Lesson Explainer: Continuity at a Point Mathematics • Higher Education

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 π‘Žβˆˆβ„. We say that a real-valued function 𝑓(π‘₯) is continuous at π‘₯=π‘Ž if limο—β†’οŒΊπ‘“(π‘₯)=𝑓(π‘Ž).

A useful property of continuity at π‘₯=π‘Ž is that we can sketch the graph of 𝑓(π‘₯) near π‘₯=π‘Ž 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 π‘₯=π‘Ž, then we say 𝑓(π‘₯) is discontinuous at π‘₯=π‘Ž.

We can then ask the question, β€œhow can a function fail to be continuous at π‘₯=π‘Ž?” For this to happen, we need the equation limο—β†’οŒΊπ‘“(π‘₯)=𝑓(π‘Ž) to not hold true. There are three ways this can happen:

  1. 𝑓(π‘Ž) is undefined;
  2. limο—β†’οŒΊπ‘“(π‘₯) does not exist;
  3. limο—β†’οŒΊπ‘“(π‘₯)≠𝑓(π‘Ž).

Therefore, to determine the continuity of a function at π‘₯=π‘Ž, we must check all three of these properties.

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

To check if the function 𝑓(π‘₯) is continuous at π‘₯=π‘Ž, we need to check whether the following three conditions hold.

  1. 𝑓 must be defined at π‘Ž (π‘Ž is in the domain of 𝑓).
  2. limο—β†’οŒΊπ‘“(π‘₯) must exist. (This is equivalent to saying both the left and right limits of 𝑓(π‘₯) at π‘₯=π‘Ž exist and are equal.)
  3. limο—β†’οŒΊπ‘“(π‘₯)and 𝑓(π‘Ž) must have the same value.

For example, let’s check the continuity of 𝑓(π‘₯)=|π‘₯| at π‘₯=π‘Ž.

First, we know that 𝑓(0)=|0|=0, so π‘₯=0 is in the domain of 𝑓.

Second, we need to determine limο—β†’οŠ¦|π‘₯|. 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: |π‘₯|=ο­βˆ’π‘₯π‘₯<0,π‘₯π‘₯β‰₯0.

We can use this to evaluate the left and right limits. First, when evaluating limο—β†’οŠ¦οŽͺ|π‘₯|, the values of π‘₯ are all negative, so |π‘₯|=βˆ’π‘₯ in this limit. We can then evaluate this limit by direct substitution: limlimο—β†’οŠ¦ο—β†’οŠ¦οŽͺοŽͺ|π‘₯|=(βˆ’π‘₯)=0.

Similarly, for the right limit, the values of π‘₯ are all positive, so limlimο—β†’οŠ¦ο—β†’οŠ¦οŽ©οŽ©|π‘₯|=π‘₯=0.

Thus, we have shown that limlimο—β†’οŠ¦ο—β†’οŠ¦οŽͺ|π‘₯|=|π‘₯|=0.

Therefore, the left and right limits of |π‘₯| at 0 are equal, so limο—β†’οŠ¦|π‘₯|=0.

Third, we have found the values of 𝑓(0) and limο—β†’οŠ¦π‘“(π‘₯) and shown that both of these are equal to the same value, 0.

Since all three conditions hold, we have shown that 𝑓(π‘₯)=|π‘₯| 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 𝑓 at π‘₯=πœ‹2, given 𝑓(π‘₯)=ο΄βˆ’7π‘₯+7π‘₯,π‘₯β‰€πœ‹2,62π‘₯βˆ’1,π‘₯>πœ‹2.sincoscos

Answer

For a function 𝑓(π‘₯) to be continuous at π‘Ž, we need three conditions to hold:

  1. 𝑓 must be defined at π‘Ž (π‘Ž is in the domain of 𝑓).
  2. limο—β†’οŒΊπ‘“(π‘₯) must exist.
  3. limο—β†’οŒΊπ‘“(π‘₯) and 𝑓(π‘Ž) must have the same value.

