Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Definition of a limit Let F of X be defined on an open interval about X naughty, except possibly at X naughty itself. We say that the limit of F of X as X approaches X naughty is the number L, and we write the limit as X approaches X naughty of F of X equaling L. If for every number epsilon greater than 0, there exists a corresponding number delta that's greater than zero such that for all X0 less than the absolute value of X -, X naughty which is less than delta implies that the absolute value of the function minus the limit is less than the epsilon. The epsilon is also thought of as the error tolerance. If we look at this pictorially, what's really happening is that if we think about having a challenge of the function minus a limit of some sort. So if here was the limit and we want the error to be less than 110th. So if here's the actual limit and we added 110th and we subtracted 110th, we could then bring these lines back and we could look to see that the response would then be here's the L -, 1/10 intersecting the function. So coming down here, here's the L plus 110th intersecting the function coming right here. Now, because of the absolute value in symmetry, we need the distance away to be the same. So what we're going to do is we're going to take the smaller of the values of the X not minus the delta 110th and the X not plus the delta 110th. Because this one, if we look at the distance here, the distance is too far out if we do symmetry because of that absolute value, if we use the shortest distance, we know that from this location to this location, the limit plus that error tolerance would definitely fall within that range. We can then make the error tolerance even more small or one 100th. So now if we're limiting the limit to positive one 100th and negative one 100th, we've made an interval that's even smaller. So if we look at the intersection here and come back to the X not lines, we'd have this one and we'd have this one. We still want to use the smaller because of the absolute value. So if we look at this blue box, we know that L plus 100th or L minus 100th would all fall within this location. As far as the error tolerance, we could go even more to do one one thousandth. And if we did 1-1 thousandth, it makes the interval of the function even smaller, so that our X not then is even smaller. We could do 100 thousandth, which would make it even smaller, and eventually we're going to do it for whatever the E is. So plus E or minus E, making it smaller and smaller and smaller till we can get to that actual location. Now, to do this algebraically, what we're going to do is if we have something like the limit as X approaches -5 of X ^2 plus six X + 5 / X + 5, so we know that this is really the limit as X approaches -5. If we factored this top, we'd realize that things cancel, so we would get it equaling just X + 1, and X cannot equal -5. So if we think about as X goes to -5, negative 5 + 1 would give us a limit of -4, so we'd have our X + 1. Our function minus our limit has got to be less than some given epsilon. Let's let our epsilon be .05%. We'd also then have X minus our X value being less than some delta. And what we want to do is we want to find that delta. So if we have X plus 1 + 4 less than .05, so we know negative .05 is going to be less than X + 5, which is less than .05. If we subtract the five, we get -5.05 less than X less than -4.95. Now what we're going to do is we're going to look over here and we're going to get negative delta, less than X + 5, less than delta. If I subtract 5 here, I get -5 minus delta less than X less than -5 plus delta. So since X is here and X is here, and these are the same XS, and we want these inequalities to match, we're going to have the -5.05 equaling -5 minus delta and solve for delta. Then we're going to do the same thing with the -4.95 equaling the -5 plus delta. So if we have -5.05 equaling -5 minus delta, we're going to get delta to equal .05. And if we have -4.95 equaling -5 plus delta here we get delta equaling .05. If those deltas were different, we would always take the smaller of the deltas. Let's look at another example. If we have the limit as X goes to 0 of sqrt 4 -, X equaling 2. So we'd have the absolute value of the function 4 -, X minus the limit of two having to be less than some epsilon. This time we're not going to be given an epsilon. The other equation is going to be X minus the given X value of 0. It's got to be less than delta. So here we're going to have negative epsilon less than sqrt, 4 -, X -, 2 less than epsilon. If we add the two, we get 2 minus epsilon less than the square root, 4 -, X less than two plus epsilon. If we square everything out to get rid of the square root, we'd have 4 -, 4 epsilon plus epsilon squared less than 4 -, X less than 4 + 4 epsilon plus epsilon squared. Subtract the four and divide by a negative X, we'd get 4 epsilon minus epsilon squared. Because we divide by a negative, we're going to flip the sign -4 epsilon minus epsilon squared. Now, I personally like my sines to be less than, so I'm going to just rewrite this as an equivalent so that perhaps I cannot get confused over here. We're going to have negative delta less than X less than delta. So we now know that -4 epsilon minus epsilon squared has got to equal negative delta and four epsilon minus epsilon squared has got to equal delta. Now we are going to solve here. So we get delta equaling 4 epsilon plus epsilon squared. And here delta was 4 epsilon minus epsilon squared. The epsilon is actually going to be a positive value, so we know that this one has got to be the smaller 1. So our delta equals 4 epsilon minus epsilon squared. If we look at another example, let's look at F of X is given as two X -, 2 and X is given as -2. So we want the limit as X approaches -2 of two X -, 2. When we stick that in, we would get -4 -, 2 negative 6. So the function two X -, 2 minus the limit of -6 has got to be less than. Let's have an error tolerance here given of .02, so it's got to be less than .02. So now we're going to get the absolute value of two X + 4 less than .02 negative .02 less than two X + 4 less than .02. Solving this out, dividing by the two, we get -2.01 less than X less than -1.99. Looking at the second part that says the X minus the X value has got to be less than some delta, and we're trying to figure out what that delta is. So we get negative delta -2 less than X, less than delta -2. So our last piece is going to be to actually solve for the delta. Make our left side of the inequality equals and our right side of the inequality equal. So we get delta equaling .01 and we get delta here equaling .01. Those are the same values. So we're going to use point O1. If we look at another one, let's have F of X equaling 120 / X and let's have X equaling 24, and let's have an epsilon value this time of 1. So if we thought about the limit as X goes to 24 of 120 / X, that would just give us 5 S The absolute value of 120 / X -, 5 has got to be less than one, and X - 24 has got to get less than delta. So -1 less than 120 / X - 5 less than one. Add the five. Now because X is in the denominator, we're going to take the reciprocal. And when we take the reciprocal, we have to flip all the signs. Now I'm going to multiply by 1:20, so I'm going to get 30 greater than X greater than 20. I'm going to rewrite it so that it's less than signs, so 20 less than X less than 30. Over here we get negative delta less than X -, 24 less than delta, adding 2424 minus delta less than X less than delta plus 24. Once again, the left sides have to equal and the right sides have to equal. So delta here is 4 and delta here is 6. So we want the smaller 1. So delta is going to be 4. Thank you and have a wonderful day.