### Video Transcript

Determine the limit as π₯ approaches three of eight π to the power of π₯ minus eight π cubed all divided by nine π₯ minus 27.

Weβre asked to determine the limit as π₯ approaches three of the quotient of two functions. And in fact, we can see both the function in our numerator and the function in our denominator are continuous functions. So we can attempt to evaluate this limit by using direct substitution. So weβll try and evaluate this limit by substituting π₯ is equal to three into the function inside of our limit. This gives us eight π cubed minus eight π cubed all divided by nine times three minus 27.

In our numerator, eight π cubed minus eight π cubed is equal to zero. And in our denominator, nine times three is 27. So our denominator is also equal to zero. So by trying direct substitution, we got the indeterminate form zero divided by zero. This means we canβt determine the value of this limit by using direct substitution. This means weβre going to need to try and evaluate our limit by using a different method. Since weβre evaluating the limit of the quotient of two functions and we know how to differentiate both of these functions, we could try using LβHΓ΄pitalβs rule. So weβll start by recalling LβHΓ΄pitalβs rule.

This tells us if π and π are differentiable functions and π prime of π₯ is not equal to zero around some constant 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 π of π₯ over π of π₯ is equal to the limit as π₯ approaches π of π prime of π₯ divided by π prime of π₯. In other words, we can use this to turn a question about the limit of the quotient of two functions into a question about the limits of the quotient of their derivatives.

To use LβHΓ΄pitalβs rule on the limit given to us in the question, weβre first going to need to determine our functions, π and π, and our value of π. We see that our limit is as π₯ is approaching three. So weβll set our value of π equal to three. In fact, we can update LβHΓ΄pitalβs rule with the value of π equal to three. Next, to use LβHΓ΄pitalβs rule, weβll set our function π to be the function in our numerator, thatβs eight π to the power of π₯ minus eight π cubed, and π of π₯ to be the function in our denominator. Thatβs nine π₯ minus 27.

We now need to check our prerequisites for using LβHΓ΄pitalβs rule. First, we need to check that π and π are differentiable around π₯ is equal to three. And we can see this is true from their definitions. First, π of π₯ is an exponential function minus a constant. So it must be differentiable for all real values of π₯. Next, π of π₯ is a linear function, so itβs also differentiable for all real values of π₯. Therefore, weβve shown both π and π are differentiable around π₯ is equal to three. Next, we need to show that π prime of π₯ is not equal to zero around the value of three. However, it can be equal to zero when π₯ is equal to three.

And the easiest way to do this is to find an expression for π prime of π₯. Thatβs the derivative of the linear function nine π₯ minus 27 with respect to π₯. Its derivative will just be the coefficient of π₯, which is nine. And this is not equal to zero for any value of π₯, so π prime of π₯ is not equal to zero around π₯ is equal to three. Next, we want to show the limit as π₯ approaches three of π of π₯ is equal to zero and the limit as π₯ approaches three of π of π₯ is equal to zero. Now, we could do this by using direct substitution. However, this is not necessary since weβve already done this.

Consider what happened when we first tried to evaluate our limit by using direct substitution. In our numerator, we directly substituted π₯ is equal to three into our function π of π₯, and we saw that this was equal to zero. We did the same in our denominator for the function π of π₯. And we also saw that this was equal to zero. So just the numerator of this calculation showed the limit as π₯ approaches three of π of π₯ was equal to zero. And just the denominator of this calculation showed the limit as π₯ approaches three of π of π₯ is equal to zero.

Therefore, all of our prerequisites for using LβHΓ΄pitalβs rule are true. And that means instead of calculating the limit as π₯ approaches three of π of π₯ over π of π₯, we can instead try evaluating the limit as π₯ approaches three of π prime of π₯ divided by π prime of π₯. However, to do this, weβre going to need to find an expression for π prime of π₯. Thatβs the derivative of eight π to the power of π₯ minus eight π cubed with respect to π₯. And we can do this term by term.

First, the derivative of the exponential function is equal to itself. And we know the derivative of a constant is just equal to zero. So we get π prime of π₯ is equal to eight π to the power of π₯. Now, we can just apply LβHΓ΄pitalβs rule to the limit given to us in the question. We get that this is equal to the limit as π₯ approaches three of π prime of π₯ divided by π prime of π₯. We then substitute our expressions for π prime of π₯ and π prime of π₯.

We get the limit as π₯ approaches three of eight π to the power of π₯ divided by nine. And this is the limit of a continuous function. So we can evaluate this by using direct substitution. So we substitute in π₯ is equal to three. This gives us eight over nine times π cubed, which is our final answer.

Therefore, by using LβHΓ΄pitalβs rule, we were able to show the limit as π₯ approaches three of eight π to the power of π₯ minus eight π cubed all divided by nine π₯ minus 27 is equal to eight over nine times π cubed.