hello welcome to another video so this is the delta epsilon proof for a quadratic limit okay so the last video that i did had to do with a linear function now this is a quadratic and you know something about quadratics there appears to be two answers okay so we might have to make two suggestions of what delta would be because it might be something on the left or something on the right so expect that to happen and i'm gonna make it as simple as possible okay let's get into the video [Music] so first things first let's write the delta epsilon statement the one that allows us to do the proof and it says that if the limit of a function is l just like this the limit of this function is 12 at a point so we're gonna call this point c and this is what it says um that the limit of f of x is equal to l as x approaches let's call it c if this is true it means that for all epsilon greater than zero which is how much the function moves away from the limit and for all delta greater than zero okay both of these are distances so they have to be greater than zero there exists a delta for all epsilon greater than zero there exists a delta greater than zero maybe i should change it that way that's the statement actually okay so whenever epsilon is whenever this function moves away from here you have to move away from here which is delta also so there exists and this is how you write it there exists a delta greater than zero such that if zero is less than the distance of x from the target point c and it's less than delta that is it would be within zero and delta then this function is not too far away from l f of x the absolute value of f of x minus the limit is also less than epsilon that's the statement that we want to show we want to show that if this is true and we know that this is positive and this is positive there is always there will always be a delta that is positive greater than zero such that if the distance of x from the target point okay is within zero and delta then the distance of the function from the limit will also be less than epsilon and that's the statement we want to make for this quadratic equation so how do we do this now this is a very easy secret let your focus be on this second statement okay let's write the second statement and see what we're gonna get remember we're gonna translate this into what we have okay so let's say basically what we're saying is zero so this is the proof okay we're basically saying zero is less than the absolute value of x minus c the c in this case is one okay that's our target and it's less than delta we don't know what delta is but we know that if this is true then this also will be true okay implies that this function the absolute value of the function what's the function we're dealing with now it's this quadratic x squared minus sorry plus five x plus six okay let's put it in parenthesis minus the limit what's the limit 12 is less than epsilon interesting okay so if this is the case then we have this let's see how we can simplify this so it becomes something we're used to so this implies that um we're gonna have if we do the subtraction remove this parenthesis we're going to end up with x squared plus 5x minus 6 actually and that's less than epsilon um this can be factored if we factor this we're going to have x plus two what do we have no we're going to have x plus six and x minus one oh that's what we're gonna have we're gonna have x plus six and x minus one okay and that's less than epsilon now this is what i was saying if this was a linear function you're gonna have something multiplying x minus one this x minus one is of interest why is it of interest because it's this x minus one that always happens so remember however we still have x here but we don't want x to be on the side because we want whatever we get as delta to not depend on x to just be a fixed distance from the target so we have to choose a value okay so we don't want this to be dependent on x because right now it's dependent on x we have to say what can x be we have to find a number that x could be so that when we put it here um it's going to make sense now let's let's just keep going okay now look at this we could say that uh we could rewrite this as the absolute value of x plus six multiplied by the absolute value of x minus one okay and that is less than epsilon so this is where this proof gets interesting because remember or what i said because this is a quadratic you'll be you'll get two situations and i need this to be a number so i need to really work on what number i could use to replace x plus six okay and how do you do that go back to the problem because the target point is one and we need to be close to one we might as well say that the distance from one is 0.1 or 0.2 or 0.3 or 0.4 we don't know but let's keep it simple let's say it's within one unit of the target okay so we can say the distance x minus say say that the distance from x to the target point is within one unit now as you can see it appears we're suggesting that delta is one no we're not saying delta is one we're just saying that this is less than one and it's also less than delta so it's possible that delta is 0.1 and this statement will still be true delta could be 0.0001 this statement will still be true because 0.0001 is still less than 1. so we're not saying delta is 1. okay so remember that we're just saying that no matter what happens we're trying to be reasonable being as close as possible to one delta i mean the distance from one at this point because i need to replace this is less than one now what does this mean from your ninth grade um algebra you know this is negative one is less than x minus one and it's less than one okay now i don't need x minus 1 right now i need to replace this guy so how do you replace it what do you do to negative 1 to make it positive 6 you add 7 so i'm going to add 7 to everything here and what does that lead me to well it tells me that negative 6 sorry that positive 6 is less than x minus 1 and x plus 6 rather x plus 6 and it's less than this is 8. so i know that x plus six is less than eight so a good number to replace this with is eight okay you get that so a sensible number to replace this with is to replace this with eight eight times x minus one absolute value is less than epsilon the constant i was looking for is now my eight this could be some side work that you do okay or just put this in a bubble and look at this this implies that absolute value of x minus one is less than epsilon over eight so with this you can now guess i'm gonna guess that my delta is epsilon over eight you could also guess that your delta is one it's possible okay in fact let's do that but again we won't be able to use this because it's a number and it's not dependent on epsilon we want it to depend on epsilon because of the statement that for every epsilon greater than zero there exists a delta such that if this is true this is true so because we want it to depend on epsilon we're just going because we can't actually do that so you want to base base it on this but what we chose here was not wrong remember okay now let's go on now that we've guessed so i guess delta is equal to epsilon over eight and that will then help you to go back to the second part of the proof part two okay if we do part two of this this one we came from here so if you do part two of this all you have to do i said this in the previous video focus on this just this alone okay just take this part leave epsilon alone leave epsilon alone in the second part so let's go here we're going to have the absolute value we're going to start with this statement the top one that x squared we just want to show that this is less than epsilon we want to show that this is less than epsilon even though we don't have epsilon in the picture we have delta so we're going to use delta to prove that epsilon exists okay or to show that if delta exists there's an epsilon but we've just guessed an epsilon at this point okay i need a delta at this point so x squared where is it plus 5x plus 6 minus 12 will be equal to so this is the part 2 will be equal to maybe i should put a line here so you see that i'm taking the second part will be equal to the absolute value if you resolve this will be x squared plus 5x minus 6. so what does this mean it means i'm at the end of my proof because all i have to do is replace this with something that i know replace this with something that i know remember we know that this is less than delta and we also know that this is less than eight so i might just say that actually this expression is less than eight delta it makes sense if this is less than eight and this is less than delta then the product of those two things definitely this product will be less than this product it makes sense in case you're confused let me show you something let's assume just small bubble here 2 is less than 3 and i tell you that 4 is less than 5 then 2 times 4 must be less than 3 times 5. there's nothing else you can do about that okay so that's exactly what has happened we know this is less than 8 we know this is less than delta the product of the two is eight delta okay so that and what does what do you think this means this actually is eight times what did we say delta was again epsilon over eight and what does this give us it's less than epsilon so this function from the beginning which is basically f of x minus l is less than epsilon so i can actually drag this down here and say that this is x squared plus 5x plus 6 minus 12 is less than epsilon that's how you do the proof go this way get a suggestion for delta and then start again and make sure you bring back delta which is in terms of epsilon and you'll be able to get that right remember it's a proof you know where you're going you're just trying to game the system i'll see you in the next video don't stop learning because those who stopped learning have stopped living bye