In our case π‘Ž=πœ‹2, we can see from the definition of 𝑓 that π‘“ο€»πœ‹2=βˆ’7ο€»πœ‹2+7ο€»πœ‹2=βˆ’7+0=βˆ’7.sincos

Therefore, πœ‹2 is in the domain of 𝑓 and π‘“ο€»πœ‹2=βˆ’7. So, the first condition for continuity at π‘₯=πœ‹2 holds.

To check the second condition for continuity, we will check whether the left and right limits of 𝑓 at π‘₯=πœ‹2 both exist and are equal. We start with the left limit and note that when π‘₯β‰€πœ‹2, we have 𝑓(π‘₯)=βˆ’7π‘₯+7π‘₯sincos, giving us limlimsincosο—β†’ο—β†’ο‘½οŽ‘οŽͺο‘½οŽ‘οŽͺ𝑓(π‘₯)=(βˆ’7π‘₯+7π‘₯).

Since this is a trigonometric expression, we can evaluate this limit by direct substitution: limsincossincosο—β†’ο‘½οŽ‘οŽͺβˆ’7π‘₯+7π‘₯=βˆ’7ο€»πœ‹2+7ο€»πœ‹2=βˆ’7+0=βˆ’7.

We can do the same for the right limit where we note that when π‘₯>πœ‹2, we have that 𝑓(π‘₯)=62π‘₯βˆ’1cos, giving us limlimcosο—β†’ο—β†’ο‘½οŽ‘οŽ©ο‘½οŽ‘οŽ©π‘“(π‘₯)=(62π‘₯βˆ’1).

We can then evaluate this limit by direct substitution: limcoscosο—β†’ο‘½οŽ‘οŽ©(62π‘₯βˆ’1)=6ο€»2ο€»πœ‹2ο‡ο‡βˆ’1=βˆ’6βˆ’1=βˆ’7.

Therefore, both the left and right limits exist and are equal to βˆ’7, so we have shown limο—β†’ο‘½οŽ‘π‘“(π‘₯)=βˆ’7.

Hence, the second continuity condition of 𝑓 at π‘₯=πœ‹2 also holds.

Finally, since this is also equal to π‘“ο€»πœ‹2, we can conclude that the third continuity condition also holds and so the function 𝑓 is continuous at π‘₯=πœ‹2.

Let’s 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 Function’s Domain

Given 𝑓(π‘₯)=π‘₯+π‘₯βˆ’2π‘₯βˆ’1, if possible or necessary, define 𝑓(1) so that 𝑓 is continuous at π‘₯=1.

Answer

For a function 𝑓(π‘₯) to be continuous at a we need three conditions to hold:

  1. 𝑓 must be defined at π‘Ž (π‘Ž is in the domain of 𝑓).
  2. limο—β†’οŒΊπ‘“(π‘₯) must exist.
  3. limο—β†’οŒΊπ‘“(π‘₯) and 𝑓(π‘Ž) must have the same value.

We are allowed to define 𝑓(1) to be any value we want, so the first condition is guaranteed to hold.

To determine limο—β†’οŠ§π‘“(π‘₯), we could try direct substitution since 𝑓 is a rational function; however, 𝑓(1)=1+1βˆ’21βˆ’1=00, 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, limlimlimο—β†’οŠ§ο—β†’οŠ§οŠ¨ο—β†’οŠ§οŽͺοŽͺοŽͺ𝑓(π‘₯)=ο€Ύπ‘₯+π‘₯βˆ’2π‘₯βˆ’1=ο€½(π‘₯βˆ’1)(π‘₯+2)π‘₯βˆ’1.

Since we are taking the limit as π‘₯ approaches 1 from the left, the values of π‘₯ are never equal to 1, so we can cancel the shared factor of π‘₯βˆ’1; this will not change the value of this limit: limlimο—β†’οŠ§ο—β†’οŠ§οŽͺοŽͺο€½(π‘₯βˆ’1)(π‘₯+2)π‘₯βˆ’1=(π‘₯+2).

Since this is a polynomial, we can then evaluate this by direct substitution: limο—β†’οŠ§οŽͺ(π‘₯+2)=1+2=3.

