Lesson Video: Limits of a Difference of Powers Mathematics

In this video, we will learn how to evaluate limits of a difference of powers.

17:59

Video Transcript

Limits of a Difference of Powers

In this video, we will discuss and prove several different results to help us evaluate the limits of a difference of powers. We will also see several different examples and applications of these results. Before we start with the general case of the limit of a difference of powers, let’s start with a case we’ve seen before, the limit of a rational function. We recall if 𝑃 of π‘₯ divided by 𝑄 of π‘₯ is a rational function β€” that means that both 𝑃 and 𝑄 are polynomials β€” then we can evaluate the limit of our rational function by using direct substitution. The limit as π‘₯ approaches π‘Ž of 𝑃 of π‘₯ over 𝑄 of π‘₯ is equal to 𝑃 of π‘Ž divided by 𝑄 of π‘Ž. And this is, of course, provided the denominator 𝑄 of π‘Ž is not equal to zero.

We can prove this directly from the properties of limits. We just use the quotient rule for limits and the fact that we can evaluate polynomials by direct substitution. And the rational function is an example of a difference of powers. For example, π‘₯ to the 𝑛th power divided by π‘₯ to the π‘šth power is equal to π‘₯ to the power of 𝑛 minus π‘š. We’ll want to generalize this even further. But for now let’s focus on the case we have with rational functions. And we’ll focus on the condition that 𝑄 of π‘Ž is not allowed to be equal to zero.

To see how we might get around this condition, let’s start by recalling the definition of a limit. We recall that we say the limit of some function 𝑓 of π‘₯ as π‘₯ approaches π‘Ž is equal to some finite value of 𝐿 if the values of 𝑓 of π‘₯ approach 𝐿 as the values of π‘₯ approach π‘Ž from both sides. In particular, we’re only interested in the output values of our function 𝑓 of π‘₯ as our values of π‘₯ approach π‘Ž. In other words, we want to know what happens as our values of π‘₯ get closer and closer to π‘Ž. We don’t actually mind what happens when π‘₯ is equal to π‘Ž.

We can use this to create a very useful theorem. What if we had a function 𝑔 of π‘₯ which was exactly equal to our function 𝑓 of π‘₯ everywhere except when π‘₯ was equal to π‘Ž? So we’ll start with a function 𝑔 of π‘₯, which is exactly equal to our function 𝑓 of π‘₯ everywhere except when π‘₯ is equal to π‘Ž. We want to use this to determine the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯. And in fact we can do this. We know that 𝑔 of π‘₯ is exactly equal to 𝑓 of π‘₯ everywhere except when π‘₯ is equal to π‘Ž. And 𝑔 of π‘₯ not being equal to 𝑓 of π‘₯ when π‘₯ is equal to π‘Ž won’t affect its limit because we don’t actually mind what happens when π‘₯ is equal to π‘Ž.

So in actual fact, their limits as π‘₯ approach π‘Ž must be equal. The limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ will be equal to the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯. And this gives us a really useful result. If we have two functions 𝑓 of π‘₯ and 𝑔 of π‘₯ which are equal everywhere except where π‘₯ is equal to π‘Ž, and we know the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ is equal to some finite value of 𝐿, then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ must also be equal to 𝐿.

We’re now ready to apply this result directly to our example involving rational functions. To do this, let’s start with an example. Suppose we want to evaluate the limit as π‘₯ approaches negative one of π‘₯ plus one multiplied by π‘₯ minus one all divided by π‘₯ plus one. Since this is a rational function, we can attempt to evaluate this limit by direct substitution. However, if we did this, in our numerator we see π‘₯ plus one would become a factor of zero. And in our denominator, π‘₯ plus one would also be equal to zero. So direct substitution gives us zero divided by zero, which is an indeterminate form, which means we can’t evaluate this limit by using direct substitution.

But it’s important to remember we can’t just conclude the limit does not exist. All this tells us is that we can’t evaluate this limit by using this method. We need to try a different method. Instead, what we might try doing is canceling the shared factor of π‘₯ plus one in the numerator and the denominator. This would then give us the limit as π‘₯ approaches negative one of π‘₯ minus one, which we can evaluate by using direct substitution. But we do need to be careful. Canceling the shared factor of π‘₯ plus one has changed the function we’re taking the limit of. Previously, negative one was not in the domain of our function. However, negative one is in the domain of π‘₯ minus one.

