### Video Transcript

Given that π of π₯ is equal to π₯ cubed and π of π₯ is equal to four π to the power of three π₯, find whether π of π₯ or π of π₯ is the dominant function.

The question gives us a polynomial function π of π₯ and an exponential function π of π₯. The question wants us to decide which of these is the dominant function. It will help us to recall what we mean by a dominant function. For eventually positive functions π of π₯ and π of π₯, we say that π dominates π if the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to β. We also say that π dominates π if this limit is equal to zero.

Finally, we can say that π and π grow at a similar rate if the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to some nonzero real constant π. So to decide which of these functions is dominant, weβre going to need to evaluate the limit as π₯ approaches β of π of π₯ divided by π of π₯. Itβs also worth noting both π₯ cubed and four π to the power three π₯ are eventually positive functions. For example, theyβre both positive when π₯ is greater than zero.

So we want to evaluate the limit as π₯ approaches β of π₯ cubed divided by four π to the power of three π₯. However, if we try to evaluate this limit directly, we see that π₯ cubed approaches β as π₯ approaches β. And four π to the power of three π₯ also approaches β as π₯ approaches β. So this gives us an indeterminate form. Since weβre calculating the limit of a quotient of two functions and it gives us an indeterminate form, we could try evaluating this by LβHopitalβs rule.

We recall the following version of LβHopitalβs rule. If we have two differentiable functions π and π, where the limit as π₯ approaches β of π of π₯ is equal to β and the limit as π₯ approaches β of π of π₯ is also equal to β. Then the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to the limit as π₯ approaches β of π prime of π₯ divided π prime of π₯, provided this limit exists or is equal to positive or negative β. In other words, if our limit as π₯ approaches β of the quotient of two functions gives us an indeterminate form, we can attempt to evaluate this limit by taking the limit as π₯ approaches β of π prime of π₯ divided by π prime of π₯.

Letβs start by checking we can use LβHopitalβs rule in this case. First, we need that π and π are differentiable functions. And we can see that this is true. π of π₯ is a polynomial and all polynomials are differentiable. Similarly, π of π₯ is an exponential function, and exponential functions are also differentiable. So our first condition of LβHopitalβs rule is true.

Next, we need both the limit as π₯ approaches β of π of π₯ and the limit as π₯ approaches β of π of π₯ to be equal to β. In fact, we chose this version of LβHopitalβs rule because we already had this case when we tried to evaluate our limit directly. We saw the limit as π₯ approaches β of π₯ cubed was equal to β and the limit as π₯ approaches β of four π to the power of three π₯ was also equal to β. So all of our conditions are true. That means we can attempt to use this version of LβHopitalβs rule to evaluate our limit.

First, we need to find expressions for π prime of π₯ and π prime of π₯. We can find π prime of π₯ by using the power rule for differentiation. We multiply by the exponent and reduce the exponent by one. This gives us three π₯ squared. Similarly, we can find π prime of π₯ by using our differentiation rules for exponential functions. We need to just multiply it by the derivative of our exponent. This gives us 12 times π to the power of three π₯.

So by this version of LβHopitalβs rule, we now need to evaluate the limit as π₯ approaches β of three π₯ squared divided by 12π to the power of three π₯. However, we can again see as π₯ approaches β, three π₯ squared approaches β and as π₯ approaches β, 12π to the power of three π₯ also approaches β. And this left us with an indeterminate form. However, we can try using LβHopitalβs rule on this new limit to help us evaluate our original limit. First, letβs update our version of LβHopitalβs rule.

Weβre now using this on the functions π prime of π₯ and π prime of π₯. If we can show that the conditions hold for LβHopitalβs rule for π prime of π₯ and π prime of π₯. Instead of calculating the limit as π₯ approaches β of π prime of π₯ divided by π prime of π₯, we could instead evaluate the limit as π₯ approaches β of π double prime of π₯ divided by π double prime of π₯. So letβs start checking our conditions. First, we need that π prime of π₯ and π prime of π₯ are differentiable. And again, we can see that this is true. π prime of π₯ is a polynomial and π prime of π₯ is an exponential function.

In fact, we can just calculate these expressions directly. π double prime of π₯ is six π₯ and π double prime of π₯ is 36π to the power of three π₯. So weβve shown that π prime and π prime are both differentiable. We now need to show the limit as π₯ approaches β of π prime and the limit as π₯ approaches β of π prime are both equal to β. In fact, we already discussed this when we tried to evaluate our limit directly. As π₯ approached β, both of these limits grew without bound, so they both approached β.

So we can now attempt to evaluate this limit by LβHopitalβs rule. This gives us the limit as π₯ approaches β of π double prime of π₯ divided by π double prime of π₯, which is the limit as π₯ approaches β of six π₯ divided by 36π to the power of three π₯. And again, if we tried to evaluate this limit directly, we see our numerator and our denominator both grow without bound. This gives us the indeterminate form β divided by β. And at this point, we might be worried that LβHopitalβs rule will never allow us to evaluate this limit.

However, remember, each time we differentiate our polynomial, weβre reducing its largest power by one. In fact, the next time we differentiate this polynomial, weβll just have a constant. This tells us we just need to use LβHopitalβs rule one more time. So letβs again update LβHopitalβs rule. This time we want to use it on our functions π double prime of π₯ and π double prime of π₯. Again, we know both of these functions are differentiable. In fact, we can just calculate π triple prime of π₯ and π triple prime of π₯. π triple prime of π₯ is six, and π triple prime of π₯ is 108 times π to the power of three π₯.

And we know the limit as π₯ approaches β of π double prime of π₯ and the limit as π₯ approaches β of π prime of π₯ are both equal to β. This is because both of these functions grow without bound as π₯ is approaching β. Now, using LβHopitalβs rule, we can write our limit as the limit as π₯ approaches β of π triple prime of π₯ divided by π triple prime of π₯. This gives us the limit as π₯ approaches β of six divided by 108 times π to the power of three π₯.

However, this time when we try to evaluate this limit, we see our numerator is a constant, six. However, our denominator grows without bound. In other words, this limit is getting closer and closer to zero. And remember, we say if the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to zero, then π of π₯ dominates π of π₯. So what weβve shown is if π of π₯ is equal to π₯ cubed and π of π₯ is equal to four π to the power of three π₯, then by using LβHopitalβs rule multiple times we can conclude that π of π₯ is the dominant function.