We can evaluate the right limit in the same way: limlimlimlimο—β†’οŠ§ο—β†’οŠ§οŠ¨ο—β†’οŠ§ο—β†’οŠ§οŽ©οŽ©οŽ©οŽ©π‘“(π‘₯)=ο€Ύπ‘₯+π‘₯βˆ’2π‘₯βˆ’1=ο€½(π‘₯βˆ’1)(π‘₯+2)π‘₯βˆ’1=(π‘₯+2)=3.

Since the left and right limits both exist and are both equal to 3, we can conclude limο—β†’οŠ§π‘“(π‘₯)=3.

For the third condition to be true, we need 𝑓(1) to be equal to limο—β†’οŠ§π‘“(π‘₯). We have shown that this limit evaluates to give us three, so we must define 𝑓(1)=3. This will then guarantee that all three conditions for continuity at this point hold.

Hence, 𝑓(1)=3 makes 𝑓 continuous at π‘₯=1.

In our next example, we will determine whether we can make the reciprocal function continuous at π‘₯=0 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 𝑓(π‘₯)=1π‘₯, if possible or necessary, define 𝑓(0) so that 𝑓 is continuous at π‘₯=0.

Answer

For a function 𝑓(π‘₯) to be continuous at π‘Ž, we need three conditions to hold:

  1. 𝑓 must be defined at π‘Ž (π‘Ž is in the domain of 𝑓).
  2. limο—β†’οŒΊπ‘“(π‘₯) must exist.
  3. limο—β†’οŒΊπ‘“(π‘₯) and 𝑓(π‘Ž) must have the same value.

We are told that we can define 𝑓(0) to be any value we want, which will allow the first condition to be true.

Since 𝑓 is a rational function, we can try evaluating the limit of 𝑓(π‘₯) as π‘₯ approaches 0 by direct substitution: limο—β†’οŠ¦π‘“(0)=10, 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 𝑦=1π‘₯.

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 π‘₯ approaches zero from the left, the outputs will approach negative infinity.

As expected, the left and right limits of 𝑓(π‘₯) at 0 do not exist andwe conclude that limο—β†’οŠ¦π‘“(π‘₯) does not exist. However, this means that the second condition for continuity of 𝑓(π‘₯) at 0 does not hold.

Hence, the function cannot be made continuous at π‘₯=0 by defining 𝑓(0) as limο—β†’οŠ¦π‘“(π‘₯) 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 𝑓(π‘₯)=⎧βŽͺ⎨βŽͺ⎩2π‘₯+4π‘₯+2π‘₯<βˆ’2,0π‘₯=βˆ’2,π‘₯+6π‘₯+8π‘₯+2π‘₯>βˆ’2ififif continuous at π‘₯=βˆ’2?

Answer

We recall that for 𝑓(π‘₯) to be continuous at βˆ’2, we need three conditions to hold:

  1. 𝑓 must be defined at βˆ’2.
  2. limο—β†’οŠ±οŠ¨π‘“(π‘₯) must exist.
  3. limο—β†’οŠ±οŠ¨π‘“(π‘₯) and 𝑓(βˆ’2) must have the same value.

From the definition of 𝑓(π‘₯), we can see that 𝑓(βˆ’2)=0, 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 π‘₯<βˆ’2, we have 𝑓(π‘₯)=2π‘₯+4π‘₯+2, so limlimο—β†’οŠ±οŠ¨ο—β†’οŠ±οŠ¨οŽͺοŽͺ𝑓(π‘₯)=ο€Ό2π‘₯+4π‘₯+2.

Attempting to evaluate this by direct substitution gives an undefined value, so instead we simplify the limit: limlimlimο—β†’οŠ±οŠ¨ο—β†’οŠ±οŠ¨ο—β†’οŠ±οŠ¨οŽͺοŽͺοŽͺο€Ό2π‘₯+4π‘₯+2=ο€½2(π‘₯+2)π‘₯+2=(2)=2.

