Video: Existence of Limits

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

16:40

Video Transcript

The Existence of Limits

In this video, we will learn how to determine whether the limit of a function at a certain value exist. There are some reasons why a limit may not exist. We will be looking at these reasons and using them to deduce whether or not certain limits exist. When we are considering the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯, there are a few reasons why the limit may not exist. The first is if the limit is not finite. In the case of this 𝑓 of π‘₯, we can see that as π‘₯ approaches zero, 𝑓 of π‘₯ is approaching positive or negative infinity depending on whether we’re approaching zero from the right or from the left. The reason why this limit does not exist is because the limit must approach a particular point. And we cannot say that infinity is a point since infinity does not actually exist; it’s just a concept.

Let’s quickly note some special cases when we have an infinite limit. In the case of 𝑔 of π‘₯, we can see that the left and right limits, as π‘₯ approaches zero, both tend to positive infinity. Since these left and right limits both tend to the same sign of infinity, we can say that the limit as π‘₯ approaches zero of 𝑔 of π‘₯ is also equal to infinity. However, we must note that this limit still does not exist since infinity is not a number; it’s just a concept. Similarly, we can note that, for β„Ž of π‘₯, the left and right limits as β„Ž approaches zero both tend to negative infinity. And so we can say that the limit as π‘₯ approaches zero of β„Ž of π‘₯ is equal to negative infinity. But again, this limit still does not exist. Though it is important to remember these as it can be useful to know whether a particular limit tends to a certain sign of infinity, rather than the case of 𝑓 of π‘₯ where the different sides of the limit tend to different signs of infinity.

Let’s now consider another case which can cause a limit not to exist. We can say that the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ does not exist if the limit does not approach a particular point. And this happens with certain oscillating functions. Here is an example of an oscillating function 𝑓 of π‘₯ where a limit does not exist. And this limit is as π‘₯ approaches zero. We can see that the closer the values of π‘₯ get to zero, the more rapidly our function oscillates up and down. This function is oscillating between the values of minus one and one. The closer π‘₯ gets to zero, the quicker it oscillates between minus one and one. And so we can say that the function does not approach a particular value as it approaches zero, since the closer and closer it gets to zero, the faster it changes between one and minus one. And from this, we can say that 𝑓 of π‘₯ does not approach a particular point as π‘₯ approaches zero. And, therefore, the limit as π‘₯ approaches zero of 𝑓 of π‘₯ does not exist.

Now, let’s consider another reason why limits may not exist. When finding the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯, it’s very important to consider the left and right limits. This is because if the left or the right limit does not exist or if both the left and right limits exist but they are not equal, then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ does not exist. If we consider the function 𝑓 of π‘₯, we can see that 𝑓 of π‘₯ is a piecewise function which is defined between the values π‘Ž and 𝑐. Let’s consider the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž. We can see that the limit as π‘₯ approaches π‘Ž from the right is equal to 𝑓 of π‘Ž. However, as π‘₯ approaches π‘Ž from the left, 𝑓 is undefined. And so, therefore, this limit from the left does not exist.

This tells us that the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ does not exist. Now let’s consider the limit as π‘₯ approaches 𝑏 of 𝑓 of π‘₯. In order to consider this limit, let’s label the two sections of our graph. We’ll call the section between π‘Ž and 𝑏 𝑔 of π‘₯ and the section between 𝑏 and 𝑐 β„Ž of π‘₯. The limit as π‘₯ approaches 𝑏 from the right will be equal to β„Ž of 𝑏 since we are coming from the right of 𝑏 and the value of 𝑓 of π‘₯ will be moving along β„Ž of π‘₯.

Now, the limit as π‘₯ approaches 𝑏 from the left will be equal to 𝑔 of 𝑏 since as the value of 𝑓 of π‘₯ gets closer and closer to 𝑏 from the left, it will be following along 𝑔 of π‘₯. And so it’ll be equal to 𝑔 of 𝑏. Since β„Ž of 𝑏 and 𝑔 of 𝑏 are not equal to one another, we must conclude that the limit as π‘₯ approaches 𝑏 of 𝑓 of π‘₯ does not exist. Now, we have covered all of the reasons why a limit may not exist. We can say that the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ exists if both the left and right limits exist and the left limit is equal to the right limit. We can also say that the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ is equal to some constant 𝐿 where 𝐿 is also equal to the left and right limits. We’re now ready to look at some examples.

