Limit of (x^5 – a^5)/(x^2 – a^2)?

This question was submitted by Sal via emailing me at admin@calculustricks.com. Thanks! I encourage everyone to ask me questions if you need help =)

Actual Question:
“Hi, this is the problem:
Find the Limit
Lt          x^5  -  a^5 / x^2  -  a^2
Read as find the limit as x approaches a of x raised to the fifth power minus a raised to the fifth power over x squared minus a squared,
How do I factorise x^5  -  a^5/x^2  -  a^2 ??

To do this problem, you must use synthetic division (or long division would work too!) I’ll show you why, and how I got there.

For any limit, the first thing you ALWAYS want to do, is simply plug in the value that x is approaching. x -> a, so were going to substitute in an “a” where ever there was an “x”. Uh oh! As you probably notice, this gives you 0/0. This is known as an indeterminate form. That means were not done yet. We need to try and change the problem just a little bit, so that we can
get a better solution. Usually, this means factoring, but sometimes thereare other tricks you have to use. Try factoring first though.

If you do not get an indeterminate form (if you get an actual number), that is your answer! You are done. =)

So, where is our trouble? What makes this an indeterminate form? This occurs when the denominator is zero (another type of indeterminate form is when you have infinity over infinity). You cannot divide by zero, so we know something isn’t right. We know that we want to try to factor first, but what are we looking for? We need to factor in order find out which one (or more) of the factors is causing the denominator to equal zero! Then, we need to try to cancel it out with something in the numerator! That is the whole trick to it!

The bottom of this fraction is fairly easy to factor. If you have the “difference of two squares,” there is an easy formula we use to factor it. This is probably the MOST important thing you learn in algebra.

In this problem, it is easy to recognize. In other problems though, it could be harder. The “a” does not have to be just a number. It can also be a variable, multiple variables, have coefficients, or even have a large exponent. If the a is to the fourth power, when you factor it, it becomes a². There are many very cool things you can do with this formula, and it is used all of the time.

So, that is how we will factor the denominator. It looks like this now: (x – a)(x + a). Now, take a closer look at it. What is making the denominator zero? When you have several terms multiplying together, if only one of them is zero, they all become zero! So, when we plug in “a” for “x”, which factor makes the entire thing zero in the denominator? The term (x – a) is what’s doing it. This becomes zero when we substitute “a” in for “x”. We need to factor the numerator now, in order to find something to cancel out the (x – a) that is in the denominator. How do we factor something out of a polynomial with a high degree? You have to use polynomial division. I prefer synthetic division, but long division also works.

If we divide  by (x – a), we will get another expression. This is basically how factoring is done. Once we know that new expression, we can say this:

We can then replace the  with the new factors, and the (x – a) on the top and bottom both cancel! Okay, so I will quickly show you how to divide these.

Using synthetic division, you need to know the root. The zero of the function. We know that we want to divide by (x – a). A root is where the function equals zero. Well, if (x – a) is a factor, we know that if x = a, the entire thing is going to be zero. I hope you follow me. =) So, we are going to use the number a. Be careful. The – is explicit. If you had (x – 3), you use the nubmer 3, not -3.

You put this number, “a”, on the left. Then, you put a line of the coefficients after the a. These are the coefficients of x^5, x^4, x^3,
x^2, x, and then the constant. If those places don’t exist, that means the coefficient is zero. You need to put the zeros in as placeholders for those spots. There was no x^4 in our problem, so that means the coefficient must have been 0, right?

This is what it looks like, when you do synthetic division. You bring the first number straight down. Then, you multiply that number on the bottom by the “a”, and add it to the next number in the top line. Then you do the same thing with that result, and you keep repeating that until you reach the end. It is good when the last number you get after adding is zero! This is always great news. If not, that is considered the remainder, and you can’t cleanly factor using that root. Okay, so what does this mean? Well, the numbers on the bottom are you new coefficients! You subtract 1 from the degree of your polynomial, and put on these coefficients. The degree was 5 before (the degree is the highest power you find in the entire polynomial), so now it becomes 4.

So after factoring out (x – a), we are left with:

This is now what the problem looks like after we have factored completely:

Now, we can cancel out those (x-a) terms! Then, we would be left with this:

We successfully factored something! So, now what? Well, we have to try to substitute again for whatever x is approaching. So, put in “a” whereever there is an x.

That is your final answer! If you get another indeterminate form, repeat this process until you get an actual number for an answer. We are done for this problem, though!

I really hope you could follow me, and that this is helpful.

Good luck!

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  1. Carmelle says:

    This is EPIC! Thank you for sharing this to us. I owe you a lot. You saved me from my project. :)

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