Video: Continuity at a Point

In this video, we will learn how to check the continuity of a function at a given point.

15:53

Video Transcript

In this video, we’re going to learn about continuity at a point. This is a step on the way to understanding continuous functions, functions like polynomial functions, exponential functions, and certain trigonometric functions. Whose graphs can be drawn in a single stroke of the pen. You may have noticed that, for a lot of functions 𝑓, the limit of 𝑓 of π‘₯ as π‘₯ approaches some value π‘Ž is just 𝑓 evaluated at π‘Ž, 𝑓 of π‘Ž. A function 𝑓 is said to be continuous at the point π‘Ž if this happens. If the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž is just 𝑓 of π‘Ž. This is the definition of continuity at a point. We can see then that the function 𝑓 is continuous at the point π‘Ž if direct substitution works to find the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž.

To understand continuity at a point, we need to understand how a function can fail to be continuous at the point π‘Ž. How this equation can fail to hold true. There are several things that can go wrong. For example, this limit on the left-hand side could not exist. For 𝑓 to be continuous at π‘Ž then, we need this limit to exist. Similarly, we need the right-hand side of the equation to exist. 𝑓 must be defined at π‘Ž. If it isn’t, then the right-hand side of our equation, that’s 𝑓 of π‘Ž, is undefined. And so our equation can’t possibly hold. Another way of saying that 𝑓 must be defined at π‘Ž is to say that π‘Ž must be in the domain of 𝑓. What else could go wrong? Well, the limit on the left-hand side could exist. And the function 𝑓 of π‘Ž could be defined. But these values could be different. We need the values of the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž and 𝑓 of π‘Ž to be equal.

These are the three things we needed to check to show that a function 𝑓 is continuous at a number π‘Ž. Let’s now apply this checklist to some examples. For convenience, we’ll always check that 𝑓 is defined at π‘Ž before checking that the limit as 𝑓 of π‘₯ approaches π‘Ž exists. And so the checklist for our examples has a slightly different order, which I’d encourage you to use too.

Given 𝑓 of π‘₯ equals π‘₯ squared plus π‘₯ minus two all over π‘₯ minus one, if possible or necessary, define 𝑓 of one so that 𝑓 is continuous at π‘₯ equals one.

So we have a function 𝑓, which is a rational function. And we’d like 𝑓 to be continuous at the point π‘₯ equals one. And we’re told that we should define 𝑓 of one to make this so, but only if it is possible or necessary. What does that mean? Well, if 𝑓 is already continuous at the point π‘₯ equals one, then it isn’t necessary to define 𝑓 of one to make this so. It’s already been done for us. On the other hand, 𝑓 could be discontinuous at the point π‘₯ equals one. But in the way that it’s not possible to things just by defining 𝑓 of one. There could be some bigger issue stopping 𝑓 from being continuous at this point.

Let’s first check whether it is necessary to define 𝑓 of one to make 𝑓 continuous at π‘₯ equals one. Is 𝑓 already continuous at π‘₯ equals one? Well, we have a checklist which allows us to find out whether a function is continuous at a certain point. 𝑓 must be defined at that point. So in our case, we need to check that 𝑓 of one is defined. And the limit of 𝑓 of π‘₯ as π‘₯ approaches that point, in our case one, must exist. And finally, these two values must be equal.

Let’s start at the top of the list. Is 𝑓 of one defined? Well, we’ll use the definition of 𝑓 of π‘₯ that we’re given in the question. Substituting one for π‘₯ gives one squared plus one minus two all over one minus one. In the numerator, one squared plus one is two. And subtracting the two gives us zero. And in the denominator, one minus one is zero. So we get the indeterminate form zero over zero. Zero over zero and hence 𝑓 of one are not defined. And so our function 𝑓 is not already continuous at π‘₯ equals one.

This isn’t necessarily a massive issue. After all, our task is to define 𝑓 of one. If this is the only thing stopping 𝑓 from being continuous at π‘₯ equals one, then we can just define 𝑓 of one. And 𝑓 will be continuous at this point, as required. We need to check then that there aren’t any other issues. We need the limit of 𝑓 of π‘₯ as π‘₯ approaches one to exist. And does it? We use the definition of 𝑓 of π‘₯ from the question. And of course, we know that direct substitution is going to give us an indeterminate form. So there must be some other way to evaluate this limit.

Well, the fact theorem tells us that as both the numerator and the denominator are zero when π‘₯ is one, both numerator and denominator must have a factor of π‘₯ minus one. And indeed, we can factor the numerator to π‘₯ plus two times π‘₯ minus one. Doing so allows us to cancel the common factor of π‘₯ minus one in the numerator and denominator. To get the limit of π‘₯ plus two as π‘₯ approaches one. This is a limit that can be evaluated using direct substitution. Substituting one for π‘₯, we get one plus two, which is three. So yes, the limit of 𝑓 of π‘₯ as π‘₯ approaches one does exist. And it equals three.

