### Video Transcript

Given that the limit as π₯ approaches negative one of π₯ squared minus π minus one times π₯ minus π all divided by π₯ plus one is equal to negative three, determine the value of π.

The question tells us that the limit as π₯ approaches negative one of a rational function is equal to negative three. It wants us to use this information to determine the value of π. Since this is the limit of a rational function, we can attempt to evaluate this by using direct substitution. Substituting π₯ is equal to negative one in the function in our limit gives us negative one squared minus π minus one times negative one minus π all divided by negative one plus one. Fully evaluating the expressions in our numerator and our denominator, we see we get the indeterminate form zero divided by zero. So evaluating our limit by direct substitution gained us no information about the value of π. So weβre going to need a different way to evaluate this limit.

We recall that LβHΓ΄pitalβs rule gives us a way of evaluating an indeterminate limit of this form. We recall the following version of LβHΓ΄pitalβs rule. If we have two differentiable functions π and π where π prime of π₯ is not equal to zero around our value of π except possibly when π₯ is equal to π, and the limit as π₯ approaches π of π of π₯ and the limit as π₯ approaches π of π of π₯ are both equal to zero. Then the limit as π₯ approaches π of the quotient of π of π₯ and π of π₯ is equal to the limit as π₯ approaches π of the quotient of π prime of π₯ and π prime of π₯. In this version of LβHΓ΄pitalβs rule, weβre told that if the limit as π₯ approaches π of π of π₯ over π of π₯ gives us the indeterminate form of zero over zero. Then, under these conditions, we can instead calculate the limit as π₯ approaches π of π prime of π₯ divided by π prime of π₯.

Letβs start by verifying that we can actually use LβHΓ΄pitalβs rule in this case. Weβll set our function π of π₯ equal to the numerator of the function in our limit. Thatβs π₯ squared minus π minus one times π₯ minus π. And weβll set our function π of π₯ to be the function in the denominator of our limit. Thatβs π of π₯ is equal to π₯ plus one. The first thing we need to use this version of LβHΓ΄pitalβs rule is that both our functions π and π are differentiable. We see the π is a linear function, and for any value π, π of π₯ is a quadratic. So both of these functions are polynomials, and therefore theyβre both differentiable.

Next, we need to show that derivative of π of π₯ is not equal to zero around π. Remember, π is the constant that π₯ is approaching in our limit. We have the limit as π₯ approaches negative one. So weβll set π equal to negative one. Since π of π₯ is the linear function π₯ plus one, we can just calculate π prime of π₯ directly by using the power rule for differentiation. We multiply by the exponent and reduce the exponent by one. In this case, we just get the constant one. And we can, of course, see that this is never equal zero.

Next, we need to show that the limit as π₯ approaches negative one of π of π₯ and the limit as π₯ approaches negative one of π of π₯ is both equal to zero. However, weβve actually already shown this. When we tried to use direct substitution to evaluate our limit, we saw that our numerator evaluated to give us zero and our denominator also evaluated to give us zero. But the numerator was just the limit as π₯ approaches negative one of π of π₯. And the denominator was the limit as π₯ approaches negative one of π of π₯. So weβve shown the last two prerequisites of this version of the LβHΓ΄pitalβs rule is true. So we can attempt to use it to evaluate our limit.

So by using this version of LβHΓ΄pitalβs rule, we have the limit given to us in the question, which is the limit as π₯ approaches negative one of π of π₯ divided by π of π₯ is equal to the limit as π₯ approaches negative one of π prime of π₯ divided by π prime of π₯. So to evaluate this limit, we need to find π prime of π₯. We recall π of π₯ is equal to π₯ squared minus π minus one times π₯ minus π. And for any value of π, this is just a quadratic. So we can evaluate this derivative by using the power rule for differentiation. We get two π₯ minus π minus one. So now that we found π prime of π₯ and π prime of π₯, weβve shown that our limit is equal to the limit as π₯ approaches negative one of two π₯ minus π minus one divided by one. Of course, dividing by one inside of our limit doesnβt change anything.

And now we see weβre just calculating the limit of a linear function. We can evaluate this by using direct substitution. We substitute π₯ is equal to negative one into our linear function to evaluate the limit. And substituting π₯ is equal to negative one into this limit gives us two times negative one minus π minus one. And if we evaluate and simplify this expression, we can see that itβs equal to negative one minus π. So weβve shown that the limit given to us in the question is equal to negative one minus π. But the question also tells us that this limit must be equal to negative three. So, negative one minus π is equal to negative three. And we can then solve this equation. We get that π is equal to two.

So weβve shown if the limit as π₯ approaches negative one of π₯ squared minus π minus one times π₯ minus π all divided by π₯ plus one is equal to negative three, then the value of π must be equal to two.