But we can justify that we’re allowed to do this by using the property we just proved. When π‘₯ is not equal to negative one, π‘₯ plus one is a nonzero number. And a nonzero number divided by itself is always equal to one. What this really means is, our original rational function and the polynomial π‘₯ minus one are equal everywhere except when π‘₯ is equal to negative one. Therefore, by using our property, their limits must be equal. And we can evaluate the limit of π‘₯ minus one as π‘₯ approaches negative one by using direct substitution. It’s equal to negative one minus one, which is of course just equal to negative two.

We can use this exact same reasoning to evaluate the limit of other rational functions. So let’s clear some space and go through one of these examples. We want to determine the limit as π‘₯ approaches π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ minus π‘Ž. And for now, we’ll assume that our value of 𝑛 is a positive integer. And this is a rational function, so we could try evaluating this limit by direct substitution. If we did this, we would get π‘Ž to the 𝑛th power minus π‘Ž to the 𝑛th power divided by π‘Ž minus π‘Ž, which simplifies to give us zero divided by zero, which is of course an indeterminate form.

So instead, to evaluate this limit, we want to use the same trick we did before. We want to cancel a shared factor of π‘₯ minus π‘Ž in the numerator and denominator. To do this, we’ll start by calling the polynomial in the numerator 𝑃 of π‘₯. That’s π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power. In particular, since 𝑃 evaluated at π‘Ž is equal to zero, the remainder theorem tells us that π‘₯ minus π‘Ž must be a factor of 𝑃 of π‘₯. In fact, by using polynomial division or otherwise, we can show π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power is equal to π‘₯ minus π‘Ž all multiplied by π‘₯ to the power of 𝑛 minus one plus π‘Ž times π‘₯ to the power of 𝑛 minus two plus π‘Ž squared multiplied by π‘₯ to the power of 𝑛 minus three. And we add terms of this form all the way up to π‘Ž to the power of 𝑛 minus one.

We can then substitute this expression directly into our limit. This then gives us the following expression for our limit. And now we can cancel the shared factor of π‘₯ minus π‘Ž in both the numerator and denominator. And it’s worth reiterating we’re allowed to do this because we’re taking the limit as π‘₯ approaches π‘Ž. Canceling the shared factor of π‘₯ minus π‘Ž will not change the value of our function anywhere except when π‘₯ is equal to π‘Ž. So we now have the limit as π‘₯ approaches π‘Ž of 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. And this is now the limit of a polynomial, so we can evaluate this limit by using direct substitution.

This gives us π‘Ž to the power of 𝑛 minus one plus π‘Ž times π‘Ž 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. And if we were to simplify each term, we would notice something interesting. Every single term in this expression is equal to π‘Ž to the power of 𝑛 minus one. And there are 𝑛 of these terms, one for each exponent of π‘₯, all the way from 𝑛 minus one to zero. So, in fact, this is just equal to 𝑛 multiplied by π‘Ž to the power of 𝑛 minus one.

And although we assumed our value of 𝑛 was a positive integer, this result is true for any value of 𝑛. We have for any real constants π‘Ž and 𝑛 their limit as π‘₯ approaches π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ minus π‘Ž is equal to 𝑛 times π‘Ž to the power of 𝑛 minus one. And that’s provided π‘Ž to the 𝑛th power and π‘Ž to the power of 𝑛 minus one both exist. Let’s now see an example of applying this formula to evaluate a limit.

Find the limit as π‘₯ approaches one of the fourth root of π‘₯ minus one multiplied by the sixth root of π‘₯ to the seventh power minus one all divided by π‘₯ minus one all squared.

In this question, we’re asked to evaluate a limit. And we can see that this is the limit of a very complicated function. However, we can see that this function is the sum, difference, quotient, product, and composition of power functions and polynomials. So we can try evaluating this limit by using direct substitution. If we substitute π‘₯ is equal to one into this function and then simplify, we see that it’s equal to zero divided by zero, which is an indeterminate form, which means we can’t evaluate the limit by using this method. We need to try a different method to evaluate this limit. Instead, we need to notice that the limit we’re asked to evaluate is very similar to one of our limit results.

