Lesson Explainer: Existence of Limits Mathematics • Higher Education

In this explainer, we will learn how to determine whether the limit of a function at a certain value exists.

The limit of a function at a point gives us useful information about the shape of the function at that point and is a fundamental tool in calculus. We recall the definition of the limit of a function at a point.

Recap: Limit of a Function at a Point

If the values of 𝑓(π‘₯) approach some value, 𝐿, as the values of π‘₯ approach π‘Ž (from both sides) but not necessarily when π‘₯=π‘Ž, then we say the limit of 𝑓(π‘₯) as π‘₯ approaches π‘Ž is equal to 𝐿 and we denote this as limο—β†’οŒΊπ‘“(π‘₯)=𝐿.

By using the notation of one-sided limits, we can rewrite this definition in terms of the left and the right limit of the function at this point.

Definition: Limit of a Function at a Point in terms of One-Sided Limits

Let 𝐿,π‘Žβˆˆβ„. If limο—β†’οŒΊοŽ©π‘“(π‘₯)=𝐿 and limο—β†’οŒΊοŽͺ𝑓(π‘₯)=𝐿, then limο—β†’οŒΊπ‘“(π‘₯)=𝐿.

When this definition holds true for a function 𝑓(π‘₯) at a value of π‘₯=π‘Ž, this gives us information about the shape of the function around this point; we would know it is arbitrarily close to the point (π‘Ž,𝐿). Therefore, we can check the limit of a function at a point by evaluating its left and right limit at this point and checking whether they are equal.

We can use this idea of the existence of limits to write another definition of the limit of a function at a point.

Definition: Limit of a Function at a Point

If the left and the right limit of a function 𝑓(π‘₯) at π‘₯=π‘Ž both exist and are equal to some value πΏβˆˆβ„, then limο—β†’οŒΊπ‘“(π‘₯)=𝐿.

It might not be possible to find the limit of a function at a point. For example, consider the graph of 𝑦=𝑓(π‘₯) shown in the diagram below.

To evaluate limο—β†’οŠ§π‘“(π‘₯), we need to evaluate both the left and the right limit at π‘₯=1 separately. First, as the values of π‘₯ approach 1 from the left, the outputs of the function stay constant at 1, so limο—β†’οŠ§οŽͺ𝑓(π‘₯)=1. Second, as the values of π‘₯ approach 1 from the right, the outputs of the function approach 2, so limο—β†’οŠ§οŽ©π‘“(π‘₯)=2. Since the left and the right limit of the function at π‘₯=1 are not equal, we cannot evaluate limο—β†’οŠ§π‘“(π‘₯).

When we cannot evaluate the limit of a function at a point, we say that its limit does not exist. In the above example, we saw that this can happen if the left and the right limit are not equal; however, there are other ways for the limit of a function at a point to not exist. Consider limο—β†’οŠ¦οŠ¨1π‘₯; one way of investigating this limit is to sketch the function.

We need to consider the left and the right limit of 1π‘₯ as π‘₯ approaches 0.

From the diagram, we can see that in both cases the outputs of the function are growing without bound. In particular, this means that the outputs are not approaching any finite value, πΏβˆˆβ„. When the outputs of the function grow without bound, we often say that the limit is equal to ∞. However, it is important to remember that this limit does not exist and saying that the limit is equal to ∞ is a shorthand for saying the limit does not exist but the outputs grow without bound.

We can say something similar for 𝑦=βˆ’1π‘₯ at π‘₯=0.

As π‘₯ approaches 0 from either side, the outputs of the function decrease without bound. We can say that limο—β†’οŠ¦οŠ¨ο€Όβˆ’1π‘₯=βˆ’βˆž, where, once again, this means the limit does not exist.

Another similar possibility is the limit of 1π‘₯ at π‘₯=0.

If our values of π‘₯ approach 0 from the right, then the outputs of the function are growing without bound. Therefore, we can say limο—β†’οŠ¦οŽ©1π‘₯=∞. However, if the values of π‘₯ approach 0 from the left, then the outputs of the function are decreasing without bound. So, we can say limο—β†’οŠ¦οŽͺ1π‘₯=βˆ’βˆž. In this case, the left and the right limits of 1π‘₯ at 0 are not equal, so we say that limο—β†’οŠ¦1π‘₯ does not exist; we do not assign it any kind of infinite value since the left and the right limits are different.

There is one final way for the limit of a function at a point to not exist. Instead of the values of the function increasing or decreasing without bound, the outputs can oscillate so that they never converge on a single value. For example, consider the following graph of 𝑓(π‘₯)=ο€Ό1π‘₯sin.

