Continuity is a relatively simple concept, but problems that require proving it can be a little tricky. Essentially, a continuous function is one that you can draw all in one motion without picking up your pencil. This is one explanation of what it means for a function to be continuous that I like because it doesn’t take any mathematical definitions or proofs to understand. Any holes or gaps in a function’s graph would be a discontinuity and would mean that the function is not continuous.

## The limit definition of continuity

By definition, a function is continuous at if $$\lim_{x \to a} f(x) = f(a).$$

Let’s think about what this equation is saying. The left side of this equation is something that we’ve already dealt with, limits. It’s asking us to find out what value we close in on as we travel along our function and close in on . Keep in mind, is a variable here which represents the input of our function, , and is a constant. This means that represents some specific number, which could be any number.

The right side of the equation is simply asking us to plug that same value into and take the value we get out.

In total, the above equation says that if we travel along a function and close in on a specific value, we should close in on the same value we would get if we simply plugged that value into the equation.

In other words, as we travel along a function toward a specific value, our value will also go toward the value of the function at that point. If that is the case, then the function is continuous at that specific point. If we can say that a function is continuous at every single possible value we could put in for , then we can say that the function is continuous for all . If this is true then we can draw our entire function in one motion without picking up our pencil!

Let’s do a couple examples.

## Example 1

Remember back in the first lesson about limits, Limits – Intro, I said I would go back to discussing the importance of our limit in the first example giving us the same value as when we plugged into the equation? I would like to go into that further.

The function we were considering was and we were finding

$$\lim_{x \to 2} x^2.$$

After looking at the graph of this function, shown in Figure 1.1, we saw that

$$\lim_{x \to 2} x^2 = 4.$$

I also pointed out that plugging in directly into the function also returns a value of . In other words, we know $$f(2) = 4.$$ Therefore, we know

$$\lim_{x \to 2} x^2 = 4 = f(2)$$

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

Notice this is exactly like the definition of what it means for a function to be continuous at a point. If you replace with , replace with , and replace with , we have just shown that is a function that is continuous at

## Example 2

The next example I want to discuss also goes back to a function we have already looked at. Referring back to Limits – Intro we will consider shown in Figure 1.2.

The question we will answer here is whether this function is continuous at or not. What do you think?

There are a few different ways we can answer this question. The simplest would be to simply look at the graph of the function and think about whether we can draw that function at and around without picking up our pencil. As you can see, there is clearly a hole at where we would need to pick up our pencil, and add a single point at .

As a result, it is probably safe to say that this function is not continuous at . However, we want to be able to show this using the actual definition of what it means for a function to be continuous.

Remember, for this function, which we are calling in this case, we need to be able to show that $$\lim_{x \to 1} f(x) = f(1).$$

If we can show this equation to be true, then is continuous at , and if it’s not true then the function is not continuous at . Luckily, back when I first used this function as an example in the Limits – Intro lesson, we found that $$\lim_{x \to 1} f(x) = 2.$$ Therefore, we just need to find out if is also and we can prove that this function is continuous or not continuous.

Looking at the graph again, we see that this function has a hole at and includes the point . In other words, if we plug into as our value, we get a value of out. This is the same as saying .

Now, at this point we have figured out $$\lim_{x \to 1} f(x) = 2 \ and$$ $$f(1) = 4.$$

Therefore, $$\lim_{x \to 1} f(x) \neq f(1)$$ and we can say that is not continuous at .

## More Examples

Find the values of and that make continuous everywhere.

$$f(x) = \begin{cases} \frac{x^2-4}{x-2} & \mbox{if } x<2 \\ ax^2-bx+3 & \mbox{if } 2\leq x<3 \\ 2x-a+b & \mbox{if } x\geq 3 \end{cases}$$

To see the solution to this problem, click here.

**I also have a calc 1 study guide that you can get now for FREE! Just enter your name and email below and I’ll send you a copy straight to your inbox!**

There are also several other lessons and problems on my limits page. It would be a good idea to get some practice with limits. A lot of other more complex topics in calculus are based around limits so they are important to understand. If you can’t find the topic you want to read about just let me know by emailing me at jakesmathlessons@gmail.com and I’ll do my best to answer your question!

How would I do a problem like this?

Find the numbers at which f is discontinuous:

f(x): x+1 if x is less than or equal to 1

1/x if 1<x<3

root of x-3 if x is greater than or equal to 3

Typically with a problem like that you want to consider this in two different parts. First figure out if each of those pieces of the piecewise function is continuous for the entire domain on which it’s defined. You want to make sure that you wouldn’t divide by 0, take an even root of a negative number, or take a logarithm of a non-positive number. Once you confirm that each of those pieces is continuous on their own little domain, you need to make sure the piecewise function is continuous at the x values where you go from one piece to the next. This would be done using the limit definition of continuity at a point and set up each one sided limit separately.

Thank you so much!!