>> Welcome to the Cypress College math review
on epsilon-delta proofs of limits.
The limit as x approaches c of f of x is equal to L means
that for every epsilon greater than zero,
there exists a delta greater than zero such that if x is
within delta of c, that implies that f of x is within epsilon
of L. Here's our diagram and the function is y equals f of x.
So no matter how small an epsilon,
which is a positive radius that we choose around our limit L,
we can find a delta, which is our positive radius around c.
Such that if x is within delta of c,
then the function values have to be within epsilon of L. Now,
this point right here does not have
to be an actual point on the function.
There could be a hole there.
For an epsilon delta proof the first thing you want to do is
to find the limit, even if the limit is given to you,
check that you wrote the problem down correctly.
Most of the proofs you will be doing will involve linear
or quadratic functions.
For both of these type
of functions to find the limit you simply substitute
in the value of c. The limit as x approaches c of f of x,
is simply f of c. Next thing you want to do is to have a copy
of a proof that your instructor did
as a format for your own proof.
Instructors vary on how they want these proofs formatted.
What is right is the way your instructor wants the proof
formatted, not the way someone else tells you how to do it.
You can learn from others the math involved,
but you must use your instructor's format
for your proof.
Alright, I'm going to show two different approaches
for each of these proofs.
Once again you're going to want to look
at your instructors formatting to see how they want it done.
So the first thing we do is substitute in the six,
and we get a limit of seven.
So here we go.
Given epsilon greater than zero we choose delta
to be some value, we don't know what that is yet.
It's going to be a function of epsilon, we'll figure
that out later, you can just leave this blank,
you can put an underline, something,
just leave it blank for now.
Then, for, and then what is x approaching?
X is approaching six, yes, x is approaching six.
So x has to be within delta of six.
We get what?
Ok, so now we look at the function,
and what does the function need to do?
Function minus limit, ok and what is that function?
That function is 4/3x minus one, right.
That's the function.
Now, what's the limit?
The limit is seven, right.
So minus seven, so there's the function, and minus the limit.
So we simplify that, we have 4/3x minus eight,
so what do we want to make it look like?
We want to look, make it look like x minus six, yes.
So how we going to make it look like x minus six?
By getting rid of the 4/3, we're going to pull out 4/3;
it's om absolute value, so we're going to have to pull it
out in absolute value.
And then you simply divide.
Will it end up being a six?
Yes, it will end up being six.
How do you figure that out?
You divide.
So we're taking eight and you're dividing by 4/3.
Well how do you divide by 4/3?
That's the same thing as multiply by 3/4, yes.
So eight times 3/4.
So eight times 3/4, what do you get?
Six, surprise, surprise.
Ok, so what do we get, so that is less than 4/3 times what?
Well, what is x minus six?
X minus six is less than delta, yes.
Ok, and what do we need delta to be then?
We need delta to be 3/4 epsilon, yes.
Why? Because we need that to be equal
to epsilon the whole thing, right?
Because eventually we need the absolute value of f of x minus L
to be, is less than epsilon, bingo.
So what goes in the box?
3/4 epsilon.
So 3/4 epsilon goes in the box and then our conclusion
at the end is therefore, which is the three dots in a triangle,
the limit as x approaches six
of our function 4/3x minus one is equal to seven.
Here's another way to write the proof.
For every epsilon greater than zero let delta equal
and then we don't know what the delta value will be right now,
such that if zero is less than absolute value of x minus,
ok now what is that x approaching?
X is approaching six is less than delta.
So here is we're telling that x is within delta of six
so if x is within delta of six,
and then we go down to the bottom.
And you kind of get an idea of how many steps it's going
to take so you go down to the bottom of the proof
and we have our final statement.
Ok, so we have our function, which is 4/3 x minus one.
And then we have our limit, minus seven,
all of that's in absolute value, is less than epsilon.
So that's 4/3x minus eight.
The advantage to this is I can go through
and I can multiply both sides by 3/4,
so I get x minus six is less than 3/4 epsilon, which tells me
that delta needs to be 3/4 epsilon.
It's just another way to do it, and it's not
like you get to choose.
You choose the way your instructor wants it done.
