Question Video: Determine the Limit of a Rational Function Raised to a Linear Exponent Mathematics

Determine lim_(π‘₯ β†’ ∞) ((π‘₯Β² βˆ’ π‘₯ βˆ’ 20)/(π‘₯Β² βˆ’ 4π‘₯ βˆ’ 5))^(3(π‘₯ + 1)).

06:18

Video Transcript

Determine the limit as π‘₯ approaches ∞ of π‘₯ squared minus π‘₯ minus 20 all divided by π‘₯ squared minus four π‘₯ minus five all raised to the power of three multiplied by π‘₯ plus one.

In this question, we’re asked to evaluate a limit, and the first thing we should always try is to see if we can evaluate this limit directly. In this case, we can see as π‘₯ is approaching ∞, inside our parentheses, we have a rational function. If we just substitute ∞ into this expression, we get ∞ divided by ∞, which we know is an indeterminate form. So instead, because this is a quotient of two rational functions where the numerator and denominator are equal degree, we know the limit as π‘₯ approaches ∞ of this expression will be determined entirely by the quotient of the leading terms. In this case, the quotient of the leading terms π‘₯ squared over π‘₯ squared is equal to one. So as π‘₯ approaches ∞, the rational function inside our parentheses is approaching one.

It’s a lot easier to see what happens to our exponent. As π‘₯ is approaching ∞, this is growing without bound. So our exponent is approaching ∞. So trying to evaluate this limit directly, we get one to the power of ∞, which is an indeterminate form, so we can’t use this to evaluate our limit. Instead, we’re going to need to try some other method. There’s a few different things we can try. For example, we could use algebraic division to simplify the rational function inside our parentheses. And this would work, and we could answer our question in this manner. However, there’s a second method we can try. We can factor fully both the numerator and denominator.

Let’s start with the numerator. We need two numbers which multiply to give negative 20 and add to give negative one. And this is negative five and four, so we can factor the numerator to give π‘₯ minus five multiplied by π‘₯ plus four. We can do the same in our denominator. We want two numbers which multiply to give negative five and add to give negative four. And we can see this is negative five and one. So we can factor our denominator to give us π‘₯ minus five multiplied by π‘₯ plus one.

So we’ve simplified the rational function inside our parentheses, and now we can notice something interesting. There’s a shared factor of π‘₯ minus five in our numerator and our denominator. We might be worried that we’re not allowed to do this. However, remember, since we’re taking the limit as π‘₯ approaches ∞, eventually our values of π‘₯ are going to be bigger than five. So because we’re taking the limit as π‘₯ approaches ∞, eventually this is just the quotient of two equal positive numbers, so we can just cancel these out. This gives us the limit as π‘₯ approaches ∞ of π‘₯ plus four all over π‘₯ plus one all raised to the power of three times π‘₯ plus one.

We might now try evaluating this by using direct substitution. However, if we were to do this, we would see we still get one to the power of ∞. So we still need to use more manipulation, and there’s a lot of different ways we could do this. We’re going to start by simplifying the rational function inside our parentheses. We could do this by using algebraic division. However, it’s easier to rewrite this as π‘₯ plus one plus three all divided by π‘₯ plus one. Then we can divide π‘₯ plus one in our numerator by π‘₯ plus one to get one. And then we just have three over π‘₯ plus one remaining. This gives us the limit as π‘₯ approaches ∞ of one plus three over π‘₯ plus one all raised to the power of three multiplied by π‘₯ plus one.

Now we can see that this limit is starting to remind us of our limit results involving Euler’s constant 𝑒. And we can use either of the two different results to evaluate this limit. It doesn’t matter which one we use. We’re going to use the fact that the limit as 𝑛 approaches ∞ of one plus one over 𝑛 all raised to the 𝑛th power is equal to 𝑒. We’re going to do this by substitution. Inside our parentheses, we want one plus one over our variable. So we’re going to use the substitution one over 𝑛 is equal to three divided by π‘₯ plus one. But by doing this, we would have a limit as π‘₯ approaches ∞ of a function involving both 𝑛 and π‘₯. So we want to rewrite our entire limit in terms of 𝑛.

To do this, we’re going to start by seeing what happens to our value of 𝑛 as π‘₯ approaches ∞. On the right-hand side of our substitution, we can see as π‘₯ approaches ∞, our numerator remains constant. However, our denominator is growing without bound. So the right-hand side of this equation approaches zero. So as π‘₯ approaches ∞, one over 𝑛 has to be approaching zero. Normally, all this would tell us is that our values of 𝑛 are growing without bound. However, the right-hand side of this equation is eventually positive. This means our values of 𝑛 must be approaching positive ∞. So as π‘₯ approaches ∞, our values of 𝑛 are approaching ∞.

The last thing we’re going to need to do is find an expression for π‘₯ in terms of 𝑛. We’ll do this by cross multiplying our substitution. We multiply both sides of our equation through by π‘₯ plus one and multiply both through by 𝑛. This gives us π‘₯ plus one is equal to three 𝑛. Normally, we would then rearrange this for π‘₯. However, we can see that π‘₯ plus one appears in our expression, so we can instead just substitute this for three 𝑛. This gives us the limit as 𝑛 approaches ∞ of one plus one over 𝑛 all raised to the power of three multiplied by three 𝑛. And of course, we can simplify this. Three multiplied by three 𝑛 is equal to nine 𝑛.

And although it will be tempting to just directly use our limit result to evaluate this expression, we can’t yet because we still have a factor of nine in our exponent. But we know how to take an exponent outside of the limit by using the power rule for limits. So by using our laws of exponents, we’ll start by rewriting our limit. π‘Ž to the power of 𝑏 times 𝑐 is equal to π‘Ž to the power of 𝑏 all raised to the power of 𝑐. By setting π‘Ž equal to one plus one over 𝑛, 𝑏 equal to 𝑛, and 𝑐 equal to nine, we get the limit as 𝑛 approaches ∞ of one plus one over 𝑛 all raised to the power of 𝑛 all raised to the ninth power.

Now we want to take our exponent outside of our limit. So to do this, we’re going to need to use the power rule for limits. One version of the power rule for limits tells us the limit as 𝑛 approaches ∞ of 𝑓 of 𝑛 raised to the π‘˜th power will be equal to the limit as 𝑛 approaches ∞ of 𝑓 of 𝑛 all raised to the π‘˜th power so long as the limit as 𝑛 approaches ∞ of 𝑓 of 𝑛 exists and π‘˜ is an integer. And the usual power rule for limits does not force our value of π‘˜ to be an integer. However, because in this case we do have π‘˜ being an integer, we can guarantee this is true. So it’s one less thing to check.

Therefore, by using the power rule for limits, we can rewrite our limit as the limit as 𝑛 approaches ∞ of one plus one over 𝑛 all raised to the 𝑛th power all raised to the ninth power provided the limit as 𝑛 approaches ∞ of one plus one over 𝑛 all raised to the 𝑛th power exists. But of course, we already know that this limit exists. This is our limit result involving Euler’s constant 𝑒, so we’ve justified our use of the power rule for limits. We can then substitute 𝑒 for this limit, giving us our final answer of 𝑒 to the ninth power.

Therefore, we were able to show the limit as π‘₯ approaches ∞ of π‘₯ squared minus π‘₯ minus 20 all divided by π‘₯ squared minus four π‘₯ minus five all raised to the power of three multiplied by π‘₯ plus one is equal to 𝑒 to the ninth power.

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