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