In the diagram, it appears that as π‘₯ approaches 0, the outputs oscillate wildly. To see more clearly what is happening around π‘₯=0 in this graph, we will try to evaluate the limit of this function from a table. We will choose our sample points to the reciprocal of odd integer multiples of πœ‹2, since these are the values that will maximize the size of the outputs.

π‘₯βˆ’299πœ‹βˆ’2101πœ‹βˆ’2103πœ‹βˆ’2105πœ‹β†’0←2105πœ‹2103πœ‹2101πœ‹299πœ‹
sinο€Ό1π‘₯1βˆ’11βˆ’1→←1βˆ’11βˆ’1

We can see that as the values of π‘₯ approach 0 from either side, the output values of the function oscillate between 1 and βˆ’1 and do not converge on any value; hence we say that this limit does not exist.

We can summarize the ways that limits cannot exist as follows.

Definition: Limit of a Function at a Point Not Existing

If the values of 𝑓(π‘₯) do not approach some value, πΏβˆˆβ„, as the values of π‘₯ approach π‘Ž, from both sides, then we say that the limit of 𝑓(π‘₯) as π‘₯ approaches π‘Ž does not exist.

In particular,

  • if the outputs of 𝑓(π‘₯) increase without bound as π‘₯ approaches π‘Ž from both sides, we say that limο—β†’οŒΊπ‘“(π‘₯)=∞;
  • if the outputs of 𝑓(π‘₯) decrease without bound as π‘₯ approaches π‘Ž from both sides, we say that limο—β†’οŒΊπ‘“(π‘₯)=βˆ’βˆž.

In other words, to determine the limit of a function at a point, we check that the left and the right limits both exist and are equal. Let’s see an example of applying this definition to a piecewise-defined function.

Example 1: Discussing the Existence of the Limit of a Piecewise-Defined Function at a Point

Discuss the existence of limο—β†’οŠ­π‘“(π‘₯) given 𝑓(π‘₯)=13π‘₯+71<π‘₯<7,14π‘₯+77≀π‘₯<8.ifif

Answer

Since this is a piecewise-defined function and π‘₯=7 is on the boundary of two subdomains, we cannot evaluate this limit by direct substitution. Instead, we recall that we can determine the limit of this function as π‘₯ approaches 7 by checking that the left and right limits of 𝑓(π‘₯) exist and are equal.

We start with limο—β†’οŠ­οŽͺ𝑓(π‘₯), since this is a left limit; we have π‘₯<7 for our input values and when evaluating this limit the values of π‘₯ will get arbitrarily close to 7, so we can assume 1<π‘₯<7, without affecting the value of the limit. When our values of π‘₯ are in this interval, 𝑓(π‘₯)=13π‘₯+7, this gives us limlimο—β†’οŠ­ο—β†’οŠ­οŽͺοŽͺ𝑓(π‘₯)=(13π‘₯+7).

We can evaluate this by direct substitution: limο—β†’οŠ­οŽͺ(13π‘₯+7)=13(7)+7=98.

We can do the same for the right limit; restricting the values of π‘₯ to be in the interval ]7,8[ does not affect the value of the limit, giving us limlimο—β†’οŠ­ο—β†’οŠ­οŽ©οŽ©π‘“(π‘₯)=(14π‘₯+7)=14(7)+7=105.

For the limit of 𝑓(π‘₯) at π‘₯=7 to exist, both the left and right limits need to be equal. However, we have shown that they are not equal.

Hence, the limit does not exist because limlimο—β†’οŠ­ο—β†’οŠ­οŽͺοŽ©π‘“(π‘₯)≠𝑓(π‘₯).

In our next few examples, we show the existence and find the value of the limit of a piecewise function at a point on the boundary of its subdomains.

Example 2: Discussing the Existence of the Limit of a Piecewise-Defined Function at a Point

Discuss the existence of limο—β†’οŠ©π‘“(π‘₯) given 𝑓(π‘₯)=|π‘₯βˆ’2|+3,βˆ’2<π‘₯<3,π‘₯+6π‘₯βˆ’27π‘₯βˆ’3π‘₯,3<π‘₯<9.

Answer

Since this is a piecewise-defined function and π‘₯=3 is on the boundary of two subdomains, we cannot evaluate this limit by direct substitution. Instead, we recall that we can determine the limit of this function as π‘₯ approaches 3 by checking that the left and right limits of 𝑓(π‘₯) exist and are equal.

To evaluate the limit as π‘₯ approaches 3 from the left, we note that when βˆ’2<π‘₯<3, we have 𝑓(π‘₯)=|π‘₯βˆ’2|+3. When evaluating this limit, the values of π‘₯ will get arbitrarily close to 3, so we can assume that βˆ’2<π‘₯<3, without affecting the value of the limit. This gives us limlimο—β†’οŠ©ο—β†’οŠ©οŽͺοŽͺ𝑓(π‘₯)=(|π‘₯βˆ’2|+3).