We know, for any real constants π‘Ž and 𝑛, the limit as π‘₯ approaches π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power divided by π‘₯ minus π‘Ž is equal to 𝑛 times π‘Ž to the power of 𝑛 minus one. And that’s provided that both π‘Ž to the 𝑛th power and π‘Ž to the power of 𝑛 minus one exist. So we need to rewrite the given limit in this form. To do this, we’ll start by using the product rule for limits to distribute the denominator over each of the factors in our numerator. First, we’ll rewrite our limit as the limit as π‘₯ approaches one of the fourth root of π‘₯ minus one all divided by π‘₯ minus one multiplied by the sixth root of π‘₯ to the seventh power minus one all divided by π‘₯ minus one.

Each of the two factors of our function are now in the form of our limit rule. And by using the product rule for limits, we can split the limit of a product of two functions into the product of the limit of those two functions. And it’s worth reiterating this will only be true provided the limit of both of our two functions exist. In fact, we’ll be able to show this by using our limit result. Before we apply this result, we need to rewrite our numerator. We’ll rewrite the fourth root of π‘₯ by using our laws of exponents as π‘₯ to the power of one-quarter and the sixth root of π‘₯ to the seventh power as π‘₯ to the power of seven over six.

We’re now ready to use our limit result to evaluate our limit. Let’s start with the first limit. We have the value of 𝑛 equal to one-quarter and the value of π‘Ž equal to one. It’s worth noting that one raised to the power of one-quarter is just equal to one. So this is in fact in the form of our limit result. Therefore, by our limit result, we can evaluate this limit to be equal to 𝑛 multiplied by π‘Ž to the power of 𝑛 minus one, which in this case is one-quarter multiplied by one raised to the power of one-quarter minus one.

We can do exactly the same for our second limit. The value of 𝑛 is seven-sixths, and the value of π‘Ž is also equal to one. And once again, by our limit result, we can evaluate this limit. It’s 𝑛 times π‘Ž to the power of 𝑛 minus one, which in this case is seven-sixths multiplied by one raised to the power of seven-sixths minus one. And of course we need to multiply these two values together. And now we can evaluate this expression directly. First, one raised to the power of any number is just equal to one. So this simplifies to give us one-quarter multiplied by seven-sixths, which is just equal to seven over 24.

Therefore, we were able to show the limit as π‘₯ approaches one of the fourth root of π‘₯ minus one multiplied by the sixth root of π‘₯ to the seventh power minus one all divided by π‘₯ minus one all squared is equal to seven divided by 24.

Using our limit result for a difference of two powers, we can actually show two other really useful limit results. First, imagine we were asked to evaluate the limit as π‘₯ approaches π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power divided by π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power. We can actually write this entirely in terms of our limit result. To do this, we’ll start by introducing a factor of π‘₯ minus π‘Ž into both the numerator and denominator. Next, instead of multiplying, we’re going to divide by the reciprocal. This gives us the following expression. And we can see that both of these two functions are in the form of our limit result. So we’re going to evaluate this limit by using the quotient rule for limits.

The quotient rule for limits tells us the limit of the quotient of two functions is equal to the quotient of the limits of those two functions. That’s provided both of the two limits exist and the limit in the denominator is not equal to zero. We can now evaluate both of these limits by using our limit result. The first limit is equal to 𝑛 multiplied by π‘Ž to the power of 𝑛 minus one. And the second limit is equal to π‘š multiplied by π‘Ž to the power of π‘š minus one. We just need to divide both of these two expressions. And when we divide these two expressions and simplify, we get 𝑛 over π‘š multiplied by π‘Ž to the power of 𝑛 minus π‘š.

And this gives us a really useful result. For any real constants π‘Ž, 𝑛, and π‘š, the limit as π‘₯ approaches π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ to the π‘šth power minus π‘Ž to the π‘šth power is equal to 𝑛 divided by π‘š multiplied by π‘Ž to the power of 𝑛 minus π‘š. And that’s 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.

There’s one more useful limit result we can show from this. We’re going to substitute 𝑦 is equal to π‘₯ minus π‘Ž into this limit result. To do this, let’s clear some space and start with our limit result. We can find an expression for π‘₯ by adding π‘Ž to both sides. We see that π‘₯ is equal to 𝑦 plus π‘Ž. We can also see as the values of π‘₯ approach π‘Ž, π‘₯ minus π‘Ž is going to approach zero. So the values of 𝑦 will approach zero. So by substituting 𝑦 is equal to π‘₯ minus π‘Ž into our limit, we get the limit as 𝑦 approaches zero of 𝑦 plus π‘Ž all raised to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by 𝑦. And this is equal to 𝑛 multiplied by π‘Ž to the power of 𝑛 minus one. We can then rewrite this limit result in terms of the variable π‘₯.

