nananana..first example we're gonna find the limit
L, then we're going to use the epsilon delta definition to prove that the limit
actually is L
and here's our first example. That linear function y is equal to 3x plus eight.
We're gonna find the limit X approaches -2
of 3x plus eight and then prove that that answer that we get actually
is the limit. Well we have just a line with the y intercept of 8 and a slope of 3.
We can look at the graph
easily see that the limit as X approaches -2
looks like it's gonna be positive two. We can do this numerically by simply
plugging
in -2 and getting again that the limit is equal to two.
But again stressing the idea that a functions limit is not necessarily the
same thing as the functions value.
That this is a piecewise function, maybe could plug this value to C into the
function to get an answer that really wasn't the limit or
maybe the functions existing at C when the limit doesn't exist, or maybe there
was a hole. I am pointing where the graph used to be
where the function doesn't exist but limit does. So,
we want to approve that the limit actually is equal to two.
Now how the heck we gonna do that? With all the Delta Epsilon proofs that we're
gonna work on,
these are the steps you are going to want to do. You wanna take
the absolute value of f(x) minus L and,
which is less than we're gonna be assuming initially that its
less than epsilon, and rewrite it to match the notation
of zero is less than the absolute of x minus C
minus Delta. Again this being greater than
zero is just the emphasis that X and C are not the same value X is approaching. So
really you're gonna see me kinda
emphasizing the right hand side here about the X minus C being less than Delta.
When we get this f(x) minus L
expression to match this X minus c notation,
that is going to allow us to create a direct relationship between
epsilon delta.
In that case you know in this line here,
you're gonna see a relationship between epsilon delta that because it was just
simply a linear function,
you can kinda see that relationship to some degree. You will it be in the
example
right there inside the linear function. Then
on to actually do the proof, see there's no way that step one can be the proof
because we're trying to prove
that well we're gonna find delta greater than 0 such that
f(x) minus L, the absolute value of that is
epsilon whenever basically the absolute value
of X minus c, I don't know why I saying basically, but is greater than 0 but really
emphasizing the absolute value of x minus c
is less and Deltas. So as long as we give an
X that is with in Delta, of
c then we're going to get a function value
that is within epsilon of the limit
right here. You know that that distance between the function's value and the
limit is less than the
epsilon, that you gave me that I gave you a delta that matched, and then we get this
function's value that's within that epsilon. Getting the whole
infinitely close idea for limits. Okay, so
we're trying to prove this part, we're trying to prove that whenever
the absolute value of x-c is less than Delta that the absolute value of
f(x) minus L is less than epsilon. We're trying to prove that so,
in step one where you see me start this work and I'm saying f(x)-L,
the absolute value of that is less than epsilon,
I'm using an assumption at the beginning of this work
that I'm trying to prove. See you gotta know that step one is not part of the
proof, that scratch work just to develop this relationship between Delta and
epsilon.
Then we actually start the proof by saying we're gonna let the absolute value
of x-c be less than that Delta that we
get through like an educated guess from Step one and
prove that the absolute value of f(x) minus
L is less than epsilon.
With the finish up that that statement that the absolute value of f(x) minus
L is less than epsilon, then that has
created our...you know that's finished our proof. That sayin
again, one more time, you give me a arbitrary distance
an arbitrarily small distance from the limit and I will give you
a matching delta's such that if we pick an x thats closer to c
than Delta then I'll move even closer to the limit than
...you know this initial arbitrary
value arbitrarily small value of epsilon.
So let's get that proof done. Okay so we are going to
what we're going to... given the fact that epsilon is greater than 0,
we're gonna start of with that absolute value
of f(x) minus L is less than
epsilon. Well
our function is 3x plus 8
the limit that we found when we let
X approach negative two was equal to
two. I think what I'm gonna do as well,
this is the goal %uh that we want to do, but let's put some notation from the
actual problem so it's not quite so generic.