Discuss the existence of the limit as π‘₯ approaches seven of 𝑓 of π‘₯ given 𝑓 of π‘₯ is equal to 13π‘₯ plus seven if one is less than π‘₯ which is less than seven and 14π‘₯ plus seven if seven is less than or equal to π‘₯ which is less than eight.

In this example, our 𝑓 of π‘₯ is a piecewise function. And we are asked to find the limit as π‘₯ approaches seven. Seven is the π‘₯ value at which our function switches between 13π‘₯ plus seven and 14π‘₯ plus seven. In order to find whether our limit exists, we need to check whether the left and right limits exist and if they are equal. We’ll start by considering the left limit. Since π‘₯ is approaching seven from below, we know that π‘₯ must be less than seven. Since π‘₯ is less than seven, we can see from our piecewise definition that 𝑓 of π‘₯ is equal to 13π‘₯ plus seven.

Since this is a polynomial, we can use direct substitution. In order to find this limit, we simply substitute π‘₯ equals seven into 13π‘₯ plus seven. And this gives us that the left limit is equal to 98. Since the limit here is equal to a real constant, we know that this limit must exist. Let’s now consider the limit as π‘₯ approaches seven from above. Since π‘₯ is approaching seven from above, we have that π‘₯ is greater than seven. Therefore, from our piecewise definition, we have that 𝑓 of π‘₯ is equal to 14π‘₯ plus seven which is again a polynomial. And so we can use direct substitution in order to find this limit. Substituting π‘₯ equals seven into 14π‘₯ plus seven, we obtain the limit as π‘₯ approaches seven from the right is equal to 105. Therefore, our limit exists.

Now, we have found that both the left and right limit exist. However, the left limit is equal to 98. And the right limit is equal to 105. Therefore, we can conclude that the limit as π‘₯ approaches seven of 𝑓 of π‘₯ does not exist because the left and the right limit are not equal to each other. In this example, we saw how the limit did not exist because the left and right limits were not equal. This is because there is a jump in the function at the point which we’re trying to take the limit. Therefore, we cannot say that the limit of 𝑓 of π‘₯ approaches a particular point since it depends upon which direction we are approaching the limit as to what the limit could equal. Now, let’s consider another example.

Discuss the existence of the limit as π‘₯ approaches three of 𝑓 of π‘₯ given 𝑓 of π‘₯ is equal to the mod of π‘₯ minus two plus three if negative two is less than π‘₯ which is less than three π‘₯ squared plus six π‘₯ minus 27 all over π‘₯ squared minus three π‘₯ if three is less than π‘₯ which is less than nine.

Let’s consider the left and right limits as π‘₯ approaches three of this piecewise function. As π‘₯ approaches three from below, we know the π‘₯ will be less than three. Therefore, 𝑓 of π‘₯ will be equal to the mod of π‘₯ minus two plus three. We can use direct substitution in order to find this limit. We find that it is equal to four. Now, let’s consider the right-hand limit. Since π‘₯ is approaching three from above, we know that π‘₯ will be greater than three. So 𝑓 of π‘₯ is equal to π‘₯ squared plus six π‘₯ minus 27 all over π‘₯ squared minus three π‘₯. If we try using direct substitution to find this limit, we will find that it’s equal to zero over zero which is undefined. However, since this is a rational function with polynomials in the numerator and denominator where both the numerator and denominator are equal to zero at three, this means we can factorise out a factor of π‘₯ minus three from the top and bottom.

We obtain π‘₯ plus nine multiplied by π‘₯ minus three over π‘₯ multiplied by π‘₯ minus three. Next, we can simply cancel the π‘₯ minus three on the top and bottom of the fraction. However, we must be careful because, in doing this, we are actually changing the domain of 𝑓 of π‘₯. Previously inputting a value of π‘₯ equals three will have given us an undefined outcome. However, now if we input π‘₯ is equal to three, we will get a number. Therefore, we should call this new function 𝑔 of π‘₯ where 𝑔 of π‘₯ is identical to 𝑓 of π‘₯ in every way except by its domain, since we can input π‘₯ equals three into 𝑔 of π‘₯, however, we cannot into 𝑓 of π‘₯. We are now able to use direct substitution on 𝑔 of π‘₯ in order to find the limit from above.

We obtain that the limit from above is equal to four. Comparing this to the limit from below, we can see that these two are equal. We can, therefore, conclude that the limit as π‘₯ approaches three of 𝑓 of π‘₯ exists. And it is equal to four. In this example, we saw how a limit of a piecewise function at the point where the definition of the function changes can exist if the left and right limit can agree on a value of which the limit is equal to. Let’s now look at another example.

