 In order to find limits algebraically, it helps to be really good at algebra, and in fact most students have more difficulties with the algebra than with the calculus, so here's a quick review of some important algebra concepts. For example, let's try to find the limit as x approaches 3 of x-grade plus 5x minus 24 divided by x minus 3. We want to find this limit algebraically, but we should also support our conclusion numerically. Since this is an algebraic function, maybe we can find the limit by evaluating the function at x equal to 3. Unfortunately, when we do that we get a zero in the denominator, and so this is undefined. However, because there's also a zero in the numerator, this suggests a possible answer, and so we actually have an indeterminate form. And remember, if we have an indeterminate form, then our limit might be found using an algebraic simplification. So there's a few useful things we can do. One of the things we might try to do is factor the expression. Now, factoring is generally a difficult task. However, we're in luck. When we need to use it, we can rely on the root theorem to find the factors. In case you've forgotten, the root theorem is that if you have a polynomial f of x and you know that f of a is equal to zero, then you know that x minus a is a factor. And the reason why it's useful is that we only ever worry about factoring is if we get a zero in the numerator and the denominator. Which means we'll have a factor immediately. Since x squared plus 5x minus 24 is zero when x is equal to 3, then we know that x minus 3 is a factor, and so I can write the factorization almost immediately. And this means my rational expression can be written as x minus 3 times x plus 8 over x minus 3. And I can simplify that. To x plus 8, provided that x is not equal to 3. What if x is equal to 3? Since we're taking a limit, we don't care. So the limit of this rational expression is going to be the same as the limit of the simpler expression. And that's because the only two places that these functions are different is at x equals 3, and the value of the function at x equals 3 is not relevant to the limit as x approaches 3. So we find our limit, which will be 11. Now remember, this is the claim that as x gets close to 3, this expression is going to be close to 11. So let's check it out. If x is equal to 2.9, our expression is equal to 10.9. And at x equals 3.1, our expression is equal to 11.1, and both of these are consistent with our claimed limit of 11. How about the limit of a function involving a square root? So again, we might just try to substitute in the value x equals 1, except if we do so, we find we have an indeterminate form 0 over 0. So we might be able to perform an algebraic simplification. Here, since we have a square root, one idea might be to rationalize one of the factors by multiplying by the conjugate over itself. So the conjugate of a set of terms involving a square root is going to be the same set of terms, except we're going to change our pluses to minuses, and our minuses to pluses. So the conjugate of square root of x minus 1 is the square root of x plus 1, and since we don't want to change the actual value of the expression, we need to multiply by square root of x plus 1 over square root of x plus 1. Now since our goal was to get rid of one of those square roots, it's worth multiplying the numerator, giving us x minus 1. Now we could multiply the denominator out, but here's a useful rule to go by. Factored form is nice. And while we wanted to multiply the numerator out so we can get rid of the square root, there's no harm in leaving the denominator in factored form. Especially because now we have a factor of x minus 1 in both numerator and denominator, and we can cancel it out and the two expressions are equal as long as x is not equal to 1. But since the value of the function at the limiting point is not relevant to the value of the limit, then the limit of our original expression and the limit of our new expression are the same because the two expressions are the same everywhere except for at x equals 1. And so we can find the limit. Another important algebraic case occurs when we have a limit of a compound fraction. And here, an algebraic simplification that we can do is to simplify a compound fraction by multiplying by a common divisor over itself. In this case, the reason that we have a compound fraction is that we have a numerator where we have fractions with denominators x plus 3 and 6. And the common denominator of x plus 3 and 6 is going to be 6 times x plus 3. So if we multiply our fraction by 6 times x plus 3 over 6 times x plus 3, we'll get a simpler expression. And as with rationalizing by multiplying by the conjugate, it's worth multiplying out the thing we want to get rid of, these fractions in the numerator, but we might not want to multiply out the denominator because factored form is nice. So if we leave the denominator in factored form but clean up the numerator a little bit, we get a much simpler form of the expression. And so the limit of the original expression is equal to the limit of a new expression and we can find that limit easily.