# Video: Determine Whether a Polynomial or Exponential Function Is Dominant

Given that π(π₯) = π₯Β³ and π(π₯) = 4π^(3π₯), find whether π(π₯) or π(π₯) is the dominant function.

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### 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, weβre going to need to evaluate the limit as π₯ approaches β of π of π₯ divided by π of π₯. Itβs also worth noting both π₯ cubed and four π to the power three π₯ are eventually positive functions. For example, theyβre 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 weβre calculating the limit of a quotient of two functions and it gives us an indeterminate form, we could try evaluating this by LβHopitalβs rule.

We recall the following version of LβHopitalβs 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₯.

Letβs start by checking we can use LβHopitalβs 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 LβHopitalβs 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 LβHopitalβs 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 LβHopitalβs 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 LβHopitalβs 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 LβHopitalβs rule on this new limit to help us evaluate our original limit. First, letβs update our version of LβHopitalβs rule.

Weβre now using this on the functions π prime of π₯ and π prime of π₯. If we can show that the conditions hold for LβHopitalβs 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 letβs 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 weβve 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 LβHopitalβs 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 LβHopitalβs rule will never allow us to evaluate this limit.

However, remember, each time we differentiate our polynomial, weβre reducing its largest power by one. In fact, the next time we differentiate this polynomial, weβll just have a constant. This tells us we just need to use LβHopitalβs rule one more time. So letβs again update LβHopitalβs 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 LβHopitalβs 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 weβve 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 LβHopitalβs rule multiple times we can conclude that π of π₯ is the dominant function.