This is just another formatting, first we find the limit,
First we find the limit
so the limit is -5, so L is -5.
So given epsilon greater than zero choose delta equal to,
we'll figure this out a little bit later,
then for absolute x minus what is x approaching four.
In other words, if x is within delta of four we get,
now we go down and we look at what our function is doing.
f of x minus L, ok what do we get?
Our function is three minus two x
and our limit is negative five,
so we have a double negative there.
So we have three minus two x plus five, which simplifies
to eight minus two x. This time, to make it into x minus four,
we're going to have to take
out the absolute value of negative two.
So that simplifies to just two because the absolute value
of negative two is two.
That is less than two times now what is the absolute value
of x minus four smaller than?
From our hypothesis it's smaller than delta, yes.
That was our hypothesis.
So what is delta?
Delta is epsilon over two, why is it epsilon over two?
Because this product needs to be equal to epsilon.
Our conclusion needs to be that f of x is within epsilon
of L. Here we go, which means that delta needs
to be epsilon over two.
Therefore, the limit as x approaches four
of our function 3-2x is equal to -5.
Let's do this with the other formatting
that I suggested earlier.
For every epsilon greater than zero let delta equal,
once again we'll figure that out later, if zero is less
than absolute x minus, what is x approaching?
Four. Is less than delta, some instructors want this zero,
some don't care about it, it's up to your instructor.
Go down to the bottom, absolute f of x,
minus L is less than epsilon.
The function was three minus 2x, the limit is -5.
So we simplify that and we get eight minus 2x.
We need x minus four, so we factor out the absolute value
of -2, you can't take our just two.
We have to take out the absolute value of -2, which leaves us
with the absolute value of x minus four.
Let's be careful with our absolute values.
But the absolute value of -2 is simply 2.
Divide both sides by two, which tells us
since the absolute value of x minus four is less than delta,
and the absolute value of x minus is less than epsilon
over two, that tells us that delta must be epsilon over two.
We've just proved that the limit as x approaches four
of the function three minus 2x is -5.
Alright let's find this limit on this quadratic function.
So we have two squared plus three times two, minus five,
so I have a limit of five.
So given epsilon greater than zero, we choose delta to equal,
now here with a quadratic function we have
to have two things going on.
So for quadratic functions you're going to have delta equal
to a minimum of the set that contains one and an unknown.
So like we did before, you'll have a blank line
or a box or something.
So delta to be the minimum of the set that contains one
and your unknown, we'll fill that in later.
Then for x within delta of two, since x is approaching two.
We get absolute value of f of x minus L equal
and our function is x squared plus 3x minus five.
Our limit is five, simplify that x squared plus 3x minus 10.
Factor then, be sure to put each factor in absolute value.
Now we need a bound on the absolute value of x plus five.
We know that x is approaching two.
So we need to take this to the side
and find a bound on the other factor.
So we draw a number line and we know that x is approaching two.
So our delta value of one is going
to be what helps us with this.
So if delta is one, that says we're between one and three.
So we go one each way from what x is approaching,
x is approaching two.
And we need a bound on the absolute value of x plus five.
Substituting in one we get the absolute value
of six, which is six.
You just do this on the end points.
The absolute value of x plus five when I substitute
in three would be the absolute value of eight which is eight.
Choose the larger value so the absolute value
of x plus five is smaller than eight.
We know that the absolute value
of x minus two is less than delta.
So we have a bound on that.
We know that we need that to be smaller than epsilon which means
that delta needs to be epsilon over eight.
That tells us what to put in our unknown quantity.
Now, if delta was the minimum of the set one and epsilon
over eight, the smaller of those two we're guaranteed
that if x is within delta of two, if x is within delta
of two then for sure both the y values will be
within epsilon of the limit.
Therefore, the limit as x approaches two
of our function is five.
Here's another format for that same proof.
We know the limit's five for every epsilon greater
than zero let delta equal the minimum of the set.
One and once again and unknown quantity such that
if zero is less than absolute x minus two is less
than delta then, go down to the bottom of the page.
Absolute value of f of x minus L is less than epsilon.
In other words f of x is within epsilon
of L. What's the function?
