>> Consider this situation, and these are very, very straightforward situations. Polynomial functions have graphs that are continuous, and they're just wavy like this, and they lend themselves well to evaluation of limit. It's real simple; all we have to do is substitute the limit as X approaches -1 of this function. This is a polynomial function. All we find is F of -1. You see, you evaluate this expression at -1, so we have -1 cubed minus 2 times -1, and that's 1. So the limit of this function is 1. And in a similar way for this one, we want the limit as X approaches 4, and 4 doesn't cause this denominator to be 0. And so since nothing funny is going on, it's a very straightforward situation, we just solve it sort of by substitution. Just replace the X's with 4's, evaluate, and we find that the numerator is 0 but that's okay. 0 over 2 is 0, a real numbered value for that limit. Here's a kind of funny situation. It's the limit as X approaches 2 for this expression. Notice that if we use substitution we have 2 squared minus 4 in the numerator, and 2 minus 2 in the denominator. Well that gives us the fraction 0 over 0. Now this is called the 0 over 0 indeterminate form, and when this situation occurs a lot of times we have an opportunity for simplification, and we can actually find the limit. You see, if X equals 2 causes the numerator and denominator to be 0... if 2 is a 0 of an expression, then X minus 2 is a factor. So it might stand to reason that we have a factorization opportunity, and certainly we do. If we take that numerator and factor it we have X plus 2 times X minus 2, we bring down the denominator, and now we take advantage of the cancellation opportunity. So now we have the limit as X approaches 2 of X plus 2, and by substitution 2 plus 2 is 4 for that limit. Here's another similar situation. It's the limit this time as X approaches 1, and once again sort of by observation we can see if X is replaced with 1, 1 squared is 1, minus 3 times 1, that's -3. 1 minus 3 is -2. -2 plus 2, oh that's 0. And then in the denominator 1 squared is 1, 1 minus 1, 0. Again, we have the 0 over 0 indeterminate form, so we look for cancellation opportunities after factoring. And in the numerator we factor to X minus 1 times X minus 2. In the denominator the difference of squares factors to X minus 1 times X plus 1, and the X minus 1's cancel so we have X minus 2 over X plus 1. And by substitution we're replacing X with 1, so we find that limit to be negative one-half. The graphs of these 2 are rather interesting. When these factors cancel, it may make some sense that... that the function are not defined at those values, and particularly on this first one when you construct the graph, it's actually a straight line on that first one. But the line has a hole in it at that place where it's undefined. If you graph it in its... you want to graph it now in its original form because that'll give you the hole in the graph at that particular point.