Module 11 Lesson 2 Limits of Rational Functions Lesson 2 Notes Part 1 Video 2 Transcript Hey everyone this is a video for module 11 lesson 2, limits of rational functions. The first example says find the limit of x cubed minus 5 x squared over x squared minus 25 as x approaches 5. So we are going to try to find the limit all 3 ways, just to show you a variety of ways to find the limit. So with the graph you notice I have 2 functions in here, this is actually going to be the second example but it is not going to show up on our graph this time, so here’s the first one that we are going to do. So we are going to hit graph there. So as you can see there is a vertical line here, and as we said before this is a vertical asymptote, this means this x-­‐value would not be in the domain. That looks like it is at -­‐5 if I hit trace -­‐5 ENTER, you are not going to get a y-­‐value there, so that’s not going to be in the domain, I will go into detail a little bit about that in just a little while about that. What I am looking at now is, as I go to x equals 5 what’s going to happen to my y-­‐value, what is it approaching? So you can see my cursor is going toward the value of x equals 5 you can see here that its climbing. So as I get closer to 5 I am going to slow down. And you can see at 4.46, 4.68, 4.89, and then I actually cross over to the other side of 5. So I am going to write these values down. 4.89 is approximately 2.42. and then at 5.1, my y value is approximately 2.58. So what we know from that is the limit is going to be between these 2 y-­‐values, because 5 is in between these y values (I should have said x values). So to verify, that looks look at the table. I am going to go to TBLSET, and TbleStart I am going to start at 5 and the delta table I am going to .01 and then I am going to look at the table. So as you notice at 5, I have an error message for the y-­‐value, I will speak to that in just a moment. But as you can see at 5.06 down to 5 what my values are doing are decreasing. On the other side you can see that they are increasing, and both of these look like they are headed towards 2.5. So, I am going to verify just a little bit more and go back to here and change this to .001 so I will have a little bit more decimals. Start it at 5 and look at the graph again. And you can see that verifies it even more that its approaching 2.5.
So based on the graph and the table, we are going to say its approximately 2.5 for now, and we will see of that is the solution indeed. So, now we are going to use substitution. So in the last video and in your notes, with substitution, we could just substitute 5 in for x. So that would become 5 cubed minus 5 times 5 squared, I put 5 in for all these x’s. And then 5 squared minus 25. Cleaning that up you would have 125 minus 5 times 25 over 25 minus 25. Now immediately I see I am going to have 25 minus 25 on the bottom which is 0. We know that we can’t have 0 on the bottom. It doesn’t matter what this number is (pointing to the numerator), if I have 0 on the bottom it just cannot happen. So we will discuss that in just a minute what that means. But, looking at this problem, you should notice, that this and this piece, the numerator and the denominator can factor. So the top can factor by taking out a GCF of x squared. So that would leave x minus 5, all of that is going to be over, and x squared minus 25 is a difference of squares, so x minus 5 times x plus 5. And what happens the x minus 5’s go away. So I reduce that fraction by a factor of x-­‐5. So now can I use direct substitution? Well lets try. So let try to put 5 into x, so we will have 5 squared on the top, and then 5 plus 5 on the bottom. This will go to 25 over 10, which reduces to 5 over 2, which is 2.5. So because I did the direct substitution I actually got a number here, these are not approximate, those are the answers. So the limit of that function as x approaches 5 is 5 over 2 or 2.5. Now, lets discuss, why we have an error message on the 5. Well when I substituted 5 directly in for x, we received 0 on the bottom, I erased it, but we had a 0 on the bottom. That’s why we received an error message here. Because that made the bottom undefined which we know we can’t have. When we looked at the graph at the very beginning we noticed we had a vertical asymptote here at -­‐5. If you actually go to the table, and you put -­‐5 in there to start with (in TBLSET), and look at the table there, you get an error message there as well. So those are the only 2 x-­‐values that you are going to get error messages at. One of these was an asymptote at x = -­‐5. And one was a hole. The vertical asymptote comes from x equals 5 because it was actually left on the bottom here (it didn’t cancel out). X = 5 a hole occurred. So what both of those mean is that those numbers are not part of the domain, so x values can be any values except x equals -­‐5 and x equals 5.