Continuity

One way to look at continuity is to start with the technical definition, and then do a bunch of examples to see what is or isn't continuous using the definition. Instead, we will begin with an intuitive understanding of continuous functions and build what we would like to see in the definition.

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Last Updated: 01.06.2022
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Intuitively Continuous

You have probably heard someone say that "a function is continuous if you can draw it without picking up your pencil". So let's look at what needs to be true for that to happen, starting at a single point p.

If you try and draw this function without picking up your pencil you can't because the function has a hole at p. In fact, it isn't even defined at p! So you are definitely going to need to assume that the function you are thinking about is defined at the point p, and this is done by saying "assume $f (p)$ exists".

Continuity function not defined StudySmarter This function is not defined at the point p | StudySmarter Originals

But is the function having a value at p enough to make sure you can draw it without picking up your pencil? Let's look at another case.

This function is defined at p, but the limit from the left and the right aren't the same. In other words

$\lim_{x \to p^{-}} f (x) \neq \lim_{x \to p +} f (x)$

$\lim_{x \to p} f (x)$

does not exist. So, the limit as x approaches p must exist as well!

Continuity limits from left and right not same StudySmarter The limit from the left and right are not the same | StudySmarter Originals

So even having the function defined isn't enough. You also need the limit from the left and right to have the same value. But is that enough to make sure you can draw it without picking up your pencil? Let's see!

The function below is defined at p. The limit exists as x approaches p. But it isn't equal to the function value! You can write these two values not being the same as

$\lim_{x \to p} f (x) \neq f (p) .$

Continuity limit not same as function value StudySmarter The function value isn't the same as the limit of the function | StudySmarter Originals

Now let's put all of that together into a definition.

Continuity Definition

The function $f (x)$ is continuous at the point p if and only if all the following three things are true:

$f (p)$ exists

2. $\lim_{x \to p} f (x)$ exists (the limit from the left and right are equal)

3. $\lim_{x \to p} f (x) = f (p) .$

If a function fails any of those three conditions, then $f (x)$ is said to be discontinuous at p, or simply not continuous at p.

The expression "if and only if" is a biconditional logic statement, meaning if A is true, then B is true, and if B is true, then A is true.

Easy Steps to check if a Function is Continuous

You can use the definition to make a step-by-step process to check and see if a function is continuous at p.

Step 1: Make sure the function is defined at p. If it isn't, stop because the function definitely isn't continuous at p.

Step 2: Make sure that $\lim_{x \to p} f (x)$ exists. If it doesn't, you can stop because the function definitely isn't continuous at p.

Step 3: Make sure that the limit and the function value are equal. If they aren't, then the function definitely isn't continuous at p.

Note that sometimes if a function is discontinuous at a point, people will say that it has a discontinuity at that point. These two phrases mean the same thing.

Continuous Function Examples

Let's practice determining if a function is continuous at a certain point!

Decide whether or not the function

f (x) = \frac{x + 2}{x - 2}

is continuous at $x = 2$ .

Answer:

If you try and evaluate the function $f (x)$ at 2, you get division by zero. So, in fact, this function is not defined at $x = 2$ , so it can't be continuous at that point either. You can graph the function to see that there is a vertical asymptote at $x = 2$ , which is why it isn't defined there.

Continuity function not continuous at vertical asymptote StudySmarter This function is not continuous at x=2 | StudySmarter Originals

So the difficulty with the previous example was that the function wasn't defined when $x = 2$ . Suppose, instead, your function is defined by

$f (x) = \{\begin{cases} \frac{x + 2}{x - 2}, & x \neq 2 \\ 7, & x = 2 \end{cases}$

which is definitely defined at $x = 2 .$ Is this function continuous at $x = 2$ ?

Answer:

In this case

$\lim_{x \to 2^{-}} f (x) = - \infty$

but

$\lim_{x \to 2 +} f (x) = \infty$ ,

so the limit doesn't exist at $x = 2$ . Therefore, even though the function is defined when $x = 2$ , it is not continuous there.

Let's rig the previous example some more. If the problem is that the limit from the left and right aren't the same, you can change the function a bit to see what happens. Take

$f (x) = \{\begin{cases} \frac{x + 2}{x - 2}, & x > 2 \\ 7, & x = 2 \\ - \frac{x + 2}{x - 2} & x < 2 \end{cases}$

which is still defined when

x = 2

. Now is the function continuous at

x = 2 ?

Answer:

Now when you look at the limit as $x$ approaches 2, you have

$\lim_{x \to 2^{-}} f (x) = \infty = \lim_{x \to 2 +} f (x) .$

But $f (2) = 7$ , which is certainly not infinity! So the function is still not continuous at $x = 2$ .

Decide whether the function

f (x) = \{\begin{cases} - x^{2} + 9, & x \leq 2 \\ x + 3, & x > 2 \end{cases}

is continuous when $x = 3$ .

Answer:

The trick here is to read the question carefully. We don't necessarily look at the point where the function changes definition, we look at what the question is asking!

