 This lesson is on the precise definition of a limit or the epsilon delta definition. We will need to know a couple of things. What is epsilon and what is delta? Well, epsilon is a small change in y. We think of delta y, but this is even smaller than that, and that is why we call it epsilon. Delta is a small change in x. Again, something like delta x, but very, very, very small. Well, we also need something else called a function. So let's create a graph of a function. And we need a point a where we are looking for our limit, which is l. So we're looking at x is equal to a and y is equal to some y value l. Now, the point in question as we have learned from limits may or may not exist. What we are doing again as you learned from the first time you looked at limits, we're going to come in from the left, come in from the right as we head towards our limiting value. And we will come close in to the left and right. And the difference between a and wherever we are as we come in from each side is called delta. So this is a delta and this is a delta. So it's a small change. So this point here is a plus delta and this point here is a minus delta. So when we come up to the y values and our limiting value is called l, then each difference here is called epsilon. And as we come out to this point, we have l minus epsilon. And as we go up to that point, we have l plus epsilon. Again, as we think of limits as we come closer and closer into our point on our deltas, we will hit that limiting value of l. So what is the precise definition using this terminology? Let f be a function on an interval that contains a and a can be omitted. It can be a whole. We can say that the limit as x approaches a of f of x is equal to l. If for every epsilon greater than zero, there exists a delta greater than zero such that f of x minus l in absolute value is less than epsilon whenever x minus a in absolute value is less than delta. What does that mean? Again, it means that as we get closer and closer in to the a from the deltas, the a plus delta, the a minus delta, we are going to hit this limit and limiting value at that point. That means epsilon will be getting closer and closer to l. Well, how do we use this? Well, let's look at an example. Consider the limit. The limit as x approaches three of x squared minus nine over x minus three. If we were to just continue and find out what the limiting value is on this, we know it does not exist at three. So there's going to be some sort of a hole here that we can factor the numerator and put that over that denominator and actually reduce this. Of course, we always want limit as x approaches three in there. And if we put a three in for x, because remember when we're looking at limiting values, we are looking for some point on the curve, even if it doesn't happen to exist at that particular place. And we put the x in for three and we get a six. So the limiting value for us is a six. So our question is, find a delta such that the limit x squared minus nine or x minus three, that's your f of x minus your l, which is your six, is less than some value, which in this case is 0.2, whenever x minus three, that's the point we are heading to. Remember, it's x minus a in that case, three is less than delta. So we have to find our delta. So we can find it two ways. We can do this algebraically, or we can do it on our calculator. So let's do this algebraically first. We have our absolute value of x squared minus nine over x minus three, our f of x minus our limiting value, which is six, is less than 0.2, which is our epsilon. We are trying to find our delta, which will be in the form of x minus three, an absolute value form, and that has to be less than some delta. So somehow we have to make what's an absolute value here into an x minus three. So the first thing we want to do is get a lowest common denominator. So it's x squared minus nine minus six x plus 18 over x minus three. And we can take off the absolute value signs too, because remember when it says absolute value is less than some number, we can put it between the positive and negative of that number. And now we will reduce what's between the two inequality signs. We have x squared minus six x plus nine over x minus three. And that again is between negative 0.2 and positive 0.2. This factors to x minus three quantity squared over x minus three between 0.2, negative 0.2, reduce this. So now we have x minus three is between negative 0.2 and positive 0.2. We can put that back in absolute value form and say the absolute value of x minus three is less than 0.2. So in this case, our delta is 0.2. So we say 0.2 is an effective delta. Or we can say delta has to be less than or equal to 0.2 or two tenths, which means the furthest we can go out with this in limiting values for our delta will be 0.2. Well, how would we do this problem on our calculator? Well, we would, first of all, have to enter in our absolute value without the absolute value in it and put it between the negative 0.2 and the positive 0.2. So now we will have x squared minus nine divided by x minus three and that's minus six. Okay, we'll put that in one y and then put it between negative 0.2 and positive 0.2. We notice how y values are from negative 0.2 to positive 0.2. So let's change our window. We can leave that negative 10 and maybe put this from negative 0.5 to positive 0.5 and graph. Now we have to find the intersecting point. So let's intersect and we'll get the first intersecting point of 2.8 for x and negative 0.2 for y. Let's do the second intersection. Okay, that'd be the first curve and let this be the second curve and we get x is equal to 3.2. So now we have x is between 2.8 and 3.2 but remember our delta is absolute value of x minus three. So we have to track three from all sides. So the absolute value of x minus three is less than 0.2 which is the same answer we got before which means delta can be less than or equal to 0.2 or 0.2 is the effective delta for that epsilon we are given. So in these kinds of problems we're usually given the epsilon and we determine the delta. Let's just try one more problem. Consider the limit as x approaches 2 of 1 over x and we know that limit to be one half and that's the L that is placed there. So now we want to find a delta again such that the absolute value of 1 over x minus one half is less than 0.1 whenever x minus 2 and 2 is your a value from your limit is less than delta. Again we're looking for that delta. So we write the inequality 1 over x minus one half is between 0.1 and negative 0.1. Now we bring the one half to each side instead of solving everything between the inequalities on this one. So we have 1 over x is less than and this is 0.6 and that one is 0.4. If we want to solve for x we can just take the reciprocal and change the inequality sign. So this is 1 over 0.4 and that's 1 over 0.6. Again when we're looking for a delta we want to subtract 2 so now we have x minus 2 is greater and less than negative 2 plus 1 over 0.4 are greater than negative 2 plus 1 over 0.6 and this evaluates to 0.5 is greater than x minus 2 is greater than negative 0.333 repeating. So now we have two values that are different. Remember on the last one both sides of our inequality had the same value. This time the value is different and what do we do? We pick the smallest one absolute value wise. So we can say now that x minus 2 is less than a delta which is 1 third or 0.333 repeating. So delta will be less than or equal to that number. Either you put the 1 third in or you put the 0.333 repeating. That is what your effective delta is. Well let's do this one on the calculator. 1 divided by x minus 1 half and we're going to go from negative 0.1 to positive 0.1. Graph that. Okay let's get our first intersecting point and our first intersecting point is at x is equal to 2.5. Let's go get our next one. The next intersecting point is x is equal to 1 and 2 thirds. So we have these two intersecting points and remember x is between 1 and 2 thirds in 2.5 but we want for our delta x minus 2 so that's 0.5 or 1 half and that's negative 1 third and remember we pick the smallest one in absolute value size. That's what we're looking for. Sides so delta is less than or equal to 1 third. That is our effective delta. This concludes your lesson on the precise definition of a limit.