# Video: Finding the Limit of Rational Functions at Infinity

Determine lim_(𝑥 → ∞) ((5𝑥² + 3)²)/((𝑥 − 5)²(4𝑥² − 3𝑥)), if it exists.

04:32

### Video Transcript

Determine the limit as 𝑥 approaches ∞ of five 𝑥 squared plus three all squared all divided by 𝑥 minus five squared times four 𝑥 squared minus three 𝑥 if it exists.

The question is asking us to evaluate the limit as 𝑥 approaches ∞ of a rational function. We call this a rational function because it’s the quotient of two polynomials. However, it might be difficult to see that this is indeed a rational function. However, remember, if we multiply polynomials together, then we just get a different polynomial. So this is just a quotient of two factored polynomials. Let’s call the polynomial in our numerator 𝑃 of 𝑥 and the polynomial in our denominator 𝑄 of 𝑥. And we know how to evaluate the limit as 𝑥 approaches ∞ of a rational function 𝑃 of 𝑥 divided by 𝑄 of 𝑥.

First, we want to find the highest power of 𝑥 which appears in our rational function. Second, we want to divide both 𝑃 of 𝑥 and 𝑄 of 𝑥 through by this power of 𝑥. We can then attempt to evaluate the limit of this new function. So let’s apply this method to the limit given to us in the question. We want to find the highest power of 𝑥 which appears in either 𝑃 of 𝑥 or 𝑄 of 𝑥.

We might be tempted to multiply out the expressions given to us in the numerator and the denominator. However, this is not necessary. We can see in our numerator the highest power of 𝑥 will be five 𝑥 squared all squared, which is 25𝑥 to the fourth power. We can do the same in our denominator. We see the highest power of 𝑥 will be 𝑥 multiplied by 𝑥 multiplied by four 𝑥 squared, which is four 𝑥 to the fourth power. So when both 𝑃 of 𝑥 and 𝑄 of 𝑥 are highest power of 𝑥 was 𝑥 to the fourth power.

Next, we need to divide both 𝑃 of 𝑥 and 𝑄 of 𝑥 through by 𝑥 to the fourth power. In other words, we want to evaluate the limit as 𝑥 approaches ∞ of five 𝑥 squared plus three squared divided by 𝑥 to the fourth power all divided by 𝑥 minus five squared times four 𝑥 squared minus three 𝑥 divided by 𝑥 to the fourth power. Let’s simplify the expression in our numerator and the expression in our denominator. We’ll start with the expression on our numerator, five 𝑥 squared plus three all squared divided by 𝑥 to the fourth power.

We’ll start by rewriting 𝑥 to the fourth power as 𝑥 squared all squared. We can simplify this by using our laws for exponents. 𝑎 to the 𝑛th power divided by 𝑏 to the 𝑛th power is equal to 𝑎 over 𝑏 all raised to the 𝑛th power. This gives us five 𝑥 squared plus three all divided by 𝑥 squared all squared. Now, what we want to do is divide through five 𝑥 squared plus three by 𝑥 squared. And this gives us, inside of our parentheses, five plus three over 𝑥 squared. So that’s the expression in our numerator simplified. Let’s do the same with the expression in our denominator.

We want to simplify 𝑥 minus five squared times four 𝑥 squared minus three 𝑥 all divided by 𝑥 to the fourth power. To do this, we want to divide each factor by the highest power of 𝑥 which it has. In this case, we’ll divide both factors by 𝑥 squared. To divide our first factor through by 𝑥 squared, we’ll use the same law of exponents. This gives us 𝑥 minus five over 𝑥 all squared. And in our second factor, we’ll just divide through by 𝑥 squared. This gives us four minus three divided by 𝑥. We can simplify our first factor by dividing through by 𝑥. 𝑥 minus five all over 𝑥 is equal to one minus five divided by 𝑥.

Now that we’ve simplified both the expression in our numerator and the expression in our denominator, we want to calculate the limit as 𝑥 approaches ∞ of five plus three over 𝑥 squared all squared divided by one minus five over 𝑥 squared times four minus three over 𝑥. And we can evaluate this limit directly. Since 𝑥 is approaching ∞, we can see that three over 𝑥 squared, five over 𝑥, and three over 𝑥 are all approaching zero. That’s because for all three of these terms, their numerator remains constant. However, their denominator is growing without bound. And we can see that our constants five, one, and four are just constants. They don’t change as 𝑥 changes, so their limit is just equal to themselves.

We have to be careful at this point since we need to remember that we’re taking the exponents of some of these factors. Combining this information, this gives us that our limit is equal to five squared divided by one squared times four. And we can calculate this expression to give us 25 divided by four. Therefore, we’ve shown the limit as 𝑥 approaches ∞ of five 𝑥 squared plus three all squared divided by 𝑥 minus five squared times four 𝑥 squared minus three 𝑥 is equal to 25 divided by four.