In this explainer, we will learn how to check the continuity of a function over its domain and determine the interval on which it is continuous.

A function is continuous at a point provided that the two one-sided limits as you approach that point agree with the value at that point (and therefore with each other). This is not hard to check in the special case where the function has a piecewise definition and we are looking at the point where two pieces meet.

A graph makes this clear. Consider the following graph of . As we move along the -axis and get closer to the value from the left (the solid blue arrow),

we see that the values of approach (the dashed green arrow). But as we approach from the right,

the values of get closer and closer to , a different number.

The actual value of the function is indicated by the solid dot at which tells us that .

In summary, we have and can conclude the following:

- The limit of this function as does not exist (because the two one-sided limits disagree).
- Therefore, the function is not continuous at .

In a single sentence, we can say this function is discontinuous at because the limit does not exist.

The fact that plays no role in this deduction.

In the next example, again from a graph, the function is not defined at all at . But, as the blue arrows indicate, moving in from either the left or the right of , it is apparent that the limiting values of are 1.5.

This time, we have

Our conclusion: firstly, we have that exists and equals 1.5, but the function is discontinuous at since is undefined.

This is because the definition of continuity at a point requires three things.

### Definition: Continuity at a Point

A function is continuous at provided that

- is defined,
- exists,
- .

In a way, this kind of discontinuity is βbetterβ than the previous example. We can repair it by making the necessary definition: if we set , then becomes a continuous function at . We say that has a removable discontinuity at the point . The previous function did not have a removable discontinuity at .

### Example 1: Finding the Values of a Variable Which Make a Piecewise-Defined Function Continuous at a Certain Point

Let Find the value of that makes continuous at .

### Answer

Instead of a graph, we are given formulas for the function on the two intervals. To decide continuity at the boundary , we would normally calculate the two one-sided limits and compare. But here, we are given that the same formula applies: whether or .

Further, since is an example of a polynomial, we know that the limit at every point is just the value there. Polynomials are continuous everywhere:

To be continuous, we need the following identity to hold:

Evidently, other values of would produce nonremovable discontinuities from the definition of above. is exactly what is needed to make continuous.

Some of the questions in this lesson exercise the computation of limits. The main result we use is the following.

### Theorem: Continuity of Standard Functions

The following classes of functions are continuous at all points in their domains:

- polynomial functions (including constants),
- rational functions,
- trigonometric functions,
- exponential functions.

### Example 2: Finding the One-Sided Limits of a Piecewise-Defined Function

Find and , where

### Answer

The definitions of to the left and right of are different. However, they are both given by polynomials, which are actually defined at , of course. By the theorem above, we have while

The correct choice is (B).

The following example asks for what can be said of the continuity/discontinuity at a point. Solving it is just a careful application of the definition and theorem above.

### Example 3: The Continuity of a Piecewise-Defined Function at a Given Point

Let

What can be said of the continuity of at ?

- The function is continuous at .
- The function is discontinuous at because is undefined.
- The function is discontinuous at because does not exist.
- The function is discontinuous at because .

### Answer

(A)

Although this function is indeed defined piecewise, the special pointβwhere the boundaries of the two intervals meetβis , not the that the question is asking about! So at and around , because . Therefore,

- the value is . The function is defined there, so (B) does not hold;
- since is a polynomial function, the limit exists at all points, so (C) is false;
- more, polynomial functions are continuous on their domainβthe whole number lineβso (D) is false;
- in fact, (A) holds.