We can evaluate the limit of absolute value functions by direct substitution, giving us limο—β†’οŠ©οŽͺ(|π‘₯βˆ’2|+3)=|3βˆ’2|+3=1+3=4.

Next, to evaluate limο—β†’οŠ©οŽ©π‘“(π‘₯), we can notice that when 3<π‘₯<9, we have 𝑓(π‘₯)=π‘₯+6π‘₯βˆ’27π‘₯βˆ’3π‘₯; this gives us limlimο—β†’οŠ©ο—β†’οŠ©οŠ¨οŠ¨οŽ©οŽ©π‘“(π‘₯)=ο€Ύπ‘₯+6π‘₯βˆ’27π‘₯βˆ’3π‘₯.

Since this is a rational function, we can attempt to evaluate this by direct substitution: 3+6(3)βˆ’27(3)βˆ’3(3)=00.

This is an indeterminate form, so we will need to simplify the rational function by factoring. Recall that we can cancel shared factors of π‘₯βˆ’3 in the numerator and denominator inside of the limit, since it only affects the value of the function when π‘₯=3, not when π‘₯ is arbitrarily close to 3 from the right: limlimlimο—β†’οŠ©οŠ¨οŠ¨ο—β†’οŠ©ο—β†’οŠ©οŽ©οŽ©οŽ©ο€Ύπ‘₯+6π‘₯βˆ’27π‘₯βˆ’3π‘₯=ο€½(π‘₯+9)(π‘₯βˆ’3)π‘₯(π‘₯βˆ’3)=ο€Όπ‘₯+9π‘₯=3+93=4.

Therefore, we have shown that the left and the right limit of 𝑓(π‘₯) at π‘₯=3 exist and they are both equal to 4.

Hence, limο—β†’οŠ©π‘“(π‘₯) exists and equals 4.

Example 3: Finding the Limit of a Piecewise-Defined Function at a Point

Find limο—β†’οŠ±οŠ―π‘“(π‘₯), where 𝑓(π‘₯)=ο­βˆ’8+|π‘₯+9|,π‘₯β‰ βˆ’9,βˆ’7,π‘₯=βˆ’9.

Answer

Since this is a piecewise-defined function and π‘₯=βˆ’9 is on the boundary of two subdomains, we cannot evaluate this limit by direct substitution. Instead, we recall that we can determine the limit of this function as π‘₯ approaches βˆ’9 by checking that the left and the right limit of 𝑓(π‘₯) exist and are equal.

We will start with limο—β†’οŠ±οŠ―οŽ©π‘“(π‘₯). When π‘₯>βˆ’9, we know that |π‘₯+9|=π‘₯+9, so βˆ’8+|π‘₯+9|=βˆ’8+π‘₯+9=π‘₯+1.

Therefore, when π‘₯>βˆ’9, we have 𝑓(π‘₯)=π‘₯+1; this means that their limits as π‘₯ approaches βˆ’9 from the right must be equal, giving us limlimlimο—β†’οŠ±οŠ―ο—β†’οŠ±οŠ―ο—β†’οŠ±οŠ―οŽ©οŽ©οŽ©π‘“(π‘₯)=(βˆ’8+|π‘₯+9|)=(π‘₯+1)=βˆ’9+1=βˆ’8.

Similarly, when π‘₯<βˆ’9, we have βˆ’8+|π‘₯+9|=βˆ’8βˆ’(π‘₯+9)=βˆ’π‘₯βˆ’17. This means limlimlimο—β†’οŠ±οŠ―ο—β†’οŠ±οŠ―ο—β†’οŠ±οŠ―οŽͺοŽͺοŽͺ𝑓(π‘₯)=(βˆ’8+|π‘₯+9|)=(βˆ’π‘₯βˆ’17)=βˆ’(βˆ’9)βˆ’17=βˆ’8.

Since the left and the right limit of 𝑓(π‘₯) both exist and are equal to βˆ’8, we can conclude that limο—β†’οŠ±οŠ―π‘“(π‘₯)=βˆ’8.

In the previous example, we saw that even though 𝑓(βˆ’9)=βˆ’7, its limit as π‘₯ approached βˆ’9 is equal to βˆ’8. This is an example of the following property.

Property: Existence of a Limit

The existence or value of 𝑓(π‘Ž) does not affect the existence or value of limο—β†’οŒΊπ‘“(π‘₯)

In our next example, we will use this method to determine the existence of the limit of a reciprocal of an absolute value function at a point.

Example 4: Discussing the Existence of the Limit of a Reciprocal Absolute Value Function at a Point

Discuss the existence of limο—β†’οŠ¨1|π‘₯βˆ’2|.

