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

Discuss the existence of lim_(π‘₯ β†’ βˆ’15) 𝑓(π‘₯) given 𝑓(π‘₯) = βˆ’15π‘₯ βˆ’ 15 if π‘₯ < βˆ’15, 𝑓(π‘₯) = π‘₯Β² βˆ’ 15 if π‘₯ β‰₯ βˆ’15. [A] The limit does not exist because lim_(π‘₯ β†’ βˆ’15⁺) 𝑓(π‘₯) exists, but lim_(π‘₯ β†’ βˆ’15⁻) 𝑓(π‘₯) does not exist. [B] The limit does not exist because lim_(π‘₯ β†’ βˆ’15⁻) 𝑓(π‘₯) exists, but lim_(π‘₯ β†’ βˆ’15⁺) 𝑓(π‘₯) does not exist. [C] The limit does not exist because both lim_(π‘₯ β†’ βˆ’15⁻) 𝑓(π‘₯) and lim_(π‘₯ β†’ βˆ’15⁺) 𝑓(π‘₯) exist, but are not equal. [D] The limit exists and equals 210. [E] The limit exists and equals βˆ’15.

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Video Transcript

Discuss the existence of the limit of 𝑓 of π‘₯ as π‘₯ approaches negative 15 given 𝑓 of π‘₯ equals negative 15 π‘₯ minus 15 if π‘₯ is less than negative 15 and π‘₯ squared minus 15 if π‘₯ is greater than or equal to negative 15.

We have five options to choose from here A, B, C, and D. Option A says that the limit does not exist because the limit of 𝑓 of π‘₯ as π‘₯ approaches negative 15 from the right exists, but the limit of 𝑓 of π‘₯ as π‘₯ approaches negative 15 from the left does not exist. Option B: the limit does not exist because the limit of 𝑓 of π‘₯ as π‘₯ approaches negative 15 from the left exists, but the limit of 𝑓 of π‘₯ as π‘₯ approaches negative 15 from the right does not exist. Option C: the limit does not exist because both the limit of 𝑓 of π‘₯ as π‘₯ approaches negative 15 from the left and the limit of 𝑓 of π‘₯ as π‘₯ approaches negative 15 from the right exist, but are not equal. Option D: the limit exists and equals 210. And option E: the limit exists and equals negative 15.

There’s quite a lot to take in here. But basically, the question is asking about the limit of a piecewise function as π‘₯ approaches the value for which the rule defining the function changes. We’re asked to discuss this and the options we’re given make it clear what this means. If the limit exists, we should say that and we should also give it a value. And if the limit does not exist, we should explain why the limit does not exist β€” which criterion fails.

With this understanding in mind, we can erase the options from the screen, which will give us space to actually answer the question. So what we need to happen for this limit to exist? Well, for a general function 𝑓, the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž exists if both the left-hand and right-hand limits exist and are equal. So there are three things which needs to be true: the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž from the left must exist, the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž from the right must exist, and these two limits must be equal.

Thinking back to the options, three of them said that our limits does not exist. And the reason given for the limits not existing was that exactly one of these three things wasn’t true β€” a different thing for each option. To answer our question then, we’re going to apply this definition to our function which is defined as above with the limit value π‘Ž as negative 15. We’re going to see if the left-hand and right-hand limits exist. And if they do, if they are equal. And the way to do this is to try to evaluate these limits.

We choose to start with the left-hand limit, where 𝑓 of π‘₯ is defined as above. This superscript minus sign, it tells us that this is a left-hand limit. So this is the limit of 𝑓 of π‘₯ as π‘₯ approaches negative 15 from the left with values of π‘₯ less than negative 15, but getting closer and closer to negative 15. And when π‘₯ is less than negative 15, 𝑓 of π‘₯ we’re told is negative 15π‘₯ minus 15. And so we can replace 𝑓 of π‘₯ inside the limit by negative 15π‘₯ minus 15. We can evaluate this limit which is just the limit of a linear function by direct substitution.

Substituting negative 15 for π‘₯, we get negative 15 times negative 15 minus 15 which is 210. We do something very similar to find the limit of 𝑓 of π‘₯ as π‘₯ approaches negative 15 from the right. Remember this superscript plus sign, it tells us that π‘₯ is greater than negative 15 as it approaches negative 15. And we can see from the definition of 𝑓, when π‘₯ is greater than or equal to negative 15 and so certainly when 𝑓 is just greater than negative 15, 𝑓 of π‘₯ is π‘₯ squared minus 15.

Therefore, inside this limit, 𝑓 of π‘₯ is π‘₯ squared minus 15. And as this is the limit of a polynomial, we can find its value by direct substitution. Substituting negative 15 for π‘₯ and then evaluating this expression to get 210, we have shown by evaluating them that both the left-hand and right-hand limits exist. And we can also see that they are equal. They are both 210. The left-hand limit exists, the right-hand limit exists, and the values of these two elements are equal. And so the limit of 𝑓 of π‘₯ as π‘₯ approaches negative 15 period exists.

And when the limit exists, its value is the value of the left-hand and right-hand limits, which are of course equal. The limit of 𝑓 of π‘₯ defined above therefore as π‘₯ approaches negative 15 exists and equals 210. This was option D in the options we had at the start.

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