 Before we embark on the complicated and somewhat messy problem of finding limits, let's see if there are some useful properties of limits that we can use. And in particular, rather than trying to find every limit from the beginning, let's see if we can use certain properties to simplify our quest. For example, suppose f of x is c, some constant function. Find the limit as x approaches a of f of x. And so remember what this notation means. We want to know the value our function is close to whenever x is close to a. If x equals 7, we have no idea if we're actually close to a, so that doesn't do us any good. So what do we mean that we're close? One possibility is that we're a little bit less than a, and we can express this as a minus some small amount. So if x is a minus 0.1, then our function applied to a minus 0.1 is c. And that's because f of x is c, so no matter what x is, the output is always going to be c. Since I'm not planning to run for a political office, it wouldn't hurt me to collect more facts. So let's take another value of x that's close to a. Again, how about a minus a small amount, say a minus 0.01? Then f of x applied to a minus 0.01 is c once again. And again, facts don't hurt. If I take another value close to a, how about a minus 0.001, then f applied to a minus 0.001 is still c. All of our x values are less than a. So let's take a look at a couple of x values that are larger than a. Again, we want to be close, so a close value is a plus a little bit. So if x equals a plus 0.1, then f of a plus 0.1 is c. If x is a plus 0.01, then our function value is c. And if x is a plus 0.001, our function value is c once again. And so now we have a bunch of function values. We can round them to the same number of decimal places. It doesn't matter how many, they're all the same. And this table suggests that as x gets close to a, the function value is going to get close to c. And so our limit as x approaches a is c. And this suggests a very useful property of limits. If c is any constant, the limit as x approaches a of the constant c is going to be the constant itself. A useful thing to be able to do in math and in life is to be able to compare two different things. So let's compare the limit as x approaches 2 of x and the limit as x approaches 2 of c times x, where c is some constant. So we might take a couple of x values that are close to but not equal to 2, 1.9, 1.99, 1.99, and then find the value of x and c times x. And similarly, we'll take a couple of x values that are close to 2 but above 2. So for example, 2.1, 2.01, 2.001, and find the value of x and c times x. So let's compile our data and then find our limits. So it appears that the limit as x approaches 2 of x is 2 and the limit as x approaches 2 of c times x is 2 times c. And it appears that the effect of this constant is to multiply the limit by the constant. And so this suggests the following theorem. Provided that the limit exists, the limit of a constant times a function is the constant times the limit of the function. One useful thing that happens is the more we find out, the more we can find out. So again, let's compare a couple of limits. Now a useful thing to remember is that the type of function is determined by the last thing that you do. So in this case, 3x is a 3 times x. This is a constant multiple. So the limit can be found using the constant multiple rule. So the limit of a constant times a function is the constant times the limit of the function. And so we just need to find the limit as x approaches 4 of x. And likewise, 6 is a constant. So the limit of a constant is just a constant. As for the limit as x approaches 4 of 3x plus 6, we don't have any rule that will tell us how to deal with the limit of a sum. So we'll have to find that limit numerically and we'll pick some x values that are close to 4 and evaluate 3x plus 6 at those values. And it appears that we're getting a limit of around 18. So putting our information together, it appears that the limit as x approaches 4 of the sum is the sum of the individual limits. And this, in fact, gives us a new limit rule provided that the limits exist. The limit of a sum is the sum of the limits. In the same way, we can discover two other important limit theorems. Provided that the limit exists, the limit of a product is the product of the limits. And likewise, provided that the limit exists and the limit of the denominator is not equal to 0, then the limit of a quotient is going to be the quotient of the limits. For example, suppose we have a table of information and we want to use this table to find the limit as x approaches 2 of an expression 5f of x g of x plus 3. So the first thing that's important to remember the value of a function at x equals a is not relevant to the limit as x approaches a. And what this means is that the value of f of x and g of x at x equals 2 is not relevant to the limit as x approaches 2. So we'll ignore these values. They are not important. On the other hand, if we look at the table, it appears that as x approaches 2, f of x approaches 8 and as x approaches 2, g of x approaches 5. And now we can use our limit properties. So the expression we're looking at is a sum. And so we know the limit of a sum is the sum of the individual limits. So we can rewrite this as the limit as x approaches 2 of 5 times f of x g of x plus the limit as x approaches 2 of 3. Next, we see that this first limit is a constant times a function. And we know that the limit of a constant times a function is the constant times the limit of the function. So I can write this first limit as 5 times the limit as x approaches 2 of f of x times g of x. Next, we see that we're trying to take the limit of a product of f of x and g of x. And we know that the limit of a product of 2 functions is the product of the individual limits. And so this will be 5 times the limit as x approaches 2 of f of x times the limit as x approaches 2 of g of x plus the limit as x approaches 2 of 3. Well, we've determined the limit as x approaches 2 of f of x and g of x and 3 is a constant. And so the limit as x approaches 2 of 3 will just be 3. So I'll substitute in the limit values that I can determine right away and find that our limit is going to be 203.