We can do the same to evaluate the right limit: limlimlimlimο—β†’οŠ±οŠ¨ο—β†’οŠ±οŠ¨οŠ¨ο—β†’οŠ±οŠ¨ο—β†’οŠ±οŠ¨οŽ©οŽ©οŽ©οŽ©π‘“(π‘₯)=ο€Ύπ‘₯+6π‘₯+8π‘₯+2=ο€½(π‘₯+2)(π‘₯+4)π‘₯+2=(π‘₯+4)=2.

Since both the left and right limits exist and are equal to 2, we have limο—β†’οŠ±οŠ¨π‘“(π‘₯)=2.

However, this is not equal to 𝑓(βˆ’2) which is 0; hence, the third condition for continuity at the point π‘₯=βˆ’2 does not hold.

Therefore, 𝑓(π‘₯) is discontinuous at π‘₯=βˆ’2 because limο—β†’οŠ±οŠ¨π‘“(π‘₯) and 𝑓(βˆ’2) 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 𝑐 that make the function 𝑓 continuous at π‘₯=𝑐 if 𝑓(π‘₯)=2+π‘₯π‘₯≀𝑐,βˆ’3π‘₯π‘₯>𝑐.ifif

Answer

We recall that for 𝑓(π‘₯) to be continuous at 𝑐, we need three conditions to hold:

  1. π‘₯=𝑐 must be in the domain of 𝑓.
  2. limο—β†’οŒΌπ‘“(π‘₯) must exist.
  3. limο—β†’οŒΌπ‘“(π‘₯) and 𝑓(𝑐) must have the same value.

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

From the definition of 𝑓(π‘₯), we can see that 𝑓(𝑐)=2+𝑐, so any real value of 𝑐 is in the domain of 𝑓 which tells us the first condition holds.

To check the second condition, we will need to evaluate the left and right limits of 𝑓 at 𝑐. First, since 𝑓(π‘₯)=2+π‘₯ when π‘₯≀𝑐, limlimο—β†’οŒΌο—β†’οŒΌοŠ¨οŠ¨οŽͺοŽͺ𝑓(π‘₯)=ο€Ή2+π‘₯=2+𝑐.

Second, since 𝑓(π‘₯)=βˆ’3π‘₯ when π‘₯>𝑐, limlimο—β†’οŒΌο—β†’οŒΌοŽ©οŽ©π‘“(π‘₯)=(βˆ’3π‘₯)=βˆ’3𝑐.

We need both of these limits to be equal, giving us the equation 2+𝑐=βˆ’3𝑐.

We can solve for 𝑐 by factoring the quadratic: 2+𝑐=βˆ’3𝑐𝑐+3𝑐+2=0(𝑐+1)(𝑐+2)=0.

Hence, 𝑐=βˆ’1 and 𝑐=βˆ’2 are the only real values of 𝑐 such that the second condition holds true.

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

If 𝑐=βˆ’1, 𝑓(βˆ’1)=2+(βˆ’1)=3𝑓(π‘₯)=2+(βˆ’1)=3.οŠ¨ο—β†’οŠ±οŠ§οŠ¨andlim

If 𝑐=βˆ’2, 𝑓(βˆ’2)=2+(βˆ’2)=6𝑓(π‘₯)=2+(βˆ’2)=6.οŠ¨ο—β†’οŠ±οŠ¨οŠ¨andlim

Hence, 𝑓 is only continuous at 𝑐 when 𝑐=βˆ’1 or 𝑐=βˆ’2.

Let’s finish by recapping some of the important points of this explainer.

Key Points

  • We say that a function 𝑓(π‘₯) is continuous at π‘Ž if limο—β†’οŒΊπ‘“(π‘₯)=𝑓(π‘Ž).
  • If a function is continuous at π‘Ž, then we can evaluate its limit at π‘Ž by direct substitution.
  • We can check the continuity of 𝑓(π‘₯) at π‘Ž by checking whether three conditions hold:
    • 𝑓 must be defined at π‘Ž (π‘Ž is in the domain of 𝑓).
    • limο—β†’οŒΊπ‘“(π‘₯) must exist (this is equivalent to saying both the left and right limits of 𝑓(π‘₯) at π‘₯=π‘Ž exist and are equal).
    • limο—β†’οŒΊπ‘“(π‘₯) and 𝑓(π‘Ž) must have the same value.

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