 In this video, we present the solution to question number six from practice exam 2 from math 1210, in which case we're asked to compute the limit as x approaches 3 from the left of the absolute value of x minus 3 over x squared minus 9. Well, given that the function is continuous on its domain, we just combine together polynomials, absolute value, quotients. That's going to be continuous on its domain. So my first inclination is just to plug in x equals 3, which case we get the absolute value of 3 minus 3 over 3 squared minus 9. In the top, you're going to get the absolute value of 0. In the bottom, you're going to get 9 minus 9. This becomes 0 over 0. In which case we have this indeterminate form. It means the limit could be anything. We actually don't know. We're going to have to try something else to figure out the limit here. Now it might be tempting to say that the limit doesn't exist on this one, which that can happen if you do get 0 over 0. Sometimes limit doesn't exist. But the limit could also exist. We don't have enough information yet. So we need to pursue this in a different way. Now the reason that we're getting 0 over 0 is because there's a common factor in the numerator and denominator. In fact, that factor is going to be x minus 3. The x minus 3 is obvious in the numerator. It's a little bit concealed in the denominator. It becomes much more apparent when we factor this thing. If we factor the denominator, you're going to end up with an x minus 3 and you get an x plus 3. I did that recognizing x squared minus 9 is a difference of squares. What happens if we pursue the factorization here? This time we're not going to plug in 3 because we'll get the exact same problem we had a moment ago. Instead I'm going to plug in 3 minus. If I take a number that's just a teeny bit less than 3, what does that do to things? In which case in the numerator we get the absolute value of 3 minus minus 3. If you take a number that's a little bit less than 3 and subtract from it 3, you're going to get something that's a little bit less than 0. The numerator will look like 0 minus in the absolute value of 0 minus. We'll come back to that in a second. 3 from the left minus 3, that's going to give you another 0 minus. Then you're going to get a number a little bit less than 3 plus 3. That's going to give you a 6 minus here. Let's pay attention to signs of these things. If I take the absolute value of a number a little bit less than 0, which means it's negative. If you're a little bit less than 0, you're negative. The absolute value always makes something positive. Therefore the numerator when you take the absolute value is going to be something positive, a little bit bigger than 0. In the denominator, you're going to get something a little bit less than 0, which is negative. If you take a number that's a little bit less than 6, that's still positive because 6 is positive. A little bit less than positive 6 is still positive 6. Whether you had a little bit less than 6, exactly 6, or a little bit greater than 6, it's always going to be positive. You want us to say just a positive 6 right here. If we just focus on the signs, you're going to end up with a plus over a minus and a plus right here. What this already tells me right here is the final answer is going to be negative. How does this thing go on from here? Notice you have this 0 plus on the top and bottom. When you cancel these things out, we're going to end up with a 1 over the minus 6. Those things cancel out giving us the negative 1, 6, which is the correct answer right there. If you find that a little unsettling, that weird arithmetic with 0 plus and 0 minus, that's okay. It really comes down to the following idea. If you take the absolute value of x minus 3 over x minus 3, this right here always gives you plus or minus 1. The difference comes down to when x is greater than 3, you're going to get a plus 1. When x is less than 3, you're going to get a minus 1. Similarly, if you take the function, the absolute value of x over x, this thing always equals plus 1 and minus 1, where you get x when you're greater than 0 and you get minus, sorry, x when it's greater than 0 will give you plus 1, and x when it's less than 0 will give you negative 1. That's what happens when you see this ratio involving the absolute value. Because if x minus 3 is positive, you're just going to get x minus 3 divided by x minus 3, it'll go away. On the other hand, if x minus 3 were negative, then the top would be positive and the denominator would be negative. The absolute value cancels out because you could just rewrite x minus 3 as the absolute value of x minus 3 negative, that'd still be negative. These things cancel out leaving you the negative sign there. So that's why we are able to do 0 plus over 0 minus. This always cancels out just to be negative 1. And it goes the other way around too. If you have 0 minus divided by 0 plus, this always equals to negative 1. If you don't like the 0 plus, 0 minus business, then the idea is just to remember this. If you have the absolute value of a quantity divided by that exact same quantity, then it'll always be plus or minus 1 depending on whether that quantity was originally positive or negative.