim}f(x)=\underset{x\to 0}{\lim}\frac{\sin x}{x}=1$

Last, compare $f(0)$ and $\underset{x\to 1}{\lim}f(x)$ . We see that

f (0) = 1 = lim x \to 0 f (x)

$f(0)=1=\underset{x\to 0}{\lim}f(x)$

Since all three of the conditions in the definition of continuity are satisfied, $f(x)$ is continuous at $x=0$ .

Watch the following video to see the worked solutions to the three previous examples.

By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.

continuity of polynomials and rational functions

Polynomials and rational functions are continuous at every point in their domains.

Proof

Previously, we showed that if $p(x)$ and $q(x)$ are polynomials, $\underset{x\to a}{\lim}p(x)=p(a)$ for every polynomial $p(x)$ and $\underset{x\to a}{\lim}\dfrac{p(x)}{q(x)}=\dfrac{p(a)}{q(a)}$ as long as $q(a)\ne 0$ . Therefore, polynomials and rational functions are continuous on their domains.

$_\blacksquare$

The domain of every polynomial function is all real numbers.

The domain of a rational functional can be found by:

Set the denominator equal to zero.
Solve to find the values of the variable that cause the denominator to equal zero.
The domain contains all real numbers except those found in Step 2.

We now apply continuity of polynomials and rational functions to determine the points at which a given rational function is continuous.

For what values of $x$ is $f(x)=\dfrac{x+1}{x-5}$ continuous?

The rational function $f(x)=\frac{x+1}{x-5}$ is continuous for every value of $x$ except $x=5$ .