# Lesson Video: Evaluating Limits Using Algebraic Techniques Mathematics • Higher Education

Evaluating Limits Using Algebraic Techniques

17:47

### Video Transcript

Evaluating Limits Using Algebraic Techniques.

In this video, we’ll learn how to use techniques such as factorisation to take the limit of a function. First, let us recall the definition of a limit, which is, for some function 𝑓 of 𝑥, which is defined near the value where 𝑥 is equal to 𝑎, as 𝑥 approaches 𝑎, 𝑓 of 𝑥 approaches 𝐿. We call this value 𝐿 the limit. And here, we have written the standard notation for the description of a limit.

Now, if 𝑓 of 𝑥 is a rational function with 𝑎 in its domain, we can simply say that the limit, as 𝑥 approaches 𝑎 of 𝑓 of 𝑥, is equal to 𝑓 of 𝑎. Since we’re inputting the value of 𝑎 into our function, this method is often described as direct substitution. An important point to note here is that even if 𝑎 is not in the domain of our function 𝑓, we can still sometimes find the limit as 𝑥 approaches 𝑎. This is because the limit concerns values as 𝑥 approaches 𝑎 but is not equal to 𝑎. And we’ll see how this comes into play in a moment.

For this video, we’ll be focusing on functions which take the form 𝑃 of 𝑥 over 𝑄 of 𝑥 where both 𝑃 and 𝑄 are polynomial functions. In the cases we’ll look at, where 𝑥 is equal to 𝑎, 𝑃 of 𝑎 over 𝑄 of 𝑎 will evaluate to the indeterminate form of zero over zero. That is to say, if we try to take the limit as 𝑥 approaches 𝑎 of our function 𝑓 of 𝑥 by directly substituting 𝑥 equals 𝑎 into the function, then our substitution fails, and we’ll need to take another approach.

Looking at our quotient, we can see that both 𝑃 of 𝑎 and 𝑄 of 𝑎 are equal to zero. By recalling the factor theorem, we can then infer that both 𝑃 of 𝑥 and 𝑄 of 𝑥 will have a common factor of 𝑥 minus 𝑎. Given this information, we can rewrite the function uppercase 𝑃 of 𝑥 as a product of 𝑥 minus 𝑎 times some other function, which we’ll call lowercase 𝑝 of 𝑥. And, of course, the same logic can be followed for uppercase 𝑄 of 𝑥.

This allows us to rewrite our original quotient as 𝑥 minus 𝑎 times lowercase 𝑝 of 𝑥 divided by 𝑥 minus 𝑎 times lowercase 𝑞 of 𝑥. In this form, we see that the common factor of 𝑥 minus 𝑎 can be cancelled in the top and bottom halves of our quotient. And we are then left with lowercase 𝑝 of 𝑥 over lowercase 𝑞 of 𝑥. Let’s define this new quotient as 𝑔 of 𝑥.

Now, a reasonable question to ask is, why are we giving this new definition when 𝑓 of 𝑥 is clearly equal to 𝑔 of 𝑥. In fact, the answer is that in cancelling the common factor of 𝑥 minus 𝑎, we have changed the domain of our function 𝑓 of 𝑥. We, therefore, have that 𝑓 of 𝑥 equals 𝑔 of 𝑥, but not at the point where 𝑥 is equal to 𝑎. The subtlety here is that 𝑔 of 𝑥 is defined when 𝑥 is equal to 𝑎, whereas 𝑓 of 𝑥 is not.

This is great, as it allows us to proceed to the following general rule. The limit, as 𝑥 approaches 𝑎 of 𝑓 of 𝑥, is equal to the limit as 𝑥 approaches 𝑎 of 𝑔 of 𝑥 if 𝑓 and 𝑔 are equal at all points over an interval except the point at which 𝑥 is equal to 𝑎. Again, it’s good to understand that this works because the limit concerns values of 𝑥 which are close to 𝑎 but not equal to 𝑎. Finally, since 𝑔 is a rational function with 𝑎 in its domain, we can simply find the limit by direct substitution, which is 𝑔 evaluated where 𝑥 is equal to 𝑎. Okay, that’s a lot of information, so let’s look at an example to illustrate this concept.

Find the limit as 𝑥 approaches negative eight of 𝑥 squared plus 13𝑥 plus 40 divided by 𝑥 cubed plus nine 𝑥 squared minus 12𝑥 minus 160.

Here, we see that we are trying to find the limit of a rational function, which we’ll call 𝑓 of 𝑥. For this type of function, if negative eight is in the domain of 𝑓, then the limit, as 𝑥 approaches negative eight, is simply 𝑓 evaluated at negative eight. The first thing we can do then is to try direct substitution of negative eight into our function.

