Video: Discussing the Existence of the Limit of a Piecewise-Defined Function at a Certain Point

Discuss the existence of lim_(π‘₯ β†’ βˆ’1) 𝑓(π‘₯) given 𝑓(π‘₯) = π‘₯ βˆ’ 4 if π‘₯ < βˆ’1, and 𝑓(π‘₯) = 20 if π‘₯ > βˆ’1.

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Video Transcript

Discuss the existence of the limit of 𝑓 of π‘₯ as π‘₯ approaches negative one given that 𝑓 of π‘₯ is equal to π‘₯ minus four if π‘₯ is less than negative one, and 𝑓 of π‘₯ is equal to 20 if π‘₯ is greater than negative one.

The question is asking us to discuss the existence of the limit of 𝑓 of π‘₯ as π‘₯ is approaching negative one. And we know that for any function 𝑔 of π‘₯, the limit of 𝑔 of π‘₯ as π‘₯ approaches π‘Ž exists if both the left-hand limit and the right-hand limit exist and are equal. In our case, we are testing the function 𝑓 of π‘₯ as π‘₯ is approaching negative one. So we can add this information into our definition here. This means that there are three parts to checking the existence of the limit in our question.

First, we need to check that the limit of 𝑓 of π‘₯ as π‘₯ approaches negative one from the right exists. Second, we need to check that the limit of 𝑓 of π‘₯ as π‘₯ approaches negative one from the left exists. Third, we need to check that if both of these limits exist, then they’re equal. So we’re going to need to check the limit from the left and the limit from the right. We will start with the limit as π‘₯ approaches negative one from the right.

Since we have that π‘₯ is approaching negative one from the right, we must have that π‘₯ is greater than negative one. We also know that if two functions 𝑓 of π‘₯ and 𝑔 of π‘₯ are equal for π‘₯ is greater than π‘Ž, then their right limits as π‘₯ approaches π‘Ž must be equal. From the question, we have that the function 𝑓 of π‘₯ is equal to 20 when π‘₯ is greater than negative one. Therefore, since 𝑓 of π‘₯ is equal to 20 when π‘₯ is greater than negative one, we can replace the 𝑓 of π‘₯ in our right-hand limit with 20. We also know that for any constant π‘˜, the limit as π‘₯ approaches π‘Ž of the constant π‘˜ is just equal to π‘˜. Therefore, we can evaluate the limit as π‘₯ approaches negative one of 20 to just be 20.

We can now use a similar method to calculate the left-hand limit. We will start by restating our rule. If two functions 𝑓 of π‘₯ and 𝑔 of π‘₯ agree everywhere, where π‘₯ is less than π‘Ž, then the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž from the left must be equal to the limit of 𝑔 of π‘₯ as π‘₯ approaches π‘Ž from the left. Since π‘₯ is approaching from the left, we must have that π‘₯ is less than negative one. And from the question, we can see that when π‘₯ is less than negative one, we must have that 𝑓 of π‘₯ is equal to π‘₯ minus four. So using our rule, we can rewrite our left limit as the limit of π‘₯ minus four as π‘₯ approaches negative one from the left.

We know that we can evaluate the limit of a polynomial function 𝑓 by using direct substitution. In our case, we want to find the limit of π‘₯ minus four which we know is a polynomial. So we can use direct substitution to find the limit of π‘₯ minus four as π‘₯ approaches negative one. So we substitute our value of negative one into the function to get negative one minus four. Which we can then evaluate as negative five. Since the left-hand limit must be equal to the limit, we must have that the limit of π‘₯ minus four as π‘₯ approaches negative one is equal to negative five. What is interesting here is if we sketch a graph of our function 𝑓 of π‘₯. So let’s clear some space.

We have when π‘₯ is greater than negative one, we have that 𝑓 of π‘₯ is equal to 20. So this gives us the line 𝑦 equals 20 for π‘₯ is greater than negative one. And when π‘₯ is less than negative one, we have that 𝑓 of π‘₯ is equal to π‘₯ minus four. So we sketch the line 𝑦 equals π‘₯ minus four. But we restrict the domain to π‘₯ being less than negative one. So we can see that our function 𝑓 of π‘₯ has no output when π‘₯ is equal to negative one. And we can also see the two limits that we have calculated on our graph. We can see that as π‘₯ approaches negative one from the left, we get closer and closer to our output value of negative five. And we can see as π‘₯ approaches negative one from the right, we stay at a constant value of 20.

From our calculations, we can see that the left-hand limit and the right-hand limit of 𝑓 of π‘₯ as π‘₯ approaches negative one are not equal. And we can see graphically that the left-hand limit and the right-hand limit of 𝑓 of π‘₯ as π‘₯ approaches negative one are not equal. Therefore, we can conclude that the limit of 𝑓 of π‘₯ as π‘₯ tends to negative one does not exist. Because the left-hand limit and the right-hand limit of 𝑓 of π‘₯ as π‘₯ approaches negative one are not equal.

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