X squared plus 3x minus five.
And what's the limit?
Five. We need that to be less than epsilon, simplify that.
That's x squared plus 3x minus 10.
Need that to be less than epsilon.
Factor it and keep each factor inside absolute value.
And now we know that we need a bound
on the absolute value of x plus five.
At this point we know that x is approaching two.
And we do a number line working
with our other delta value of one.
So we go one to the left of two and one to the right of two
since we know that x is approaching two.
Now we come up with a bound
on the absolute value of x plus five.
Absolute value of x plus five if x was one would be six.
And the absolute value of x plus five
if x was three would be eight.
Therefore the absolute value of x plus five is less than eight.
So we have that the absolute value
of x plus five times the absolute value
of x minus two is less than eight times.
And what are we going to put in for delta?
Because we know that the absolute value
of x minus two is less than delta.
We're going to put in epsilon over eight, yes.
We're going to put in epsilon over eight because epsilon
over eight times eight gives us epsilon.
And then write your conclusion at the very bottom of the proof.
Therefore the limit as x approaches two equals five.
Let's find our limit.
So we have the square of -1 minus three times -1.
So we have one plus three is four so the limit is four.
So given epsilon greater than zero we choose delta
to be the minimum of the set that contains one
and an unknown quantity that we'll figure out later.
Then for x within delta of -1 we get absolute value
of the quantity f of x minus L equaling absolute value
of x squared minus 3x minus the limit of four, factor that,
be sure to put each factor in absolute value.
And now we know that we need a bound
on the absolute value of x minus four.
If we come over here this simplifies
as the absolute of x plus one.
So we know we have a bound on the absolute value
of x plus one, that bound is delta.
We know that we have a bound on x plus one; we need a bound
on the absolute value of x minus four.
So now we need to do our number line.
So delta equals one puts us where?
Between -2 and zero.
So the absolute value of x minus four would be what?
-2 minus four is -6, absolute value of -6 is six.
Absolute value of x minus four would be four.
So the absolute value of x minus four is smaller than six,
and that's the bound that we needed.
So this is smaller than six times and what's the bound
on the absolute value of x plus one?
Delta. So that's six times what was delta was smaller than?
Well delta is the smaller of the two,
one or our unknown quantity there and that needs
to be epsilon over six because we need
that product to be epsilon.
So that's the unknown quantity that goes in our box.
So as long as delta was the smaller of those two then
if x is within delta of -1, that guarantees that f of x will be
within epsilon of the limit of four.
Which guarantees that the limit as x approaches -1
of our function is four.
Alright we know the limit's four, so we're going
to do this proof again from a different formatting approach.
For every epsilon greater than zero let delta equal the minimum
of the set that contains one and an unknown quantity such that
if zero is less than absolute value of x minus -1 is less
than delta, then and we go down to the bottom of the proof.
Absolute value of f of x minus L is less than epsilon.
Our function is x squared minus 3x.
And our limit's four.
We factor that, being careful to put each factor
in its own absolute value.
We have a bound on the absolute value x plus one, that's delta.
We need a bound on the absolute value of x minus four.
So now we go to our number line.
So delta equal one to the left, and delta equal one
to the right, that's our radius.
So we need the absolute value of x minus four.
So that would be -2 minus four, that's the absolute value
of -6, which gives us six.
We're substituting in zero
so we get the absolute value of -4, which is four.
So the absolute value of x minus four is smaller than six.
The absolute value of x plus one is smaller than delta.
We need this product; we need this product to be smaller
than epsilon which tells us that what needs to go
in the box is epsilon over six.
In other words the other value for delta is epsilon over six.
That completes our proof after we write our conclusion,
the limit as x approaches -1
of our function x squared minus 3x is four.
So as long as delta is the minimum of the set one
and epsilon over six, if x is within delta of -1,
that forces f of x to be within epsilon of the limit four.
Here's some additional problems
that I encourage you to try on your own.
Pause the video, write out the proofs following your
instructor's format and then restart the video and check
with the delta values that I have at the end of the video.
Here are the delta values for the extra practice problems.
Although the correct answer is a well written proof
that matches the format laid out by your instructor.