Here, we have performed the substitution. And following through with our working, we find that our answer evaluates to zero over zero. And this is an indeterminate form, which means we cannot evaluate the limit using direct substitution. We can also conclude that negative eight is not in the domain of our function 𝑓. Here, we’re gonna need to use a different approach. And we can do so first by recognising that our function 𝑓 of 𝑥 is in the form 𝑃 of 𝑥 over 𝑄 of 𝑥, where both 𝑃 and 𝑄 are polynomial functions.

Looking at the failed direct substitution that we just tried, we can see the both 𝑃 of negative eight and 𝑄 of negative eight are equal to zero. From this information, we can use the factor theorem to conclude that 𝑥 plus eight is a factor of both 𝑃 of 𝑥 and 𝑄 of 𝑥. Since direct substitution failed, let’s instead try to factorise both the top and bottom halves of our quotient, given the fact that both of them have a common factor of 𝑥 plus eight.

For the numerator, factoring a quadratic equation should be a familiar skill to us. And with some inspection, we see that 𝑥 squared plus 13𝑥 plus 40 factorises to 𝑥 plus eight and 𝑥 plus five. Now, for the denominator, ordinarily, factoring a cubic is a much more difficult task. However, given the fact that we already know one of the factors is 𝑥 plus eight, we can use techniques such as polynomial division or comparing coefficients to find the other factor.

For this video, we’re going to compare coefficients. And we begin by recognising that our other factor will take the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. To find 𝑎, we note that we have an 𝑥 multiplied by an 𝑎𝑥 squared. And this is equal to 𝑥 cubed. It, therefore, follows that 𝑎 is equal to one. And we then see that our second factor begins with the term 𝑥 squared.

Next, to find 𝑐, we note that we have eight multiplied by 𝑐 is equal to negative 160. And we can calculate that 𝑐 is equal to negative 20. Finally, to find 𝑏, we’ll choose to look at the nine 𝑥 squared term. Going by the previous coefficients we have, we note that we first have an eight multiplied by an 𝑥 squared. This gives us eight 𝑥 squared. We also have an 𝑥 multiplied by 𝑏𝑥 and this gives us 𝑏𝑥 squared. The sum of these two terms is nine 𝑥 squared. And it then follows that 𝑏 is equal to one.

The denominator of our quotient is now fully factorised. And our missing factor was 𝑥 squared plus 𝑥 minus 20. Following this factorisation, we’re able to cancel the common factor of 𝑥 plus eight from the top and bottom half of our quotient. And we are then left with 𝑥 plus five over 𝑥 squared plus 𝑥 minus 20.

Now, here, we should remember that in cancelling the common factor of 𝑥 plus eight, we have changed the domain of our original function 𝑓 of 𝑥. We’re able to say that 𝑓 of 𝑥 is equal to the right-hand side of our equation, which we’ll call 𝑔 of 𝑥, at all values where 𝑥 is not equal to negative eight. It now follows that the limit, as 𝑥 approaches negative eight of 𝑓 of 𝑥, is equal to the limit as 𝑥 approaches negative 𝑥 of 𝑔 of 𝑥. Crucially, this is because the limit concerns values where 𝑥 is close to negative eight but not equal to negative eight.

An important point is that 𝑔 of 𝑥 is defined when 𝑥 is equal to negative eight. And so, we can, therefore, find the limit by direct substitution. Here, we have performed the substitution of 𝑔 of negative eight. And if we follow through with our calculations, we see that this evaluates to negative three over 36. This fraction can be simplified to negative one over 12. We have now answered the question. And we have found our limit.

This example illustrates that when our functions 𝑃 of 𝑥 and 𝑄 of 𝑥 evaluate to zero, we can use the factor theorem to help us with potentially tricky factorisations such as cubics. Let’s now look at another technique which can help us avoid potentially tricky factorisations first by focusing on expressions of the form 𝑥 to the 𝑛 minus 𝑎 to the 𝑛, or the difference of two 𝑛th powers.

Since inputting 𝑥 equals 𝑎 into this equation will always give zero, our friend the factor theorem, again, tells us that all expressions of this form will evaluate to zero. And therefore, all of them will have a factor of 𝑥 minus 𝑎. Using this general rule, we see that our second factor will take the form of a polynomial with 𝑛 terms having decreasing powers of 𝑥 and increasing powers of 𝑎 up to the powers of 𝑛 minus one. Some examples of this have been shown below.

This general rule can be used to derive a very useful formula in the following way. Consider the case of a difference of two 𝑛th powers, 𝑥 to the 𝑛 minus 𝑎 to the 𝑛, divided by the difference of two 𝑚th powers, 𝑥 to the 𝑚 minus 𝑎 to the 𝑚. We can use our general rule to express the top and bottom halves of our quotient as a product of two factors. In both cases, one of these factors is 𝑥 minus 𝑎.

