 Intuitively, the limit as x approaches a of f of x is equal to l, when f of x is close to l, as long as x is close to a, but not equal to a. And in fact, mathematicians were comfortable with this informal definition for a couple hundred years, because, well, it worked. But some non-mathematicians were quite critical, and in the 19th century a more formal definition of limit emerged. So the key step in this mathematician process is to put a definition on what we mean by close to. So we might remember that x and y are close if the absolute value of their difference is small. And so we get our formal definition of limit. Now, we want f of x to be close to l. So we can make the difference between f of x and l less than some error, epsilon. We also want x to be close to a. So we can take about x and a being some distance delta apart, absolute value of x minus a less than delta. We also want x not to be equal to a. And so we can guarantee that by also requiring this absolute difference be greater than zero. Now, we also have these epsilons and deltas floating around. And so we need to determine which came first, the epsilon or the delta. And so a basic principle in life is start with what you want. In this case, since we want f of x to be close to l, then we want this epsilon to be our starting point. So we might phrase it this way, for any epsilon greater than zero, there is a delta where if x is close but not equal to a, then f of x is close to l. For example, let's prove the limit as x approaches two of x squared is equal to four. So definitions are the whole of mathematics. All else is commentary. We'll pull in our definition. So we know our starting point is we want the absolute value of x minus two to be between zero and delta. And we want to end with absolute x squared minus four to be less than epsilon. So we can work a few steps back from the end and we see that we have to do something with this absolute value of x plus two. So remember, you can assume anything you want as long as you're explicit about it. So let's see what we can do with this absolute value of x plus two. Since we're looking at x close to two, let's consider some interval around two, say how about between one and three. Then on this interval we have... So in this interval, we can replace the absolute value of x plus two with something smaller or something greater. Now, since our proof has to read forward, we want x plus two to be less than what we replaced. In other words, on the previous line we had something. We replaced it with absolute value of x plus two and because that was smaller, we maintained the inequality. So we'll use the upper bound five in the previous line. So we assumed x was between one and three and in this interval the absolute value of x plus two is less than five. So if we had a five, we could replace it with something smaller and maintain the inequality. And now let's take one last step backwards and now the only difference between our two lines is that one has a delta and the other one has epsilon fifths. Again, if we want to replace delta with epsilon fifths, because this is on the greater than side, we need to make sure that delta is less than epsilon fifths. Now, we typically make one last simplification. Note that if we pick a small enough epsilon, we'll be in this interval between one and three. So we can replace this requirement with the fairly standard phrase sufficiently small and say that for sufficiently small epsilon and delta less than epsilon fifths, we have absolute x squared minus four less than epsilon whenever x minus two is between zero and delta. And that completes our proof.