### Video Transcript

Determine the limit as π₯ approaches β of negative five π₯ plus four times two π₯ plus four cubed all divided by π₯ squared plus two times five π₯ plus one squared.

We see the question wants us to evaluate the limit as π₯ approaches β of a rational function. And we call this a rational function because itβs the quotient of two polynomials. In fact, we can see the polynomial in our numerator, which we will call π of π₯, is factored. And the polynomial in our denominator, we will call π of π₯, is also factored. And we know how to evaluate the limit of a rational function in three steps. First, we need to find the highest power of π₯ which appears in either the polynomial in our numerator or the polynomial in our denominator. Weβll call this π₯ to the πth power. We then want to factor this π₯ to the πth power from both the polynomial in our numerator and the polynomial in our denominator inside of our limit. Doing this will give us a new limit which we can then evaluate.

Letβs try doing this to the limit given to us in the question. First, we need to find the highest power of π₯ which appears in either π of π₯ or π of π₯. We might be tempted to try and expand our parentheses and rewrite these as polynomials. However, this is not necessary. If we were to multiply out the parentheses in our numerator, we see the highest power of π₯ would be negative five multiplied by two π₯ all cubed, which is, of course, a term of π₯ to the fourth power. We can do the same in our denominator. We would have π₯ squared multiplied by five π₯ all squared. This gives us 25π₯ to the fourth power, again, a term of π₯ to the fourth power.

So, our highest power of π₯ which appeared anywhere inside of our rational function was four. So, we need to take a factor of π₯ to the fourth power out of both of our polynomials π of π₯ and π of π₯. Weβll start by taking a factor of π₯ to the fourth power out of π of π₯. Weβll do this by treating each factor separately. First, our first factor, the highest power of π₯, is just π₯. So, weβll divide this through by π₯. And in our second factor, we can see the highest power of π₯ is π₯ cubed. So, weβll divide this through by π₯ cubed.

And we can simplify this by using our laws of exponents. We know that π to the πth power divided by π to the πth power is equal to π over π all raised to the πth power. And we can use this to simplify our second factor. Instead of having two π₯ plus four all cubed divided by π₯ cubed, we can have two π₯ plus four divided by π₯ all cubed. Now, we just need to divide through each of our factors. This gives us that π of π₯ is equal to π₯ to the fourth power times negative five plus four over π₯ multiplied by two plus four over π₯ all cubed.

We now want to do the same for π of π₯. We need to take out a factor of π₯ to the fourth power. This time we can see both of our factors have a highest power of π₯ squared. So, weβll divide each of our factors through by π₯ squared. And in our second factor, just as we did before, we can simplify this by using our laws for exponents. And just as we did before, we can divide through by these powers of π₯. This gives us that π of π₯ is equal to π₯ to the fourth power times one plus two over π₯ squared multiplied by five plus one over π₯ squared.

Weβre now ready to see how this affected our limit. We want to know the limit as π₯ approaches β of π of π₯ divided by π of π₯. And we rewrote π of π₯ and π of π₯ by taking out factors of π₯ to the fourth power. This gives us the following expression for our limit. And we can now see that both the numerator and the denominator share a factor of π₯ to the fourth power. We can cancel this. And now this limit is in a form which we can evaluate. We see that our limit is as π₯ is approaching β. This tells us that four over π₯ approaches zero. Two over π₯ squared approaches zero. One over π₯ approaches zero. And four over π₯ also approaches zero. Because in all of these terms, their numerators remain constant; however, their denominators are growing without bound.

We can also see that negative five, two, one, and five all remain constant. They donβt change as π₯ changes. So, their limits will just be equal to themselves. So, this tells us our limit is equal to negative five times two cubed divided by one times five squared, which we can evaluate to give us negative eight divided by five. So, weβve shown the limit as π₯ approaches β of negative five π₯ plus four times two π₯ plus four cubed divided by π₯ squared plus two times five π₯ plus one squared is equal to negative eight divided by five.