Question Video: Discussing the Existence of the Limit of a Piecewise-Defined Function at a Point Mathematics • Second Year of Secondary School

Video Transcript

What can be said of the limit as 𝑥 approaches negative six of 𝑓 of 𝑥 for the function 𝑓 of 𝑥 is equal to negative nine 𝑥 minus nine if 𝑥 is less than negative six and 𝑓 of 𝑥 is equal to 45 if 𝑥 is greater than negative six?

The question gives us a piecewise-defined function 𝑓 of 𝑥. And the question wants us to discuss the limit of 𝑓 of 𝑥 as 𝑥 approaches negative six. At first, we might be worried about this. Since our limit is as 𝑥 is approaching negative six, we can see that our function 𝑓 of 𝑥 is not actually defined when 𝑥 is equal to negative six. However, we don’t need to worry about this. When we say that our limit is as 𝑥 is approaching negative six, we want to know what happens as 𝑥 gets closer and closer to negative six. We don’t actually need to know what happens when 𝑥 is equal to negative six.

But now, we can see our second problem. When 𝑥 is less than negative six, we have that our function 𝑓 of 𝑥 is negative nine 𝑥 minus nine. But when 𝑥 is greater than negative six, our function 𝑓 of 𝑥 is the constant 45. So how would we calculate the limit as 𝑥 approaches negative six of this function? And we can get around this by instead of calculating the limit of 𝑓 of 𝑥 as 𝑥 approaches negative six, we can look at the left-hand and right-hand limit of 𝑓 of 𝑥. We say that the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 exists if the following three properties hold. First, the limit as 𝑥 approaches 𝑎 from the left of 𝑓 of 𝑥 must exist. Second, the limit as 𝑥 approaches 𝑎 from the right of 𝑓 of 𝑥 must exist. And third, these two limits must be equal.

If all of these conditions are satisfied, then our left-hand and right-hand limits are equal to some finite value 𝐿. And we then say the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 is equal to 𝐿. Since we want the limit as 𝑥 approaches negative six of 𝑓 of 𝑥, we’ll set 𝑎 equal to negative six. So we now need to check if each of these three properties are true. Let’s start by checking the limit as 𝑥 approaches negative six from the left of 𝑓 of 𝑥. Since we’re taking the limit as 𝑥 approaches negative six from the left, all of our values of 𝑥 are less than negative six. And we can see from our piecewise definition of the function 𝑓 of 𝑥, if 𝑥 is less than negative six, then our function 𝑓 of 𝑥 is exactly equal to negative nine 𝑥 minus nine.

So when we’re taking the limit as 𝑥 approaches negative six from the left, our function 𝑓 of 𝑥 is exactly equal to negative nine 𝑥 minus nine. That means that their limits will be equal in this case. But now, we can see we’re just trying to evaluate the limit of a linear function. We can do this by direct substitution. Substituting 𝑥 is equal to negative six gives us negative nine times negative six minus nine, which we can calculate to give us 45. So we’ve shown that the limit as 𝑥 approaches negative six of 𝑓 of 𝑥 exists. It’s equal to 45. We now need to check the limit as 𝑥 approaches six from the right of 𝑓 of 𝑥. We can do this in a very similar way. Since 𝑥 is approaching negative six from the right, our values of 𝑥 will be bigger than negative six.

And from our piecewise definition of the function 𝑓 of 𝑥, when 𝑥 is greater than negative six, our function 𝑓 of 𝑥 is exactly equal to the constant 45. So when 𝑥 is greater than negative six, our function 𝑓 of 𝑥 is exactly equal to 45. Therefore, their limits when 𝑥 approaches negative six from the right will be equal. But now, we see we’re just trying to evaluate the limit of a constant. And we know a constant does not change as our value of 𝑥 changes. So the limit of 45 as 𝑥 approaches negative six from the right will just be equal to 45.

So we’ve shown the limit as 𝑥 approaches negative six from the right of 𝑓 of 𝑥 also exists. It’s also equal to 45. Our final condition was that our left-hand limit and right-hand limit were equal. And we showed that both of these were equal to 45. So our third condition is also true. And remember, if all of these three conditions are true, we say the limit as 𝑥 approaches negative six of 𝑓 of 𝑥 is equal to the shared value of our left-hand and right-hand limit, which in this case is 45. So not only have we shown the limit as 𝑥 approaches negative six of 𝑓 of 𝑥 exists, we’ve also shown that it’s equal to 45.