Explainer: Continuity of Functions

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 𝑦=𝑓(π‘₯). As we move along the π‘₯-axis and get closer to the value π‘₯=1 from the left (the solid blue arrow),

we see that the values of 𝑓(π‘₯) approach 𝑦=1 (the dashed green arrow). But as we approach from the right,

the values of 𝑓(π‘₯) get closer and closer to 𝑦=2, a different number.

The actual value of the function is indicated by the solid dot at (1,3) which tells us that 𝑓(1)=3.

In summary, we have limlimο—β†’οŠ§ο—β†’οŠ§οŽͺοŽ©π‘“(π‘₯)=1,𝑓(π‘₯)=2,𝑓(1)=3 and can conclude the following:

  1. The limit of this function as π‘₯β†’1 does not exist (because the two one-sided limits disagree).
  2. Therefore, the function is not continuous at π‘₯=1.

In a single sentence, we can say this function is discontinuous at π‘₯=1 because the limit limο—β†’οŠ§π‘“(π‘₯) does not exist.

The fact that 𝑓(1)=3 plays no role in this deduction.

In the next example, again from a graph, the function 𝑔 is not defined at all at π‘₯=1. But, as the blue arrows indicate, moving in from either the left or the right of π‘₯=1, it is apparent that the limiting values of 𝑔(π‘₯) are 1.5.

This time, we have limlim:undefinedο—β†’οŠ§ο—β†’οŠ§οŽͺοŽ©π‘”(π‘₯)=1.5,𝑔(π‘₯)=1.5,𝑔(1).

Our conclusion: firstly, we have that limο—β†’οŠ§π‘”(π‘₯) exists and equals 1.5, but the function 𝑔 is discontinuous at π‘₯=1 since 𝑔(1) is undefined.

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

Definition: Continuity at a Point

A function 𝑓 is continuous at π‘₯=π‘Ž provided that

  1. 𝑓(π‘Ž) is defined,
  2. limο—β†’οŒΊπ‘“(π‘₯) exists,
  3. limο—β†’οŒΊπ‘“(π‘₯)=𝑓(π‘Ž).

In a way, this kind of discontinuity is β€œbetter” than the previous example. We can repair it by making the necessary definition: if we set 𝑔(1)=1.5, then 𝑔 becomes a continuous function at π‘₯=1. We say that 𝑔 has a removable discontinuity at the point π‘₯=1. The previous function 𝑓 did not have a removable discontinuity at π‘₯=1.

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

Let 𝑓(π‘₯)=ο­π‘Žπ‘₯+18π‘₯β‰ 9,βˆ’6π‘Žπ‘₯=9.ifif Find the value of π‘Ž that makes 𝑓 continuous at π‘₯=9.

Answer

Instead of a graph, we are given formulas for the function on the two intervals. To decide continuity at the boundary π‘₯=9, we would normally calculate the two one-sided limits and compare. But here, we are given that the same formula applies: 𝑓(π‘₯)=π‘Žπ‘₯+18 whether π‘₯<9 or π‘₯>9.

Further, since π‘Žπ‘₯+18 is an example of a polynomial, we know that the limit at every point is just the value there. Polynomials are continuous everywhere: limlimο—β†’οŠ―ο—β†’οŠ―οŠ¨οŠ¨π‘“(π‘₯)=π‘Žπ‘₯+18=π‘Žο€Ή9+18=81π‘Ž+18.

To be continuous, we need the following identity to hold: limorο—β†’οŠ―π‘“(π‘₯)=𝑓(9)81π‘Ž+18=βˆ’6π‘Ž87π‘Ž=βˆ’18π‘Ž=βˆ’1887=βˆ’629.

Evidently, other values of π‘Ž would produce nonremovable discontinuities from the definition of 𝑓 above. π‘Ž=βˆ’629 is exactly what is needed to make 𝑓 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 limο—β†’οŠ±οŠ―οŽͺ𝑓(π‘₯) and limο—β†’οŠ±οŠ―οŽ©π‘“(π‘₯), where 𝑓(π‘₯)=78π‘₯<βˆ’9,βˆ’9π‘₯βˆ’7π‘₯β‰₯βˆ’9.ifif

  1. limlimο—β†’οŠ±οŠ―ο—β†’οŠ±οŠ―οŽͺοŽ©π‘“(π‘₯)=74,𝑓(π‘₯)=78
  2. limlimο—β†’οŠ±οŠ―ο—β†’οŠ±οŠ―οŽͺοŽ©π‘“(π‘₯)=78,𝑓(π‘₯)=74
  3. limlimο—β†’οŠ±οŠ―ο—β†’οŠ±οŠ―οŽͺοŽ©π‘“(π‘₯)=78,𝑓(π‘₯)=78
  4. limlimο—β†’οŠ±οŠ―ο—β†’οŠ±οŠ―οŽͺοŽ©π‘“(π‘₯)=74,𝑓(π‘₯)=74

Answer

The definitions of 𝑓 to the left and right of π‘₯=βˆ’9 are different. However, they are both given by polynomials, which are actually defined at π‘₯=βˆ’9, of course. By the theorem above, we have limlimconstantfunctionο—β†’οŠ±οŠ―ο—β†’οŠ±οŠ―οŽͺοŽͺ𝑓(π‘₯)=78=78() while limlimpolynomialfunctionο—β†’οŠ±οŠ―ο—β†’οŠ±οŠ―οŽ©οŽ©π‘“(π‘₯)=βˆ’9π‘₯βˆ’7=βˆ’9(βˆ’9)βˆ’7()=74.

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 𝑓(π‘₯)=7π‘₯π‘₯β‰€βˆ’3,7π‘₯+3π‘₯>βˆ’3.ifif

What can be said of the continuity of 𝑓 at π‘₯=βˆ’7?

  1. The function is continuous at π‘₯=βˆ’7.
  2. The function is discontinuous at π‘₯=βˆ’7 because 𝑓(βˆ’7) is undefined.
  3. The function is discontinuous at π‘₯=βˆ’7 because limο—β†’οŠ±οŠ­π‘“(π‘₯) does not exist.
  4. The function is discontinuous at π‘₯=βˆ’7 because limο—β†’οŠ±οŠ­π‘“(π‘₯)≠𝑓(βˆ’7).

Answer

(A)

Although this function is indeed defined piecewise, the special pointβ€”where the boundaries of the two intervals meetβ€”is π‘₯=βˆ’3, not the π‘₯=βˆ’7 that the question is asking about! So 𝑓(π‘₯)=7π‘₯ at and around βˆ’7, because βˆ’7β‰€βˆ’3. Therefore,

  1. the value is 𝑓(βˆ’7)=(βˆ’7)(βˆ’3)=21. The function is defined there, so (B) does not hold;
  2. since 𝑓(π‘₯)=βˆ’7π‘₯ is a polynomial function, the limit exists at all points, so (C) is false;
  3. more, polynomial functions are continuous on their domainβ€”the whole number lineβ€”so (D) is false;
  4. in fact, (A) holds.

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