The limit that we did find was again equal two
and the c that were concerned about
in this particular problem is, I'm not sure gonna
that I am going to fit this in here, is negative two
so you know I'm basically gonna try and get this
expression, this f(x) minus L expression to look like
the absolute value of
f(x) plus two. And if I do
then I will be able, to you know, i mean it absolute value and it's absolute value,
it's less than an epsilon and delta. So I will get that direct relationship
Well, eight minus two
is six, we have the absolute value of
3x plus six is less than
epsilon. We can take a two out of that absolute value. The absolute value
3x plus six is the same as the absolute value of two
times the absolute value of x
plus 2. Wait a minute, I am not taken a 2 I'm taking a 3. I'm thinking of the answers.
So let's just change that. We are factoring a
three out. There we go.
So the absolute value 3x plus 6 is equal to the absolute value three times
the absolute value
of x+2 and of course the absolute value of 3 is just 3. So we are going to move that three over to the
other side and get the absolute value
of x plus 2 is less than
epsilon over three. Well
so I've changed my f(x) minus L
expression to have that
matching x+2 on the inside, so if x+2..the absolute value of x+2
was less than Delta,
and the absolute value of x+2 is less than epsilon
over three, then I can say
that we're going to let...
or that d
is...not d, delta
is equal to epsilon. Epsilon.
There we go, over
three. That really shouldn't be as a shocking revelation
that Delta is gonna be one-third basically of epsilon, because the slope
of this line is 3 for every
one unit along the x-axis and
delta is sort of... it is a horizontal move along the x-axis
from c. That the function, the slope is
three, so for every one unit on the x-axis you move the vertical distance is
gonna change
by a multiple of three. So, if you want to put it in terms of delta, delta
is the horizontal distance along the number line, is one-third of epsilon.
Now you are not always going to be able to see that, because
many of your problems not gonna be linear functions, but there's that relationship.
Now what you wanna do,
this is our scratch work. This is that
direct relationship between Delta and epsilon that we were talking about.
So now,
just make sure to write this the wrong way. You want to actually start the proof
by writing this,
for every Delta
greater than 0
let Delta
equal epsilon over three.
And 0
is less than the absolute value ok x minus
-2, which is less than Delta
or just the idea that but we're going to be
letting the X approach that negative two. So you can write X minus -2 or just simply
X plus two. Now
again our goal is to prove that f(x) minus L
is less than epsilon when you have this relationship
that Delta is one-third, equal to one-third
of epsilon. So we're going to start off with
so we have f(x)
which is going to be, I am pausing cuz I'm
realize I'm not going to have enough room to write all this. f(x) is 3x+8
minus our limit
which is equal to 2.
I am not going to write that this is less than epsilon, I'm trying to prove.
So let's see we got here, we have 8 -2 which is going to be of course six.
And actually I want to pause here for second. let's say that
that this expression is our
absolute value f(x) minus L.
I just had a written there so we have it to reference at the
end of our proof. So
the absolute value of f(x)-L is equal to the absolute value of (3x-8) minus 2.
8 -2 of course is equal to 6. We can
the... Now this is equal to. Now we have our 3x+6. We are going to take out
our factor of three, and get x plus
two. This f(x) minus L is still
equal to that expression of 3x plus 3. But now,
see for every Delta greater than 0, let Delta equal epsilon over three
and the assumption that we're gonna be picking x's, X minus
two is less than Delta.
So I can bring this in here in here and now say that this expression
is less than 3
times delta. Because it says right there,
and absolute value x plus 2 two basically
is less than Delta. Well Delta we have set up to be
equal to 3x
or equal to e over three. So now that
is, now this inequality is gonna be in reference to
our absolute value of f(x) minus L. S that
is going to be less than three...
I hope my next problem does not have 3's and d's together.
3 times delta which we set
we're setting equal to epsilon over three. And of course
you know multiplying by three and divided by three gonna cancel out
and we have that when those can't allow we have less than
epsilon. So we have proven that the absolute value
of f(x) minus L
is less than epsilon when we have this direct relationship between delta which
is equal to one-third epsilon.
That direct relationship we got from our initial scratch work.
Getting our f(x) minus L expression to match that X minus see expression
That's the end of our first proof. So that's saying that as long as
you give me an arbitrarily small distance
epsilon, I can give you a delta which is one-third
the value of epsilon. And then when you give me....
When then when I come up with an x inside that that
that delta which is one-third epsilon
I can give you an X, go to the function
and get a value from that function that is within
the distance of epsilon to
the limit. And that is the end of our first example.
BAM!!!