Now the last thing on our checklist is that the limit of 𝑓 of π‘₯ as π‘₯ approaches one must be 𝑓 of one. We found that the left-hand side, the limit of 𝑓 of π‘₯ as π‘₯ approaches one, is three. But the right-hand side 𝑓 of one is undefined. But our task is to define 𝑓 of one. If we define 𝑓 of one to be three, that’s the value of the limit of 𝑓 of π‘₯ as π‘₯ approaches one. Then our third item on the checklist will be satisfied. And of course, defining 𝑓 of one to be three also sorts out the first item on our checklist. 𝑓 of one is now defined. So defining 𝑓 of one to be three makes 𝑓 continuous at π‘₯ equals one. Because 𝑓 of one is now defined, the limit of 𝑓 of π‘₯ as π‘₯ approaches one, as we saw, is three. And now that we’ve defined 𝑓 of one to also be three, these two values are equal.

It might be helpful to look at the graph of 𝑓 of π‘₯ to see what we’ve done. As we saw with π‘₯ not equal to one, 𝑓 of π‘₯ is just π‘₯ plus two. And so the graph of 𝑓 of π‘₯ is just the straight line graph of 𝑦 equals π‘₯ plus two, with this hole here when π‘₯ is one. This hole in the graph comes because 𝑓 of one is not defined. And as a result, 𝑓 is not continuous at π‘₯ equals one. There is then this link between the technical definition of continuity and our intuitive understanding of what continuity means. There should be no gaps. We plug this gap and make the function continuous by defining 𝑓 of one to be three. Now, there’s no hole in the graph. And 𝑓 is continuous at π‘₯ equals one. Let’s now see an example where we can’t simply plug the gap.

Given 𝑓 of π‘₯ equals one over π‘₯, if possible or necessary, define 𝑓 of zero so that 𝑓 is continuous at π‘₯ equals zero.

So we have the reciprocal function here, which hopefully we know and love. And we know what its graph looks like. The graph has two pieces, one in the first quadrant and the other in the third. And these pieces are separated by a vertical asymptote at π‘₯ equals zero. Intuitively then, it feels like this function is not continuous at π‘₯ equals zero. Because the graph of our function is not one continuous curve but is formed of two curves with the break at that asymptote π‘₯ equals zero. Just to the left of this line when π‘₯ is small but negative, 𝑓 of π‘₯ is larger magnitude but negative. And to the right of this line when π‘₯ is small but positive, 𝑓 of π‘₯ is larger magnitude and positive. So as we pass π‘₯ equals zero, the right of 𝑓 of π‘₯ changes from a value which is larger magnitude and negative to a value which is larger magnitude and positive.

Let’s see if our intuition agrees with the technical definition by going through our checklist to see if the function is continuous. For 𝑓 to be continuous at π‘₯ equals zero, we need 𝑓 of zero to be defined. The limit of 𝑓 of π‘₯ as π‘₯ approaches zero to exist. And finally, we need these values to be equal. Okay, so is 𝑓 defined at zero? Well, if we substitute zero into the definition of 𝑓 of π‘₯, 𝑓 of π‘₯ is one over π‘₯, we get 𝑓 of zero is one over zero. And this is not defined. Zero is not in the domain of our function. So 𝑓 is not defined at zero. And hence, 𝑓 is not continuous at π‘₯ equals zero. But this isn’t necessarily a surprise as our task is to define 𝑓 of zero to make 𝑓 continuous at π‘₯ equals zero. We wouldn’t have any work to do if 𝑓 were continuous at π‘₯ equals zero and 𝑓 of zero was defined.

We check the second criterion. The limit of 𝑓 of π‘₯ as π‘₯ approaches zero must exist. With the idea that if this limit does exist, then we can just define 𝑓 of zero to be the value of this limit. And, then the function will be continuous at π‘₯ equals zero as required. Does this element exist? Well, if we look at the left-hand limit as π‘₯ approaches zero, 𝑓 of π‘₯ approaches negative infinity. And it gets worse. The limit from the right is positive infinity. As π‘₯ approaches zero from the right, 𝑓 of π‘₯ gets larger and larger without bound.

The limit of 𝑓 of π‘₯ as π‘₯ approaches zero therefore does not exist. And this is a problem. We can define 𝑓 of zero to be whatever we like. But whatever we define it to be, this doesn’t change the fact that the limit of 𝑓 of π‘₯ as π‘₯ approaches zero does not exist. And so 𝑓 cannot be continuous at π‘₯ equals zero. So this is our answer then. The limit of 𝑓 of π‘₯ as π‘₯ approaches zero doesn’t exist. And so 𝑓 cannot be made continuous at π‘₯ equals zero by defining 𝑓 of zero. The fact that 𝑓 of zero is undefined is not really the problem here. And looking at the graph, we can kind of see why this is true. There isn’t just a small gap on the graph, which can be plugged by a single point. There is a huge chasm. While 𝑓 of π‘₯ is not continuous at π‘₯ equals zero and cannot be made so just by defining 𝑓 of zero, it’s worth pointing out that the function 𝑓 of π‘₯ is actually continuous at other values of π‘₯.

