 In this video, we provide the solution to question number 15 for practice exam two for math 1210. We're asked to compute the limit as x approaches one of x cubed minus one over x squared minus one. Now the tempting thing to do is just to plug in x equals one and you get one cubed minus one over one squared minus one. That's equal to, well, one minus one over one minus one which gives us zero over zero. That's an indeterminate form. So that's not going to give us the limit. That just tells us we gotta do something else. It does tell us though that since we got zero on top and bottom, there's some common factor to the top and bottom which is necessarily going to be x minus one. The only way that the top could be zero is because the top's factorable by x minus one. Same thing with the denominator as well. So we're gonna have to factor these things to, we're gonna have to factor these things in order to compute the limit. Try to simplify it here. Take the limit as x approaches one of x cubed minus one x squared minus one. So many of us will probably see how to factor the denominator as it's a difference of squares. We can factor it as x minus one and x plus one. So we can see that factor right there. The numerator can also be factored. This one is a little bit less known but it still has very special factorizations. We're putting on a note card, I would say. You know, you should study this one. The difference of cubes factorization. So we're familiar that a squared minus b squared can factor as a minus b and a plus b. But some other factorizations worth remembering if you have a cubed minus b cubed, the difference of cubes, this factors as a minus b and you get a squared plus ab plus b squared. So the way I like to think of it is if you have the difference of cubes, well, one of the factors is a difference but it's like you forgot the cube. Yikes, I forgot to bring the ice cubes to the party guys, I forgot them. And then the other one is a squared plus ab plus b squared for which, if you multiply out, you see why this thing works but you can kind of compare this to like the perfect square trinomial. If you have a plus b squared, this multiplies out as a squared plus two ab plus b squared right here or right. And so you see like there's the two that's missing in this factorization. So the first factor is you forgot the cubes. On the second factor, you forgot the two. And so in order to remember the difference of cubes, you need to forget, right? So remember the formula, forget the cubes, forget the two. And that also applies if you wanna do the sum of cubes. There's not a sum of squares formula because that would involve imaginary coefficients but for cubes, you can get away with it. If you take a cube plus b cube to factors as a plus b, you forgot the cubes and then you get a squared minus ab plus b squared which looks like the perfect square trinomial factorization a minus b squared, which would look like a squared minus two ab plus b squared but you forgot the two. So the same thing, forget the cubes, forget the two, that's how you remember. The other thing to remember is that this negative sign will be the same here, but you'll see an opposite sign there. Instead, you have a plus here, plus here, you get a minus there. So these are some special factorizations. If you don't remember them, maybe put them on a no card. It could be very helpful in the situation because if you take x cube minus one with this factorization in mind, specifically this one, you'll get x minus one times x squared plus x plus one. So with that factorization in mind, you can simplify this thing very effectively like so. And so now you see that as x goes to one, these other things don't have any problems. You don't have the zero over zero anymore. You have x squared plus x plus one over x plus one as x approaches one. So now we can just plug in x equals one. We get one squared plus one plus one over one plus one, which will end up giving us three over two, which is the correct limit of this value. Now, before ending this video, I didn't wanna make mention to you that if by some horrible chance you forgot this formula, but you need a factor, it's like, oh, yikes, x cube minus one, I don't know how to factor it. Well, factoring is just dividing, right? If you're trying to factor this thing, you can just divide it instead. And like I told you, the reason why you got zero over zero when you take the limit as x goes to one, it's because x minus one is a factor. And so let's factor, let's divide x cube minus one by x minus one. Again, it's not the most ideal way to do it, but if you're in some type of desert island calculus scenario, it's better than not being able to solve the problem. So x goes into x cubed, x squared times, that is just to say, just take x cubed divided by x, you get x squared. Then you're gonna take x minus one times it by x squared, you're gonna get x cubed minus x squared. You're gonna subtract it from above, the x cubes cancel, and then you have a double negative, right? You're taking a negative negative x squared, so you get a negative x squared, bring down the negative one. You rinse and repeat, x goes into x squared x many times, for which, again, where that x come from, it just took x squared divided by x, you get an x right there. Then we're gonna take x times x minus one here, that gives us x squared minus x, subtract it from above. Again, the x squares will cancel, and then you're gonna get a negative negative x, so you get negative, negative negative x is positive x minus one, and then x minus one goes in x minus one exactly one time, and the remainder there is actually gonna be zero, so you get zero. And so there we go, we get the quotient right here. And so that's long division of polynomials, you can use it if you have to the factor, but it turns out even easier than that is if you remember how synthetic division works, since we know, again, the limit, we're taking the limit as x approaches one right here, we know that since we're approaching one, our polynomial, we have to divide by one, that's what we're doing here. And with synthetic division, you typically just write the coefficients here, so one x cubed, zero x squared, zero x minus one, you can do the synthetic division, bring down the one, one times one is one, plus zero is one, times one is one, plus zero is one, times one is one, minus one is zero, in which case then, we see these are the coefficients of the quotient, x squared plus x plus one. So I don't want you to get in a situation where you feel like you're stuck because you don't remember one of these special factorizations because there are solutions to do, to get around this. One, write down the special factorizations either in your brain or on a note card that you can use on a test. And if per chance you don't have them but you need to factor, factor is just dividing. If you get zero over zero, which is very probable for question number 15, that means there's a common factor in these situations. And so, factored using division if you have to. Now, these are techniques that are gonna work if you have a rational function. We had x cubed minus one over x squared minus one. If you have something a little bit more convoluted, if there's like fractions involved, you might have to clear the denominators. If there's square roots involved, you might have to rationalize by multiplying by conjugates. If you combine these algebraic techniques, then you should be able to calculate this limit.