Question Video: Finding the Limit of a Rational Function

Find lim_(π‘₯ β†’ 5) (π‘₯⁴ βˆ’ 625)/(π‘₯Β³ βˆ’ 125).

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Video Transcript

Find the limit as π‘₯ approaches five of π‘₯ to the fourth power minus 625 all divided by π‘₯ cubed minus 125.

In this question, we’re asked to evaluate the limit of a rational function. And we recall we can try doing this by using direct substitution. We substitute π‘₯ is equal to five into our rational function to get five to the fourth power minus 625 divided by five cubed minus 125. However, if we evaluate the numerator and denominator of this expression, we see it’s equal to zero divided by zero. This is an indeterminate form. This does not mean that we can’t evaluate this limit. All it means is we can’t evaluate this limit by using direct substitution alone. So we’re going to need to use a different method.

And there are a few different ways of evaluating this limit. The easiest way is to notice that this limit is given in the form of the limit of a difference between powers. This means we can evaluate this limit by recalling the following result. The limit as π‘₯ approaches some constant π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power is equal to 𝑛 divided by π‘š multiplied by π‘Ž to the power of 𝑛 minus π‘š. And this is provided π‘š is not equal to zero and π‘Ž to the 𝑛th power, π‘Ž to the π‘šth power, and π‘Ž to the power of 𝑛 minus π‘š all exist.

We know 625 is five to the fourth power and 125 is five cubed. So we can rewrite our limit as the limit as π‘₯ approaches five of π‘₯ to the fourth power minus five to the fourth power all divided by π‘₯ cubed minus five cubed. Now we can see we’ve rewritten this in the form of a limit of a difference of powers. Our value of π‘Ž is five, our value of 𝑛 is four, and our value of π‘š is three. Therefore, we can evaluate our limit by using our limit result. We substitute these values in to get four-thirds multiplied by five to the power of four minus three. And we can evaluate this expression. Four minus three is equal to one, and five to the first power is equal to itself. It’s equal to five. Five times four over three is 20 divided by three, which is our final answer. And this is the easiest way of answering this question.

However, it can be useful to recall algebraically why this result holds true. So instead, we’ll also go through the algebraic way of evaluating this limit. This will help reiterate why our limit result in general holds true. To evaluate this limit algebraically, we recall when we substituted π‘₯ is equal to five into our rational function, we got zero divided by zero. By the remainder theorem, this means the polynomial in our numerator and the polynomial in our denominator are divisible by π‘₯ minus five. To evaluate this limit algebraically, we want to cancel all of the shared factors of π‘₯ minus five in the numerator and denominator. So we’ll start by factoring both of these polynomials.

Let’s start with the polynomial in our numerator. We can see this is a difference between squares since π‘₯ to the fourth power is π‘₯ squared all squared and 625 is 25 squared. Therefore, we can rewrite our numerator as π‘₯ squared plus 25 multiplied by π‘₯ squared minus 25. And we can even notice something interesting. π‘₯ squared minus 25 is also a difference between two squares. So we can already factor this further. It’s equal to π‘₯ minus five multiplied by π‘₯ plus five.

Now, let’s move on to factoring our denominator. There’s a few different ways of doing this. For example, we could use algebraic division to take out a factor of π‘₯ minus five. However, the easiest way is to recall the following result. For a positive integer 𝑛, π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power is equal to π‘₯ minus π‘Ž multiplied by π‘₯ to the power of 𝑛 minus one plus π‘Ž times π‘₯ to the power of 𝑛 minus two. And we add terms of this form all the way up to π‘Ž to the power of 𝑛 minus one. Applying this result, we get the denominator is equal to π‘₯ minus five multiplied by π‘₯ squared plus five π‘₯ plus 25. And this is exactly the same result we would have got by using polynomial division.

And now, we set the limit of these two rational functions to be equal. And now we see we’re taking the limit of this function as π‘₯ approaches five. And remember, this means we want to know what happens to the outputs of our function as our values of π‘₯ get closer and closer to five. So the value of our function when π‘₯ equals five will not change the value of this limit. This means we can cancel the shared factor of π‘₯ minus five in the numerator and denominator of this function. If π‘₯ is not equal to five, π‘₯ minus five divided by π‘₯ minus five is equal to one. This then gives us the limit as π‘₯ approaches five of π‘₯ squared plus 25 multiplied by π‘₯ plus five all divided by π‘₯ squared plus five π‘₯ plus 25.

And now, we can just evaluate this limit by using direct substitution. Substituting π‘₯ is equal to five into our rational function, we get five squared plus 25 multiplied by five plus five all divided by five squared plus five times five plus 25. We can then evaluate this expression. We get 50 times 10 divided by 75, which, if we simplify, is equal to 20 divided by three. Therefore, we were able to show the limit as π‘₯ approaches five of π‘₯ to the fourth power minus 625 all divided by π‘₯ cubed minus 125 is equal to 20 divided by three.

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