Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College definition of a limit: Let f(x) be defined on an open interval about x-naught except possibly at x-naught itself we say that the limit of f(x) as x approaches x-naught is the number L and we write the limit as x approaches x-naught of f(x) equaling L if for every number epsilon greater than zero, there exist a corresponding number delta that's greater than zero such that for all x, zero less than the absolute value of x minus x-naught 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 wanted the error to be less than one tenth, so if here's the actual limit and we added one tenth and we subtracted one tenth we could then bring these lines back and we could look to see that the response would then be here's the L minus one tenth intersecting the functions so coming down here here's the L plus one tenth intersecting the function coming right here now because the absolute value and symmetry we need the distance away to be the same so what we're gonna do is we're gonna take the smaller of the values of the x-naught minus the delta one tenth and the x-naught plus the delta one tenth 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 1/100 so now if we're limiting the limits to positive 1/100 and -1/100 we've made an interval that even smaller so if we look at the intersection here and come back to the x-naught lines we'd have this one and we have this one. We still want to used the smaller because of the absolute value so if we look at this blue box we know that L plus 1/100 or L minus 1/100 would all fall within this location as far as the error tolerance we could go even more to do 1/1000 and if we did 1/1000 it makes the interval of the function even smaller so that our x-naught then is even smaller we could do 1/100,000 which would make it even smaller and eventually we're gonna do it for what ever the E is, so plus E or minus E making it smaller and smaller and smaller so 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 negative five of x squared plus six x plus five divided by x plus five so we know that this is really the limit as x approaches negative five if we factored this top we'd realize that things cancel so we would get it equaling just x plus one and x can not equal negative five so if we think about as x goes to negative five, negative five plus one would give us a limit of negative four so we'd have our x plus one, our function minus our limit has got to be less than some given epsilon, let's let our epsilon be 0.05% we'd also then have x minus our x value being less than some delta and what we wanna do is we want to find that delta so if we have x plus one plus four less than 0.05 so we know -0.05 is gonna be less than x plus five which is less than 0.05 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 gonna get negative delta less than x plus five less than delta if I subtract five here I get negative five minus delta less than x less than negative five plus delta so since x is here and x is here and these are the same x's and we want these inequalities match. We're gonna have that -5.05 equaling -5 minus delta and solve for delta then we're gonna do the same thing with the negative 4.95 equaling the negative 5 plus delta so if we have -5.05 equaling -5 minus delta we're gonna get delta to equal 0.05 and if we have -4.95 equaling -5 plus delta here we get delta equaling 0.05 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 zero of the square root of four minus x equaling two so we'd have the absolute value of the function four minus 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's gonna be x minus the given x value of zero's gotta be less than delta so here we're gonna have negative epsilon less than the square root of four minus x minus 2 less than epsilon if we add the two, we get two minus epsilon less than the square root four minus x less than two plus epsilon if we square everything out to get rid of the square root, we'd have four minus four epsilon plus epsilon squared less than four minus x less than four plus four epsilon plus epsilon squared subtract the four and divide by a negative x, we'd get four epsilon minus epsilon squared because we divide by a negative, we're gonna flip the sign negative four epsilon minus epsilon squared now I personally like my signs to be less than so I'm gonna just rewrite this as an equivalent so that perhaps I cannot get confused over here we're gonna have negative delta less than x less than delta so we now know that negative four epsilon minus epsilon squared has got to equal negative delta and four epsilon minus epsilon squared has got to equal delta now we are gonna solve here so we get delta equaling four epsilon plus epsilon squared and here delta was four 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 one so our delta equals four epsilon minus epsilon squared If we look at another example let's look at f(x) is given as two x minus two and x is given as negative two so we want the limit as x approaches negative two of two x minus two when we stick that in we would get negative four minus two, negative six so the function two x minus two minus the limit of -6 has gotta be less than- let's have an error tolerance here given of 0.02 so it's gotta be less than 0.02 so now we're gonna get the absolute value of two x plus four less than 0.02 -0.02 less than two x plus four less than 0.02 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 gotta be less than some delta and we're trying to figure out what that delta is so we get negative delta minus two less than x less than delta minus two so our last piece is gonna 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 0.01 and we get delta here equaling 0.01 those are the same values so we're gonna use 0.01 if we look at another one let's have a f(x) equaling 120 divided by x and let's have x equaling 24 and let's have an epsilon value this time of one so if we thought about the limit as x goes to 24 of 120 divided by x, that would just give us five so the absolute value of 120 over x minus five has got to be less than one and x minus 24 has got to get less than delta so negative one less than 120 over x minus five less than one add the five now because x is in the denominator, we're gonna take the reciprocal and when we take the reciprocal we have to flip all the signs now I'm gonna multiply by 120, so I'm gonna get 30 greater than x greater than 20. I'm gonna 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 minus 24 less than delta adding 24, 24 minus delta less than x less than delta plus 24 once again the left sides have to equal in the right sides have to equal so delta here is four and delta here is six so we want the smaller one, so delta is gonna be four thank you and have a wonderful day