This gives us the following result. For any real constants π‘Ž and 𝑛, the limit as π‘₯ approaches zero of π‘₯ plus π‘Ž all raised to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ is equal to 𝑛 times π‘Ž to the power of 𝑛 minus one. And that’s provided both π‘Ž to the 𝑛th power and π‘Ž to the power of 𝑛 minus one exist.

Let’s now see an example of applying one of these limit results.

Find the limit as π‘₯ approaches two of π‘₯ minus four all cubed plus eight all divided by π‘₯ minus two.

In this question, we’re asked to evaluate the limit of a function. We can see in our numerator we have a polynomial and in our denominator we have a polynomial. So this is a rational function. And we could always try to evaluate the limit of a rational function by direct substitution. Substituting π‘₯ is equal to two into the function, we get two minus four all cubed plus eight all divided by two minus two, which, if we evaluate, we see is zero divided by zero, which is an indeterminate form. Since this gives an indeterminate form, we can’t evaluate this limit by using direct substitution; we’re going to need to use a different method.

We need to notice the limit given to us in the question is very similar to one of our limit results. That is, the limit as π‘₯ approaches zero of π‘₯ plus π‘Ž all raised to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ is equal to 𝑛 times π‘Ž to the power of 𝑛 minus one. And that’s provided π‘Ž to the 𝑛th power and π‘Ž to the power of 𝑛 minus one both exist. But this limit result has π‘₯ approaching zero, and the limit we’re asked to evaluate has π‘₯ approaching two. So we’re going to use the substitution 𝑦 is equal to π‘₯ minus two. Then, as our values of π‘₯ approach two, π‘₯ minus two is going to be approaching zero. So our values of 𝑦 approach zero.

In our denominator, we have π‘₯ minus two, which is just going to be equal to 𝑦. However, in our numerator, we have π‘₯ minus four. So we need to find an expression for π‘₯ minus four. And we can find this by subtracting two from both sides of our equation for 𝑦. We get 𝑦 minus two is equal to π‘₯ minus four. Therefore, by using the substitution 𝑦 is equal to π‘₯ minus two, we were able to rewrite our limit as the limit as 𝑦 approaches zero of 𝑦 minus two all cubed plus eight all divided by 𝑦.

And this is now almost exactly in the form of our limit result. We can write it in the exact form of our limit result by noting that 𝑦 plus negative two is the same as 𝑦 minus two and eight is the same as negative one times negative two all cubed. So our value of π‘Ž is negative two and our value of 𝑛 is three. Therefore, our limit result tells us that this limit is equal to 𝑛 multiplied by π‘Ž to the power of 𝑛 minus one. Substituting π‘Ž is equal to negative two and 𝑛 is equal to three, we get three multiplied by negative two to the power of three minus one, which we can evaluate is equal to 12. Therefore, we were able to show the limit as π‘₯ approaches two of π‘₯ minus four all cubed plus eight all divided by π‘₯ minus two is equal to 12.

Let’s now go over some of the key points we found in this lesson. First, if we have two functions 𝑓 of π‘₯ and 𝑔 of π‘₯ which are equal everywhere except when π‘₯ is equal to π‘Ž, and we know the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ is equal to 𝐿, then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ must also be equal to 𝐿. This is a really useful result. One thing it allows us to do is cancel shared factors of π‘₯ minus π‘Ž when we’re evaluating the limit of rational functions.

We also showed three useful limit results which hold for any real constants π‘Ž, 𝑛, and π‘š. First, the limit as π‘₯ approaches π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ minus π‘Ž is equal to 𝑛 times π‘Ž to the power of 𝑛 minus one. And that’s provided π‘Ž to the 𝑛th power and π‘Ž to the power of 𝑛 minus one both exist. Second, we showed the limit as π‘₯ approaches π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ to the π‘šth power minus π‘Ž to the π‘šth power is equal to 𝑛 divided by π‘š multiplied by π‘Ž to the power of 𝑛 minus π‘š. And that’s provided π‘š is nonzero 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. And our final limit result told us the limit as π‘₯ approaches zero of π‘₯ plus π‘Ž all raised to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ is equal to 𝑛 times π‘Ž to the power of 𝑛 minus one. And that’s provided π‘Ž to the 𝑛th power and π‘Ž to the power of 𝑛 minus one both exist.

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