the epsilon-delta definition of a limit (ultimate introduction) - YouTube

i am pretty sure this is the hardest topic in  calculus one yes we are talking about the epsilon  

delta definition for proving limits and the first  thing that we have to do is just to calm down  

it's not so bad i'm going to show you guys the  definition and we'll explain the definition with  

an actual example an actual number and a picture  and i'll show you guys how to write a proof word  

with the four keyword method so let's  go ahead and get started right here  

here's the definition if we have the limit  as x approaching some number a and that's  

that we have a function f of x and let's say  we do end up with a limit that's kodak to be l  

well this right here means that  we are going to get the following  

for all epsilon greater than zero there exists a  delta greater than zero such that if the absolute  

value of x minus a it's in between of zero and  delta so let me just write that down right here  

then we must have the absolute value of f of  x minus l is less than epsilon so what exactly  

is this right because yeah where is the number  well no zero but come on okay here is the deal

it's hard to explain this we will just have to  use an example and the best way is use the example  

with picture with actual number that's how i'm  going to do it so for now don't worry about this  

let's take a look and perhaps i'll box this  right here all right so let's look at an example  

if we have the limit as x approaching 4 and let's  say we have the square root function 2x plus 1.  

to figure this out it's not bad at all right  because square root of 2x plus 1 is continuous  

so we can just put the 4 inside so we get square  root of 2 times four and then plus one worked out  

we get three done deal right yeah but now this  is the time that all the mathematicians will  

say how do we know that this is in d equal  to three where's the proof so this is the  

time that we are going to use the epsilon delta  definition to prove this right here and based on  

what we saw earlier when we have this right here  this means we are going to get the following  

so let me write that down for all epsilon greater  than zero there exists another number called delta  

that's greater than zero such that and the word  such that just means that the following statement  

is going to choose so we will have the following  condition we will have for the following property  

such that what well in this case the a is equal  to 4 so let me just put the 4 inside here so we  

will have 0 less than the absolute value and then  x minus 4 and then we have to have this less than  

delta if this inequality holds then we will  get the inequality at our function which is  

that so i'll just put down square root of  2x plus 1 and then minus the limit that we  

got earlier which is 3 this right here has to be  less than delta so here is the quick explanation  

the idea is that if x is not too far away from  four how far is too far well the distance between  

x and four it has to be less than delta and delta  is supposed to be a small number if x is really  

close to 4 then the corresponding value of the  function it must be really close to our limit  

and how close is close epsilon distance  close so that's pretty much the idea so  

this shows that you can get to three you can  approach three as close as possible now let me  

give you guys a picture for this right here and  we see that the starting is at negative one half  

and then here is our square root function and  let's say four is right here and we go up and  

we see that we have three right here perfect  now i'm going to break this down for you guys  

you see that we have for all epsilon greater  than 0 what's epsilon epsilon is the distance  

between the function and the limit all right and  to make this more clear i'm going to work with  

an actual value for epsilon and because epsilon  can be any positive number i'm just going to say  

i want to use epsilon to be 0.2 can we choose  0.1 yes can we choose 0.17 yes up to you okay  

once we have this value we have 2  right here again it's the y value  

distance right so that means we can go up by  0.2 i'm just going to put it down right here  

so that will give us 3.2 but don't forget we  can also go down by 0.2 so that will give us  

2.8 cool and once we have the epsilon value you  can just see that we can create a region like this  

and you see that because epsilon can be any  positive number i put down 0.2 right here  

but imagine if you have 0.001 you can kind of  narrow it down and you see that the value of  

the functions will approach to three perfect huh  and here's the thing based on the limit you really  

don't have to have a closed circle here if when x  is four and you happen to have an open circle here  

even though the value of the function is right  here guess what when you close it right when  

you get close and close to 3 the value of the  function is close to 3 the limit still equal to 3.  