In this case, the function changes definition at x=2, but we are asked if it is continuous at x=3. So all you need to consider is if the function $g (x) = x + 3$ is continuous at $x = 3$ . But this is just a line, so you know

\lim_{x \to 3} g (x) = 6 = g (3)

and the function $f (x)$ is continuous at $x = 3$ .

Let's work it out step by step:

Step 1: Make sure the function is defined at $x = 3$ . $g (3) = 6$ . Therefore, it is defined.

Step 2: Make sure the limit at $x = 3$ exists. That is, check if the limit from the left and right of $x = 3$ are equal.

$\lim_{x \to 3^{-}} g (x) = 6$ and $\lim_{x \to 3^{+}} g (x) = 6$ .

The limits from the left and right of $x = 3$ are indeed equal.

Step 3: Lastly, check if the limit is equal to the function value at $x = 3$ . That is:

$\begin{array}{rcl} \lim_{x \to 3} g (x) & = & g (3) \\ 6 & = & 6 \end{array}$

This last condition is satisfied. Therefore, the function is continuous at $x = 3$ .

In the previous example, the point p wasn't where the function had a switch in which formula was used. What if, instead, the point you cared about was $p = 2$ ?

Decide whether the function

$f (x) = \{\begin{cases} - x^{2} + 9, & x \leq 2 \\ x + 3, & x > 2 \end{cases}$

is continuous at the point $p = 2$ .

Answer:

Let's follow the same steps as the previous example.

Step 1: Check if the function is defined at $p = 2$ .

$f (2) = - 2^{2} + 9 = - 4 + 9 = 5$ .

Therefore, the function is defined at $p = 2$ .

Step 2: Now you check if the limit exists. The limit from the left gives

$\lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{-}} (- x^{2} + 9) = 5,$

and the limit from the right is

$\lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} (x + 3) = 5,$

which means that

$\lim_{x \to 2} f (x) = 5 .$

Step 3: Lastly, check if the function value from step 1 and limit from step 2 are equal. They both equal 5!

So, you've checked:

That the function is defined at the point,
The limit of the function exists at that point, and
The function value at that point has the same value as the limit.

Therefore, the function $f (x)$ is continuous at $p = 2$ .

What if we changed the function in the previous example slightly?

Decide whether the function

$f (x) = \{\begin{cases} - x^{2} + 9, & x \leq 2 \\ x + 1, & x > 2 \end{cases}$

is continuous at the point $p = 2$ .

Answer:

Step 1:

Just like in the previous example, $f (2) = 5$

Step 2: Check left and right-hand limits:

The limit from the left:

$\lim_{x \to 2^{-}} f (x) = 5 .$

But now the limit from the right is

$\lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} (x + 1) = 2,$

$\lim_{x \to 2} f (x)$

does not exist. Therefore the function is not continuous at the point $p = 2$ .

Here, since the criterion in step 2 is not met, we don't have to continue on to step 3!

Continuity in Calculus

Why should you care whether or not a function is continuous? Suppose that you are modeling a population with x measured in years, and you find that the formula for it is given by

$f (x) = \{\begin{cases} - x^{2} + 9, & x \leq 2 \\ x + 1, & x > 2 \end{cases}$

which from the work you did above is not continuous at $p = 2$ . Taking a look at the graph of this function,

Continuity piecewise function which is not continuous StudySmarter A piecewise function which is not continuous at p=2 | StudySmarter Originals

Knowing that the function is not continuous at $p = 2$ lets you know that something drastic happened to the population you're studying. In this case, it is a sudden die-off, which is a problem you would want to investigate.

Types of Continuity

Here you have looked explicitly at continuity at a point. But what about continuity over an interval, or even the whole real line? For information on intervals, see Continuity over an Interval, and for more theorems on continuity, see Theorems of Continuity.

Continuity - Key takeaways

Intuitively, a function is continuous if you can draw it without picking up your pencil.
The function $f (x)$ is continuous at the point $x = p$ if and only if the function is defined at $x = p$ , the limit of the function exists at $x = p$ , and the function value and the limit at $x = p$ both have the same value.
A function that is not continuous at the point $x = p$ is said to be discontinuous.
A function fails to be continuous at the point x = pif
- the function isn't defined there, or
- if the limit doesn't exist there,
- or if the limit and the function value aren't the same there

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Frequently Asked Questions about Continuity

What is continuity?

Continuity is when the limit as x approaches p of a function is the same as the function value at p.

What is an example of continuity?

Polynomials are functions which are continuous everywhere.

What is the concept of continuity?

The idea of continuity is that you can draw the function without picking up your pencil. In other words the function doesn't have a gap or a jump at the point in question.

What does continuous mean in math?

Continuity is when the limit as x approaches p of a function is the same as the function value at p.

What is a continuous graph in math?

Intuitively it means you can draw it without picking up your pencil. In other words the limit as x approaches p of a function is the same as the function value at p.

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