We can cancel this common factor of 𝑥 minus 𝑎 in the top and bottom halves of the quotient. In cancelling our common factor, we must remember that the left- and right-hand sides of our equations are equal as long as the value of 𝑥 is not equal to 𝑎. Since the limit, as 𝑥 approaches 𝑎, concerns values of 𝑥 close to 𝑎 but not equal to 𝑎, we’re able to say that the limit of the left-hand side is equal to the limit of the right-hand side.

Considering the numerator of our quotient for a moment, we now use the following trick of direct substitution, where 𝑥 takes the value of 𝑎. With this substitution, we’re left with the sequence 𝑛 terms long of which all of the terms are equal to 𝑎 to the power of 𝑛 minus one. And this is simply equal to 𝑛 times 𝑎 to the power of 𝑛 minus one. Although, we were using the numerator for an example, the same logic follows for the denominator of our quotient and gives us 𝑚 times 𝑎 to the power of 𝑚 minus one.

We then proceed with a few simplifications of the powers of 𝑎. We then find that our limit, as 𝑥 approaches 𝑎, is equal to 𝑛 over 𝑚 multiplied by 𝑎 to the power of 𝑛 minus 𝑚. This is a really useful result for equations of this form since even when the powers 𝑛 or 𝑚 are large, we can avoid lengthy factorisations when taking the limit. Let’s now see how this technique can be used in an example.

Find the limit as 𝑥 approaches two of eight 𝑥 cubed minus 64 divided by 𝑥 squared minus four.

Here, we have a function, which we’ll call 𝑓 of 𝑥. Given that this is a rational function, the first thing we may try is a direct substitution of 𝑥 equals two into our function. Here, we have performed the substitution. And when we evaluate our answer, we find that we’re left with the indeterminate form of zero over zero. Instead, we’re gonna need to move on to a different method based on factorisation.

For this method, we first note that our function 𝑓 of 𝑥 is in the form 𝑃 of 𝑥 over 𝑄 of 𝑥, where both 𝑃 and 𝑄 are polynomial functions. By inspecting the numerator of our quotient, we noticed that both the terms have a factor of eight. We can, therefore, factorise our numerator as eight times 𝑥 cubed minus eight. And given that this eight is a constant, we can take it outside of our limit as follows. If instead we find the limit as 𝑥 approaches two of 𝑥 cubed minus eight divided by 𝑥 squared minus four and multiply this entire thing by our constant eight, this will give us the same answer.

To proceed, we can then notice that the eight in our numerator and the four in our denominator can both be expressed as powers of two, which are two cubed and two squared, respectively. After doing this, we see that our limit now takes the following form. Here, we’ll say that the top half of our quotient is equal to the difference of two 𝑛th powers, with 𝑛 being three, and the bottom half of our quotient is equal to the difference of two 𝑚th powers, with 𝑚 being two.

Given this form, we’re able to use the following general rule, which tells us that the limit will be equal to 𝑛 over 𝑚 times 𝑎 to the power of 𝑛 minus 𝑚. At this point, you may notice that both the top and bottom half of our quotient would have a common factor of 𝑥 minus two. We could, instead, cancel this common factor and proceeded by refactorising. However, this general rule allows us to move directly to our limit.

In cases, where 𝑛 or 𝑚 are large, this helps us potentially avoid a lengthy or time-consuming refactorisation. Inputting our values into our general rule, where 𝑎 is two, 𝑛 is three, and 𝑚 is also two, we find that our limit is three over two multiplied by two to the power of three minus two. We also mustn’t forget to multiply this entire thing by the eight which we took out of our limit.

Three minus two is, of course, just one. And so, we can simplify this by cancelling the two and the one over two. We then find that our answer is equal to eight times three, which 24. We have now found that the limit, as 𝑥 approaches two of our function 𝑓 of 𝑥, is equal to 24. And we have answered our question.

The general rule that we used in this video provides us with a useful shortcut that can be used when our function 𝑓 of 𝑥 can be expressed in this form. Again, although for our question the powers involved are relatively low, when 𝑛 and 𝑚 are sufficiently large, you may be able to save yourself much more time. An interesting point worth noting is that some of the techniques that we have shown in this video can also be used to take the limits of functions which include radical. Specifically, where 𝑓 of 𝑥 is equal to 𝑃 of 𝑥 over 𝑄 of 𝑥, where either 𝑃 or 𝑄 include a radical term, such as the square root of 𝑥.

Although these would not be classed as rational functions, some of our tools will still work. An example would be if we were to take the limit as 𝑥 approaches eight of the cube root of 𝑥 minus two divided by 𝑥 minus eight. If we were to try a direct substitution approach, we would again be left with the indeterminate form of zero over zero.