Answer

We recall that we can determine the limit of this function as π‘₯ approaches 2 by checking that the left and the right limits of 𝑓(π‘₯) at π‘₯=2 exist and are equal. To determine the value of this limit, we can sketch a graph of 𝑦=1|π‘₯βˆ’2|. First, we can sketch 𝑦=1π‘₯βˆ’2 as a translation of 𝑦=1π‘₯ two units to the right, then we notice that 1|π‘₯βˆ’2|=|||1π‘₯βˆ’2|||, so we reflect the negative parts of the curve in the π‘₯-axis, giving us the following.

The graph will have a vertical asymptote at π‘₯=2. If we look at the outputs of the function to the left and right of π‘₯=2, we can see that as the values of π‘₯ approach 2 from either side, the outputs of the function will grow without bound.

In terms of limits, this tells us limandlimο—β†’οŠ¨ο—β†’οŠ¨οŽ©οŽͺ1|π‘₯βˆ’2|=∞1|π‘₯βˆ’2|=∞.

Recall that saying a limit is equal to infinity means that the limit does not exist. However, since the left and right limits agree, we can say that limο—β†’οŠ¨1|π‘₯βˆ’2|=∞.

Hence, limο—β†’οŠ¨1|π‘₯βˆ’2| does not exist, but limο—β†’οŠ¨1|π‘₯βˆ’2|=∞.

In our final example, we will investigate the limit of a function that has oscillating behavior.

Example 5: Understanding Limits and Oscillating Behavior

Investigate the behavior of 𝑓(π‘₯)=2ο€Ό1π‘₯cos as π‘₯ tends to 0.

  1. Complete the table of values of 𝑓(π‘₯) for values of π‘₯ that get closer to 0.
    π‘₯199πœ‹1100πœ‹1999πœ‹11000πœ‹19999πœ‹110000πœ‹
    𝑓(π‘₯)
  2. What does this suggest about the graph of 𝑓 close to 0?
    1. That it increases without bound
    2. That it decreases without bound
    3. That it changes randomly
    4. That it approaches 2
    5. That it oscillates rapidly between βˆ’2 and 2
  3. Hence, evaluate limο—β†’οŠ¦π‘“(π‘₯).

Answer

Part 1

We find each entry in the table by substituting the given value of π‘₯ into the function 𝑓(π‘₯)=2ο€Ό1π‘₯cos. For example, 𝑓199πœ‹οˆ=21=2(99πœ‹).coscosοŠ§οŠ―οŠ―οŽ„

The cosine function is periodic with a period of 2πœ‹, so we can simplify 2(99πœ‹)=2(98πœ‹+πœ‹)=2(πœ‹)=βˆ’2.coscoscos

Similarly, 𝑓1100πœ‹οˆ=21=2(100πœ‹)=20=2.coscoscosοŠ§οŠ§οŠ¦οŠ¦οŽ„

We can follow this process for all of the entries in the table to get the following.

π‘₯199πœ‹1100πœ‹1999πœ‹11000πœ‹19999πœ‹110000πœ‹
𝑓(π‘₯)βˆ’22βˆ’22βˆ’22

Part 2

In the table, we can see that, from left to right, the values of π‘₯ approach 0. However, the outputs of the function do not approach any value; instead, the outputs oscillate between βˆ’2 and 2.

Therefore, the table suggests that the graph rapidly oscillates between βˆ’2 and 2, which is option E.

Part 3

The table in part 2 suggests that limο—β†’οŠ¦οŽ©π‘“(π‘₯) does not converge to any value; instead, the outputs oscillate between βˆ’2 and 2. This means that the right limit of 𝑓(π‘₯) at π‘₯=0 does not exist. For the limit of 𝑓(π‘₯) at π‘₯=0 to exist, we need both the left and the right limit to exist and be equal.

Hence, the limit does not exist.

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

Key Points

  • To determine if the limit of 𝑓(π‘₯) at π‘₯=π‘Ž exists, we check three things:
    • if the left limit of 𝑓(π‘₯) at π‘₯=π‘Ž exists,
    • if the right limit of 𝑓(π‘₯) at π‘₯=π‘Ž exists,
    • if these two limits are equal.
    If all three of these conditions hold, we say that the limit of 𝑓(π‘₯) at π‘₯=π‘Ž exists and is equal to its left and right limits; otherwise, we say that the limit does not exist.
  • If the outputs of 𝑓(π‘₯) increase without bound as π‘₯ approaches π‘Ž from both sides, we say that limο—β†’οŒΊπ‘“(π‘₯)=∞; this is an example of the limit not existing.
  • If the outputs of 𝑓(π‘₯) decrease without bound as π‘₯ approaches π‘Ž from both sides, we say that limο—β†’οŒΊπ‘“(π‘₯)=βˆ’βˆž; this is an example of the limit not existing.

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