you really don't need to have the value b exactly  as 3 because we have this part of the inequality  

x does not have to be equal to a x does not  have to be equal to 4. now here's the question  

once we have the epsilon this is the inequality  that we have what do we do next well you see that  

for all epsilon greater than zero there exists a  delta greater than zero what is the delta theta if  

you look at this inequality you see that we have  the absolute value of x minus 4 is less than delta  

so that's the distance between the x and 4.  yeah so how do we do it well we have the y  

values right here already we just have to kind  of trace back down like this so i'll just put  

it down like this and then of course i put it down  like that cool now we actually just have to solve  

some equations once we have epsilon greater than  once we have epsilon equal to 0.2 the question is

delta is equal to what well this  is our function which is y equals  

square root of 2x plus 1. we know the  y values we can just go ahead and put  

3.2 for the y and solve for x so that we can  figure this out so let's go ahead and do that

and of course just solve this equation square  both sides minus 1 divided by 2 and i'll tell  

you x will give us 4.62 yes i have the answer  over the transparency so this x value is  

4.62 and of course let's do a similar scene right  here this is when y is equal to 2.8 put it there  

solve the equation we get x is equal to 3.42 so  of course i will come here and write down 3.42  

i know the picture is horribly wrong but i  just have to do this so that we can now see  

the picture of course all right now we have  the x values so we can talk about the distance  

x minus four has to be inside of the  delta neighborhood technical term  

okay so how big is the delta  well let's see from here to here

let's just compute the distance so we can  do this minus that which is going to be  

0.62 and then from here to here we can  do 4 minus that which will get 0.58

great now you see when epsilon is  equal to 0.2 we have this and that so  

which number do we choose to be delta let me tell  you a secret pick the smaller one that you have  

pick the smaller one that you have  so the answer is going to be 0.58

yeah why again what this means is that if x is  inside of the delta neighborhood right delta  

region right here when you go up you can see  that your y value is going to be inside of the  

epsilon neighborhood the epsilon region if you  pick 0.61 for example 0.61 is right here okay you  

go up you can see that the value of the function  is inside of this part that's good but if you  

go from 4 and then you move to the left by  0.61 you are a little bit outside when you  

go up you see that this point it's outside of  the absolute region so pick the smaller one yeah  

here is the truth if you have  a question with the actual  

epsilon value most likely you  can say delta is equal to 0.00003

and most likely it's going to work why because  if you say delta is this number that means you  

are only allowing yourself to move out away from  four by just a tiny bit like that just a tiny bit  

if you are inside here if you go up of course you  will be really close to street of course you will  

be inside of the absalom neighborhood but if you  do this your teacher will get really mad at you  

usually the direction is  going to say find the biggest  

the author that will make this statement true  so hopefully this right here gives you the idea  

and now as i promised it i will show you guys how  to write a proof for this limit with the epsilon  

delta definition and the key is know these four  words but first of all write down pf for proof  

here we go first word you see that right here  huh the upside down a is the for o what you do  

is you write given yeah always if you don't have  this let me tell you it's wrong right when you  

are writing the epsilon delta definition proof  always write down given epsilon greater than  

zero all right next you see that we have the  there exists and earlier you see that we have to  

do the work to find it and then sometimes you may  have to choose it right so the next way is choose  

choose delta to be um to be water well  unfortunately i don't know yet because  

you see that epsilon here is just arbitrary  and i cannot do any computation yet right  

so don't worry just leave it right just leave it  just leave it next what do we have next such that  

your satchel doesn't really matter but you see  that we want to have this condition to be useful  

so we're going to suppose that this condition  to be true so i'm just going to put down suppose  

so given choose suppose this is the third word  suppose this right here is true so we will have  

the absolute value of x minus 4 meaning that x is  in between of delta and of course x does not have  

to be equal to 4 so we put that down so that's  pretty much what we're saying earlier x has to be  

in this region here on this interval so and then  later on when you go up you want to make sure  

uh the value of the function is in this interval  but anyway third word suppose next we want this  

right here to be true we want so we have to check  it that's actually true so the last word is check  

right given choose suppose check and let's go  ahead and write that down we have the absolute  

value and we have the square root of two  x plus one and then minus three and do  

not just write down less than epsilon because  we haven't done anything right here all right  

this is the part that requires some computations  and some logical reasoning so check this out  

here we have a square root case so what we're  going to do is multiply by the conjugate and also  

