Video: Finding the Value of the Variable That Makes a Piecewise-Defined Function Have a Limit at a Certain Point

The function 𝑓(π‘₯) = π‘₯Β² βˆ’ 9π‘˜, if π‘₯ < βˆ’5 and 𝑓(π‘₯) = 2π‘₯ + 116, if π‘₯ > βˆ’5 has a limit as π‘₯ β†’ βˆ’5. What is the value of π‘˜?

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Video Transcript

The function 𝑓 of π‘₯ is equal to π‘₯ squared minus nine π‘˜ if π‘₯ is less than negative five and 𝑓 of π‘₯ is equal to two π‘₯ plus 116 if π‘₯ is greater than negative five has a limit as π‘₯ approaches negative five. What is the value of π‘˜?

The question gives us a piecewise-defined function 𝑓 of π‘₯. And we’re told that this function 𝑓 of π‘₯ has a limit as π‘₯ approaches negative five. We need to find the value of π‘˜ that makes this possible. We recall that we say that the limit as π‘₯ approaches π‘Ž of a function 𝑓 of π‘₯ exists if the following conditions hold. We need the limit as π‘₯ approaches π‘Ž from the left of 𝑓 of π‘₯ and the limit as π‘₯ approaches π‘Ž from the right of 𝑓 of π‘₯ to both exist. And these two limits must be equal. And we’re told that this is true for our function 𝑓 of π‘₯ when our value of π‘Ž is equal to negative five. So let’s set π‘Ž equal to negative five in our definition.

Since we know the limit as π‘₯ approaches negative five of 𝑓 of π‘₯ exists, all of our conditions must be true. The limit as π‘₯ approaches negative five from the left of 𝑓 of π‘₯ and the limit as π‘₯ approaches negative five from the right of 𝑓 of π‘₯ must both exist, and they must both be equal. So let’s try evaluating these limits. We’ll start with the limit as π‘₯ approaches negative five from the left of 𝑓 of π‘₯. Since π‘₯ is approaching negative five from the left, our values of π‘₯ will be less than negative five. And we can see from our piecewise definition of the function 𝑓 of π‘₯, if our values of π‘₯ are less than negative five, our function 𝑓 of π‘₯ is exactly equal to the function π‘₯ squared minus nine π‘˜.

So because our function 𝑓 of π‘₯ is exactly equal to the function π‘₯ squared minus nine π‘˜ for all values of π‘₯ less than negative five, their limits as π‘₯ approaches negative five from the left will be the same. But now we can see this is just the limit of a polynomial. We can evaluate this by direct substitution. Substituting π‘₯ is equal to negative five, we get negative five squared minus nine π‘˜. And since negative five squared is equal to 25, we can simplify this expression to 25 minus nine π‘˜. Let’s do the same to evaluate the limit as π‘₯ approaches negative five from the right of 𝑓 of π‘₯.

Since π‘₯ is approaching negative five from the right, our values of π‘₯ will be greater than negative five. And again, from our piecewise definition of the function 𝑓 of π‘₯ for values of π‘₯ greater than negative five, our function 𝑓 of π‘₯ is equal to two π‘₯ plus 116. So because we’re taking the limit as π‘₯ approaches negative five from the right, we can rewrite 𝑓 of π‘₯ as two π‘₯ plus 116. And we can now see we’re just evaluating the limit of a linear function. We can do this by direct substitution.

Substituting π‘₯ is equal to negative five, we get two times negative five plus 116. And we can calculate this to give us 106. Now, remember, since this limit exists, we know the limit as π‘₯ approaches negative five from the left of 𝑓 of π‘₯ and the limit as π‘₯ approaches negative five from the right of 𝑓 of π‘₯ must both be equal. This tells us that 25 minus nine π‘˜ is equal to 106. So we now just have to solve the equation 25 minus nine π‘˜ is equal to 106. We’ll subtract 25 from both sides of the equation. This gives us negative nine π‘˜ is equal to 81. Then, we’ll just divide both sides of the equation through by negative nine. This gives us that π‘˜ is equal to negative nine.

So we’ve shown if the function 𝑓 of π‘₯ is equal to π‘₯ squared minus nine π‘˜ if π‘₯ is less than negative five and 𝑓 of π‘₯ is equal to two π‘₯ plus 116 if π‘₯ is greater than negative five has a limit as π‘₯ approaches negative five. Then the value of π‘˜ must be equal to negative nine.

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