 We can reason our way through many limits. However, there are a couple of limits that we have to do a little bit more work, and these generally fall into the category of what's known as an indeterminate form. So, for example, suppose we're trying to find the limit as x approaches a, and it appears that our limit is something we can't evaluate. For example, the limit of x squared minus 5x plus 6 over x squared plus 5x minus 14 is x gets close to 2. As x gets close to 2, our numerator looks like it's getting close to 2 squared minus 5 times 2 plus 6. That's 4 minus 10 plus 6. That's 0. And our denominator, x squared plus 5x minus 14, also looks like it's going to 2 squared plus 5 times 2 minus 14, and after a little test settles, it looks like that's going to 0 as well. And so it appears that our rational expression is going to 0 over 0, but we can't divide by 0. And so what we have here is one of a number of different types of what are called indeterminate forms. You'll see other types as the course goes on, but for right now, our main concern is these indeterminate forms that look like 0 divided by 0. So what can you do? Well, if you were in pre-calculus or any previous course, you'd throw your hands up and say, can't divide by 0, we're done with the problem. However, part of the goal of calculus is to be able to handle indeterminate forms, and so we have to do a little bit of extra work. We must do some additional analysis, and that's usually in the form of an algebraic simplification. So let's consider that problem again. So again, as x goes to 2, this looks like it's going to 0 over 0. It's an indeterminate form, and we have to do some additional analysis. Now one thing that we might notice here is that part of the reason that we're having problems is that numerator and denominator are both equal to 0 when x is equal to 2. Now the fact that they're both equal to 0 when x equals 2 is actually a good thing, because there is a theorem that you learned in algebra, which is that if a value of x makes the polynomial equal to 0, then that value of x corresponds to a factor of the polynomial. In particular, because x equals 2 makes them both 0, they both have a factor of x minus 2, which means that both of these can be factored and both of these can be simplified. Now generally speaking, factoring is a very difficult problem, but we already know a factor. So let's go ahead and make use of that. So I know the numerator has a factor of x minus 2, and the denominator also has a factor of x minus 2. So once I know the one factor, it's a lot easier to figure out what the other factor is, and here we can do what we've always wanted to do. We can cancel out our common factor of x minus 2, and we just have x minus 3 over x plus 7, and as x gets close to 2, this thing gets close to 2 minus 3 over 2 plus 7, and after all the dust settles, that's minus 1 over 9. Now the problem didn't ask us to do this, but it's always worth checking our answer numerically. So again, this asks us what happens to this as x gets close to 2. So you might check this answer by substituting in a value that's very close to 2, 1.99, 2.01, and seeing if the value of this expression is someplace in the vicinity of minus 1 over 9. Part of the reason you do that is there's a lot of work between this line and that line, and it's very easy to make a mistake someplace along the way. Doing the factorization wrong, substituting in the values wrong, there's a lot of places for mistakes, and it's worth doing a quick and easy check. Does a numerical value support my claim? The limit is minus 1, 9. So here's another one, square root of x minus 3 over x minus 9. As x goes to 9, this expression goes to root 9 minus 3 over 9 minus 9. That's going to be 0 over 0, so we have another indeterminate form, and we have to do some additional work. So this time, because we have a square root, we might try to multiply numerator and denominator by the conjugate and simplify. So remember, the radical conjugate just consists of the same terms with the operator change. So if it was subtract, it becomes plus. So I'm going to multiply by the square root conjugate and simplify the expression. Now, I'll want to multiply the numerator out. The denominator I'll leave because maybe we don't need to do the simplification. We'll multiply the numerator out because the reason behind multiplying by the conjugate is it gets rid of the square root up here. So we'll multiply that out. That's a difference times a sum. That's going to simplify x minus 9, and good thing we didn't multiply out the denominator because those x minus 9s will cancel with each other. And we're left with this expression, and let's check it out as x gets close to 9. 1 over square root of x plus 3 gets close to something we can actually evaluate, 1 over square root of 9 plus 3, or 1, 6.