However, if instead we were to factorise, specifically taking out a factor of the cube root of 𝑥 minus two from our denominator, we would then be able to proceed in a familiar fashion. By cancelling out the common factor in the top and bottom halves of the quotient and then directly substituting 𝑥 equals eight into the remaining expression to finally find an answer of one over 12. Although, here, we’ve shown an example where our current techniques work, in some cases, we’ll need a different method based on multiplication by a conjugate. This technique can be illustrated using an example.

Determine the limit as 𝑥 approaches eight of 𝑥 squared minus eight 𝑥 divided by the square root of 𝑥 plus one minus three.

Here, we see that we’re taking the limit of a function 𝑓 of 𝑥, which takes the form 𝑃 of 𝑥 over 𝑄 of 𝑥. When we try a direct substitution approach to solve this, our expression evaluates to the indeterminate form of zero over zero. And instead, we’re gonna need to use a different approach. The first thing we can notice is that our numerator can be factorised as 𝑥 times 𝑥 minus eight. Given our knowledge of the factor theorem, we may have even expected this factor of 𝑥 minus eight given that the function 𝑃 evaluated where 𝑥 equals eight gave us a zero.

Unfortunately, the denominator of our quotient is less straightforward to evaluate. But we can understand how to move forward by viewing it as a binomial in the form 𝑎 minus 𝑏, where 𝑎 is equal to the square root of 𝑥 minus one and 𝑏 is equal to three. The conjugate of any binomial is found by flipping the symbol in between the two terms. The conjugate of our binomial would then be 𝑎 plus 𝑏. And in this case, that is the square root of 𝑥 plus one plus three.

Now, multiplying a binomial by its conjugate is a useful tool because it leaves us with the difference of two squares. Given that our denominator contains a square root, we can utilise this relationship to evaluate our function. Now, multiplying by the conjugate over itself is the same as multiplying by one. Nonetheless, the bottom half of our quotient then simplifies to a difference of two squares.

And with some working, we see that our denominator then becomes 𝑥 minus eight. Following this, we’re able to cancel the common factor of 𝑥 minus eight in the top and bottom half of the quotient. And on the right-hand side of our equation, we are left with 𝑥 multiplied by the square root of 𝑥 plus one plus three.

Now, since our original function 𝑓 of 𝑥 is equal to the right-hand side of our equation at all points where 𝑥 is not equal to eight, we’re able to say the limit, as 𝑥 approaches eight of our original function 𝑓 of 𝑥, is equal to the limit as 𝑥 approaches eight of our new function, which we’ll call 𝑔 of 𝑥. This is because the limit concerns values of 𝑥 which are arbitrarily close to eight but not equal to eight.

In this form, we can perform a direct substitution of 𝑥 equals eight into our function 𝑔 to find the limit. Performing the substitution and then evaluating our answer leaves us with a value of 48. And this is, in fact, the limit that we were looking to find.

Now, it’s worth recapping here that, originally, we were not able to find our limit using direct substitution because we were left with the indeterminate form of zero over zero. Instead, after factorising and using our conjugate method to cancel a common factor, direct substitution was possible. We should, therefore, be on the lookout to use this method for questions of this form when we observe a radical in our quotient.

Let’s now recap a few key points to summarise. When taking the limit of a rational function 𝑓 of 𝑥, which is expressed as 𝑃 of 𝑥 over 𝑄 of 𝑥, and evaluating by direct substitution, we may be left with the indeterminate form of zero over zero. In these cases, if the function 𝑓 of 𝑥 is almost equal to 𝑔 of 𝑥. That is to say 𝑓 of 𝑥 is equal to 𝑔 of 𝑥 at all values of 𝑥 aside from where 𝑥 is equal to 𝑎. Then it follows that the limit, as 𝑥 approaches 𝑎 of 𝑓 of 𝑥, is equal to the limit as 𝑥 approaches 𝑎 of 𝑔 of 𝑥.

Assuming that 𝑎 is in the domain of our new function 𝑔, that is to say 𝑔 of 𝑥 is defined where 𝑥 is equal to 𝑎, we can then find our limit by direct substitution, which is to say taking the value of 𝑔 of 𝑎. We can find such a function 𝑔 of 𝑥 via a number of methods such as factorisation or multiplication by a conjugate of the numerator or denominator.

The methods that we have shown in this video involve cancelling a common factor of 𝑥 minus 𝑎 from the top and bottom half of our quotient. Since taking the limit involves values of 𝑎 which are close to but not equal to 𝑎, it does not matter that in cancelling the common factor we have changed the domain of our original function 𝑓 of 𝑥. Finally, when our function takes certain forms, there are some useful shortcuts that we can take to find the limit directly, potentially avoiding some lengthy factorisations.