Video: Determine Which of Two Polynomials Is Dominant

Given that 𝑓(π‘₯) = 3π‘₯Β³ + 7π‘₯ and 𝑔(π‘₯) = 5π‘₯Β² + 4, use lim_(π‘₯ β†’ ∞) 𝑓(π‘₯)/𝑔(π‘₯) to determine whether 𝑓(π‘₯) or 𝑔(π‘₯) is dominant.

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Video Transcript

Given that 𝑓 of π‘₯ is equal to three π‘₯ cubed plus seven π‘₯ and 𝑔 of π‘₯ is equal to five π‘₯ squared plus four, use the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ to determine whether 𝑓 of π‘₯ or 𝑔 of π‘₯ is dominant.

The question gives us two polynomial functions, 𝑓 of π‘₯ and 𝑔 of π‘₯. The question wants us to determine whether 𝑓 of π‘₯ or 𝑔 of π‘₯ is dominant by evaluating the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯. First, let’s recall what it means for one function to dominate another function. For eventually positive functions 𝑓 and 𝑔, if the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to ∞, then we say that 𝑓 of π‘₯ dominates 𝑔 of π‘₯. However, if the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to zero, then we say that 𝑔 of π‘₯ dominates 𝑓 of π‘₯. Finally, if the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to some finite, nonzero, real constant π‘Ž, then we say neither of these functions are dominant.

So, we can see we can decide which of our functions is dominant by evaluating the limit as π‘₯ approaches ∞ of three π‘₯ cubed plus seven π‘₯ divided by five π‘₯ squared plus four. However, if we try to evaluate this limit directly, we see that both our numerator and our denominator are growing without bound. We get the indeterminant form ∞ divided by ∞. And we have a few choices of methods that we could use to try and evaluate this limit. For example, we could use L’HΓ΄pital’s rule. And this would work. However, in this case, we’re evaluating the limit as π‘₯ approaches ∞ of a rational function. So, we can also evaluate this limit by dividing our numerator and our denominator through by the highest power of π‘₯ which appears in our rational function.

Let’s start by finding the highest power of π‘₯ which appears in our rational function. We can see that this is π‘₯ cubed. So, we’re going to divide our numerator and our denominator through by π‘₯ cubed. Let’s start with the numerator of this limit. That’s three π‘₯ cubed plus seven π‘₯ divided by π‘₯ cubed. We divide through by π‘₯ cubed, and we get three plus seven over π‘₯ squared. We can do the same with our denominator. We get five π‘₯ squared plus four all divided by π‘₯ cubed is equal to five over π‘₯ plus four over π‘₯ cubed. This gives us that we want to evaluate the limit as π‘₯ approaches ∞ of three plus seven over π‘₯ squared all divided by five over π‘₯ plus four over π‘₯ cubed.

And we can now see since our limit is as π‘₯ is approaching ∞, seven over π‘₯ squared, five over π‘₯, and four over π‘₯ cubed are all approaching zero. Their numerators remain constant; however, their denominators are growing without bound. And of course, our constant three does not vary as π‘₯ varies. So, this tells us that our limit is equal to three divided by zero. And it’s worth noting that this does not necessarily mean that our limit is approaching positive or negative ∞. It could be fluctuating between the two.

To help us see what case we’re in, we’ll multiply our numerator and our denominator through by π‘₯. Of course, this is the same as if we had originally just divided through our numerator and our denominator by π‘₯ squared. This gives us the limit as π‘₯ approaches ∞ of three π‘₯ plus seven over π‘₯ all over five plus four over π‘₯ squared. And this time our result is much clearer. As π‘₯ is approaching ∞, both seven over π‘₯ and four over π‘₯ squared are approaching zero. Their numerators remain constant; however, their denominators are growing without bound. So, now we can see as π‘₯ approaches ∞, our denominator approaches the constant five; however, our numerator is growing without bound. In other words, this limit is equal to positive ∞.

And remember, if the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to ∞, then we say that 𝑓 of π‘₯ dominates 𝑔 of π‘₯. Therefore, what we’ve shown is if 𝑓 of π‘₯ is equal to three π‘₯ cubed plus seven π‘₯ and 𝑔 of π‘₯ is equal to five π‘₯ squared plus four π‘₯, then the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to positive ∞. And this tells us that 𝑓 of π‘₯ is the dominant function.

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