Discuss the existence of the limit as π‘₯ approaches two of one over the mod of π‘₯ minus two.

Let’s start by considering the limit as π‘₯ approaches two of mod π‘₯ minus two which is the denominator of our fraction. We can find this by using direct substitution. And it gives us zero. What this tells us is that, as π‘₯ approaches two, the modulus of π‘₯ minus two gets smaller and smaller as it approaches zero. And since mod of π‘₯ minus two is getting smaller, this means that the reciprocal one over mod of π‘₯ minus two will be getting larger and larger. And this tells us that here we have an infinite limit. Therefore, the limit does not approach a particular value and so the limit does not exist. This is because infinity is not a number; it’s simply a concept. Now, there is more that we can deduce other than the limit simply does not exist. So let’s draw the graph of one over the mod of π‘₯ minus two.

Our graph will go to infinity at π‘₯ is equal to two. And we know that the graph must be positive everywhere on π‘₯ since our graph has a modulus or absolute value within it. Since everything outside of the modulus is positive, this means that our graph must be positive everywhere on π‘₯. We can see for our graph that, as π‘₯ approaches two from above and below, our function is tending to positive infinity. So we can say that both the left and right limits as π‘₯ approaches two are equal to infinity. Since these are both equal to positive infinity, we can conclude that the limit as π‘₯ approaches two of one over mod of π‘₯ minus two is also equal to positive infinity.

It’s important to remember that this limit still does not exist since infinity is not a number. Therefore, we can conclude that the limit as π‘₯ approaches two of one over the mod of π‘₯ minus two does not exist. But the limit is equal to infinity. In this example, we saw what happens when our limit tends to infinity. Now, let’s look at one final example.

Investigate the limit of 𝑓 of π‘₯ is equal to two cos of one over π‘₯ as π‘₯ tends to zero. a) Complete the table of values 𝑓 of π‘₯ for values of π‘₯ that get closer to zero.

In order to solve this first part, we simply substitute the values of π‘₯ into 𝑓 of π‘₯. We find that 𝑓 of one over 99 is equal to negative two. 𝑓 of one over 100πœ‹ is equal to two. Continuing in this way, we can complete the rest of the table and we find that this is the solution to part a).

Part b), what does this suggest about the graph of 𝑓 close to zero?

Our table suggests that the closer and closer we get to zero, the more rapidly the graph of 𝑓 of π‘₯ will oscillate between negative two and two. Let’s quickly consider why this is the case. We know that as π‘₯ approaches zero from above, one over π‘₯ would approach infinity. And as π‘₯ approaches zero from below, one over π‘₯ would approach negative infinity. What this means is as π‘₯ gets smaller, one over π‘₯ will get very very large in either the positive or negative direction depending on which way we’re approaching zero. Now, the closer and closer we get to zero, the rate at which one over π‘₯ will get larger increases.

Now, we know that the cos function oscillates between negative one and one at a constant rate. However, since the rate at which one over π‘₯ is increasing is increasing, this means that cos of one over π‘₯ will oscillate between negative one and one at an increasing rate. So this is why our graph will oscillate rapidly between negative two and two.

Part c), hence evaluate the limit as π‘₯ approaches zero of 𝑓 of π‘₯.

In part b), we saw that as π‘₯ approaches zero, 𝑓 of π‘₯ will oscillate more and more rapidly between negative two and two. So as π‘₯ approaches zero, we cannot deduce a particular value of which 𝑓 of π‘₯ will approach since the graph will be changing between negative two and two more and more quickly. Since the function does not tend to a particular point as π‘₯ goes to zero, we can deduce that the limit does not exist. We have now seen why the limit of an oscillating function may not exist. And so we have covered a few examples of why a limit may or may not exist. Let’s recap some of the key points of the video.

If a function tends to infinity at a point then the limit of the function at that point does not exist. In order for a limit to exist it must approach a particular point. So in the case of some oscillating functions, they may begin to oscillate incredibly rapidly as they approach a point. And, therefore, the limit will not exist. If the limit as π‘₯ approaches π‘Ž from below of 𝑓 of π‘₯ or the limit as π‘₯ approaches π‘Ž from above of 𝑓 of π‘₯ does not exist or if the limit from below is not equal to the limit from above, then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ does not exist. If the limit as π‘₯ approaches π‘Ž from below of 𝑓 of π‘₯ and the limit as π‘₯ approaches π‘Ž from above of 𝑓 of π‘₯ both exist and the limit from below is equal to the limit from above which is equal to some constant 𝐿, then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ exists. And we can say that the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ is equal to 𝐿.

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