Given 𝑓 of π‘₯ equals one over π‘₯, if possible or necessary, define 𝑓 of one so that 𝑓 is continuous at π‘₯ equals one. We go through our checklist with this new question. We need 𝑓 of one to be defined. The limit of 𝑓 of π‘₯ as π‘₯ approaches one to exist. And the limit of 𝑓 of π‘₯ as π‘₯ approaches one to equal 𝑓 of one. Is 𝑓 of one defined? Yes, 𝑓 of one is one over one, which is one. 𝑓 of one is defined. One is in the domain of 𝑓. Now, does the limit of 𝑓 of π‘₯ as π‘₯ approaches one exist? Let’s look at the graph. As π‘₯ approaches one from the left and as π‘₯ approaches one from the right. The value of 𝑓 of π‘₯ approaches the same at finite value, the value one. If we wanted to find the value of this limit without using the graph, we’d probably just use direct substitution. We can see then that the third item on the checklist at the limit of 𝑓 of π‘₯ as π‘₯ approaches one equals 𝑓 of one is also satisfied. The values of the left-hand side and right-hand side are both one. Our answer then is that 𝑓 is already defined and continuous at π‘₯ equals one. It’s not necessary to define 𝑓 of one again to make this so. Let’s see one more quick example before we conclude.

Is the function 𝑓 of π‘₯, which is piecewise defined to be two π‘₯ plus four all over π‘₯ plus two if π‘₯ is less than negative two. Zero if π‘₯ is equal to negative two. And π‘₯ squared plus six π‘₯ plus eight all over π‘₯ plus two if π‘₯ is greater than negative two, continuous at π‘₯ equals negative two.

To find out, we go through our checklist. For the function 𝑓 to be continuous at π‘₯ equals negative two, 𝑓 of negative two must be defined. The limit of 𝑓 of π‘₯ as π‘₯ approaches negative two must exist. And these values must be equal. So we start by checking that 𝑓 of negative two is defined. Well, that’s easy to see from the question. We’re told that 𝑓 of π‘₯ is zero if π‘₯ is negative two. So 𝑓 of negative two is defined. 𝑓 of negative two is equal to zero. Now, we need to check that the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two exists. There are different rules for the values of 𝑓 of π‘₯ when π‘₯ is less than negative two and when π‘₯ is greater than negative two. Which we’ll use to check the left-hand and right-hand limits, making sure that they exist and are equal.

What’s the left-hand limit? Well, to the left of negative two, 𝑓 of π‘₯ is defined by two π‘₯ plus four all over π‘₯ plus two. So we have to find the limit of two π‘₯ plus four all over π‘₯ plus two, as π‘₯ approaches negative two from the left. If we try direct substitution, we get the indeterminate form zero over zero. So we have to find this limit some other way. We do this by factoring the numerator, getting two times π‘₯ plus two. And this factor of π‘₯ plus two cancels with the factor of π‘₯ plus two in the denominator. And so our limit is that of the constant function two whose value must be two. So the left-hand limit exists. What about the right-hand one?

To the right of π‘₯ equals negative two, 𝑓 of π‘₯ is this algebraic fraction. And we can find the value of this limit in a similar way. Factoring the numerator and canceling the common factor of π‘₯ plus two, to get the limit of π‘₯ plus four as π‘₯ approaches negative two from the right. This can be solved using direct substitution. We get negative two plus four, which is two. So this right-hand limit also exists and is equal to the left-hand limit. Both of them have the value two. Hence the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two from either direction also exists and has the value two.

All that’s left to check is this third point. For 𝑓 of π‘₯ to be continuous at π‘₯ equals negative two, we need the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two to be equal to the value of 𝑓 at negative two, 𝑓 of negative two. But we’ve just seen that the value of this limit is two, whereas 𝑓 evaluated at negative two is zero. The limit of 𝑓 of π‘₯ as π‘₯ approaches negative two is not equal to 𝑓 of negative two. Although the first two conditions on our checklist are satisfied, the last one isn’t. And so our answer is no. The function 𝑓 is not continuous at π‘₯ equals negative two, as the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two is not equal to 𝑓 of negative two.

Let’s now conclude with the key points that we’ve covered in this video. A function 𝑓 is said to be continuous at a number π‘Ž if the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž equals 𝑓 of π‘Ž. To check the continuity at π‘Ž, we must check that 𝑓 of π‘Ž is defined. That is, that π‘Ž is in the domain of 𝑓 and that the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž exists. And only then does it make sense to ask if they are equal. So if 𝑓 of π‘Ž is not defined or if the limit doesn’t exist, then 𝑓 is not continuous at π‘Ž. However, if 𝑓 of π‘Ž is undefined, but the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž exists, we can define 𝑓 of π‘Ž to be the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž to make 𝑓 continuous at π‘Ž. If the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž does not exist, we can’t make 𝑓 continuous at π‘Ž by defining or redefining 𝑓 of π‘Ž.

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