In this explainer, we will learn how to check the continuity of a function at a given point.

Evaluating the limit of a function at a point is a useful way to gain information about what happens to the outputs of that function near (but not at) the point and is an integral part of calculus. In other words, a limit tells us the value the function approaches as the inputs get closer and closer to some point or number.

For some functions, evaluating these limits can be difficult and may require rewriting the limit in a different form. However, we can often evaluate these limits by directly substituting this point into our function. When this holds true, we call the function continuous at this point.

### Definition: Continuity of a Function at a Point

Let . We say that a real-valued function is continuous at if

A useful property of continuity at is that we can sketch the graph of near without lifting the pen off the paper.

To study the continuity of functions at a point, we first introduce the notion of discontinuity. If a function does not satisfy the definition of continuity at , then we say is discontinuous at .

We can then ask the question, βhow can a function fail to be continuous at ?β For this to happen, we need the equation to not hold true. There are three ways this can happen:

- is undefined;
- does not exist;
- .

Therefore, to determine the continuity of a function at , we must check all three of these properties.

### How To: Checking Whether a Function Is Continuous at a Point

To check if the function is continuous at , we need to check whether the following three conditions hold.

- must be defined at ( is in the domain of ).
- must exist. (This is equivalent to saying both the left and right limits of at exist and are equal.)
- and must have the same value.

For example, letβs check the continuity of at .

First, we know that , so is in the domain of .

Second, we need to determine . To evaluate this limit, we recall that we can determine whether a limit exists by checking if the left and right limits at this point both exist and are equal. We need to recall the piecewise definition of the modulus function:

We can use this to evaluate the left and right limits. First, when evaluating , the values of are all negative, so in this limit. We can then evaluate this limit by direct substitution:

Similarly, for the right limit, the values of are all positive, so

Thus, we have shown that

Therefore, the left and right limits of at 0 are equal, so

Third, we have found the values of and and shown that both of these are equal to the same value, 0.

Since all three conditions hold, we have shown that is continuous at 0.

In our first example, we will determine the continuity of a piecewise-defined function at the endpoints of its subdomains.

### Example 1: Discussing the Continuity of a Piecewise-Defined Function Involving Trigonometric Ratios at a Point

Discuss the continuity of the function at , given

### Answer

For a function to be continuous at , we need three conditions to hold:

- must be defined at ( is in the domain of ).
- must exist.
- and must have the same value.

In our case , we can see from the definition of that

Therefore, is in the domain of and . So, the first condition for continuity at holds.

To check the second condition for continuity, we will check whether the left and right limits of at both exist and are equal. We start with the left limit and note that when , we have , giving us

Since this is a trigonometric expression, we can evaluate this limit by direct substitution:

We can do the same for the right limit where we note that when , we have that , giving us

We can then evaluate this limit by direct substitution:

Therefore, both the left and right limits exist and are equal to , so we have shown

Hence, the second continuity condition of at also holds.

Finally, since this is also equal to , we can conclude that the third continuity condition also holds and so the function is continuous at .

Letβs consider an example where we use the definition of continuity at a point to determine how we can extend a function to be continuous at a value outside of its domain.

### Example 2: Using Continuity to Find the Value Needed to Extend a Functionβs Domain

Given , if possible or necessary, define so that is continuous at .

### Answer

For a function to be continuous at a we need three conditions to hold:

- must be defined at ( is in the domain of ).
- must exist.
- and must have the same value.

We are allowed to define to be any value we want, so the first condition is guaranteed to hold.

To determine , we could try direct substitution since is a rational function; however, so we cannot evaluate this limit by direct substitution. Instead, we will check that the left and right limits both exist and are equal. First,

Since we are taking the limit as approaches 1 from the left, the values of are never equal to 1, so we can cancel the shared factor of ; this will not change the value of this limit:

Since this is a polynomial, we can then evaluate this by direct substitution:

We can evaluate the right limit in the same way:

Since the left and right limits both exist and are both equal to 3, we can conclude

For the third condition to be true, we need to be equal to . We have shown that this limit evaluates to give us three, so we must define . This will then guarantee that all three conditions for continuity at this point hold.

Hence, makes continuous at .

In our next example, we will determine whether we can make the reciprocal function continuous at by extending its domain.

### Example 3: Determining Whether It Is Possible to Extend the Domain of the Reciprocal Function So That It Is Continuous

Given , if possible or necessary, define so that is continuous at .

### Answer

For a function to be continuous at , we need three conditions to hold:

- must be defined at ( is in the domain of ).
- must exist.
- and must have the same value.

We are told that we can define to be any value we want, which will allow the first condition to be true.

Since is a rational function, we can try evaluating the limit of as approaches 0 by direct substitution: since this is undefined, we cannot evaluate this limit by direct substitution. Instead, we will check whether the left and right limits both exist and are equal. We can see why these limits do not exist by considering the graph of .

As the input values of our function approach zero from the right, the denominator of the fraction becomes a smaller and smaller positive number; it grows without bound. Similarly, as approaches zero from the left, the outputs will approach negative infinity.

As expected, the left and right limits of at 0 do not exist andwe conclude that does not exist. However, this means that the second condition for continuity of at 0 does not hold.

Hence, the function cannot be made continuous at by defining as does not exist.

In our next example, we will check the continuity of a piecewise-defined function at a point.

### Example 4: Determining the Continuity of a Function at a Point

Is continuous at ?

### Answer

We recall that for to be continuous at , we need three conditions to hold:

- must be defined at .
- must exist.
- and must have the same value.

From the definition of , we can see that so the first condition holds.

To check whether the second condition holds, we will show that both the left and right limits exist and are equal. First, when , we have , so

Attempting to evaluate this by direct substitution gives an undefined value, so instead we simplify the limit:

We can do the same to evaluate the right limit:

Since both the left and right limits exist and are equal to 2, we have

However, this is **not** equal to which is 0; hence, the third condition for continuity at the point does not hold.

Therefore, is discontinuous at because and have different values.

**Solution**: No

In our final example, we will use the definition of continuity to determine the possible values of a variable that allow the function to be continuous at that point.

### Example 5: Finding the Values of a Variable That Make a Piecewise-Defined Function Continuous

Find the values of that make the function continuous at if

### Answer

We recall that for to be continuous at , we need three conditions to hold:

- must be in the domain of .
- must exist.
- and must have the same value.

All three of these conditions must be true for the values of we are looking for. We can check each condition separately, starting with the first condition.

From the definition of , we can see that so any real value of is in the domain of which tells us the first condition holds.

To check the second condition, we will need to evaluate the left and right limits of at . First, since when ,

Second, since when ,

We need both of these limits to be equal, giving us the equation

We can solve for by factoring the quadratic:

Hence, and are the only real values of such that the second condition holds true.

We still need to check whether these values allow the third condition to hold true.

If ,

If ,

Hence, is only continuous at when or .

Letβs finish by recapping some of the important points of this explainer.

### Key Points

- We say that a function is continuous at if
- If a function is continuous at , then we can evaluate its limit at by direct substitution.
- We can check the continuity of at by checking whether three conditions hold:
- must be defined at ( is in the domain of ).
- must exist (this is equivalent to saying both the left and right limits of at exist and are equal).
- and must have the same value.