divided by the conjugate so i'm going to write  this down right here i'm going to put on bigger  

absolute value and i will write this down again  square root of 2x plus 1 and then minus 3 and  

then i will multiply the top and bottom by the  conjugate which is square root of 2x plus 1  

and then change the minus to a plus and then we  still have the stream so that's the conjugate  

and then of course do the same thing on the bottom  plus 3. perfect now on the top multiply this out  

square will cancel so we just have 2x plus  1 and then minus let me just put this on red  

three times two which is nine so i'll just  put down minus nine like that all right so on  

the top we see that it's two x minus eight we  can factor out the two so just do the algebra  

and that would be equal to factor out the 2  and then the x minus 4 will still be instead  

of the absolute value and you see i just put  the absolute value on the top and that's the  

property of the absolute value next we'll just  put the absolute value on the bottom but notice  

the output of the square root is always positive  and when we ask you to it it's always positive  

so for the bottom here the absolute value doesn't  matter i'll just write down square root of two x  

plus one and then of course we have  the plus three so what's next then well

here's the key when we're writing proofs we must  use our assumptions which is this right here if we  

don't use the assumption you know something's  wrong the whole totally strong actually  

here we have x minus 4 in the absolute value is  less than delta so you can actually replace this  

with less than delta that's good but before we  can do that you see that we have the bottom here  

ah man it's trouble i want to play with some  inequality thing check this out here's the deal  

this is always greater than 0 and then we  add 3 to it so the bottom is always greater  

than 3. let me tell you we can actually just  ignore the bottom and we can just say this is  

always going to be less than just the numerator  namely 2 times the absolute value of x minus 4.  

why can't we do that well again let me give  you an example let's say we have 10 and 10  

of course they're equal but right here if  i divide this by 3 which number is bigger  

of course this is bigger right so you see i just  ignore the bottom right if you compare this and  

that this is going to be less than just the top  because square root of two x plus one plus three  

do not mention is greater than zero you actually  have to mention that this is greater than one  

of course you can also say it's greater than  three that will also work but the key is you have  

to make sure that the bottom is greater than 1  because if you have 10 and 10 if you divide it by  

0.1 in fact this right here will be bigger this  part is exactly our assumption so we can say  

this right here is less than right this is  less than the delta that we have right here  

and then we have the 2 in front like that  cool now remember in the very end we want  

to end up with just an epsilon so i really  want to just end with an epsilon right here  

now pop quiz for you guys 2 times what will  give us epsilon epsilon over 2 is the answer  

right so you see we can just put on the two and  suppose we choose delta to be epsilon over two aha  

two and two cancel nicely so that means we just  have to go here and choose delta to be epsilon  

over two and you see delta is equal to epsilon  over two and if you read the whole thing now given  

epsilon greater than zero that's this part we  found a delta that's epsilon over two and notice  

because epsilon is greater than zero epsilon over  two is of course still greater than zero and if  

we have this condition right then we see that the  absolute value of this is less than is less than  

epsilon so that means we are done so of course  in the end we can just put on a box and shade  

the ink because this is a very nice proof what do  you guys think cool now before we go i really want  

to talk about this right here you see how earlier  when we have the actual epsilon like 0.2 you can  

just do the computation and we have found that  the delta in this case was 0.58 cool right um  

you see right here we only have a  formula for the delta and the delta is  

epsilon over 2. keep this in mind when you are  doing the proof most likely you just end up with  

a formula in terms of epsilon right here for the  delta and let's see if epsilon was 0.2 right here

okay will the delta work if epsilon is  equal to 0.2 like what we got earlier  

delta based on this formula is going  to be 0.2 divided by 2 which is 0.1

earlier we found the delta was 0.58 and right  here if we use this formula we get 0.1 is there  

anything wrong no don't worry about it right  here it's just because we are doing the proof  

as i said earlier when you have a specific  epsilon value that's 0.2 you can just put  

up delta to be 0.0061 that will work right most  likely but you are just going to make people mad  

don't do that as long as this delta is less  than this then you know this is going to work  

check out my other videos if you need help with  writing the proofs hope for this help that's it