Getting on to that quadratic example. Find the limit, then use the Epsilon Delta definition
to prove that the limit actually that L, that real limit that you believe it to be. We are
going to find the limit as x approaches five of x squared minus x minus 6. Ok, well let's
just get the limit out of the way. The limit as x approaches five of x squared minus x
minus six, well I kind of gave it away there, is going to be equal to five squared minus
five minus six. Well that is 25 minus 5 minus 6 which is going to be equal to 14. So, we
are going to try and prove that the limit of this expression is 14 when x approaches
5. We are going to that like the last example. The first thing that we need to do is some
scratch work to develop a direct relationship between the variables of epsilon and delta.
We are going to do that again by starting off with this expression here the absolute
value of f(x) minus 14 is less than epsilon and rewrite it into this format where we have
the absolute value of x - c alone and then on the right hand side it is less than something,
and we hopefully get that direct relationship between epsilon and delta like we had in the
last example. We have the absolute value of, oops...I dropped this chalk earlier and now
it is all busted up... the absolute value of our function is x squared minus x minus
6, so there is our f(x), minus the limit that we found to be 14. That probably should have
been L there originally and then fill it in. That is going to be initially in our scratch
work assumed to be less than epsilon. We are really trying to prove that in the end. Combine
like terms. We have the absolute value of x squared minus x minus 20 is less than epsilon.
That left hand side is factorable and when you have a expression inside of an absolute
value symbol that is factorable you can still factor it, just make sure you leave each of
those factors in their own absolute value symbol. We have the absolute value of x minus
5 times the absolute value of x plus 4 is less than epsilon. Trying to isolate that
absolute value of x-5 and manipulate that original f(x)-L statement to match the x-c
expression like we did in the last problem. We did both sides by the absolute value of
x+4. That should be a minus sign. Ok, well we can see now that there is a problem with
this and I forgot my given that epsilon is greater than zero. I will sneak
that in a little bit neater later. We can see that we do not, yes I have the absolute
value of x-5 by itself but less then delta and less than epsilon over the absolute value
of x+4. We don't have a direct relationship between these two variables and we need that
to...you know...do our proof. Oh, man. Ok, how the heck are we going to do that? We need
to remember that we are finding limits. We are trying to let x approach c and in this
case that c value is 5. We are trying to let x approach five. We don't need to be 10 or
15 units away from 5, we want to let x approach 5. To create a direct relationship between
delta and epsilon we need to eliminate this absolute value of x plus 4 or at least rewrite
it in a form that does not have a variable in it. What we want to do, what we need to
do, is set a limit on delta...set a maximum distance from the value of c, some books call
that a by the way, that we are willing to us for the proof...put a condition on it.
Then we want to find the maximum value that this absolute value of x plus 4 is going to
take on. So, this little diagram, I just took my quadratic and put into standard form or
vertex form or whatever your textbooks call it. I have a sketch here. Again, I am just
trying to show you that delta being a horizontal distance away from c. We are going to set
that a maximum of one. I will write all that out and then from that idea of making sure
that the absolute of value of x minus 5 does not take on or compare to a delta that is
greater than one, get some kind of relationship...some kind of maximum value that the absolute value
of x plus 4 will take on. So just writing out what I said, we need that direct relationship
between epsilon and delta. Now, if we do maximize that delta or allow it to be only a maximum
of one or smaller, we have delta is less than or equal to one. We are going to again whenever
the absolute value of x-5 is less than delta from that statement, we have the absolute
value of x-5 is less than delta and then setting that to be a maximum of one, we have the absolute
value of x-5 is less than one and I wrote it as a compound inequality because obviously
of a lack of room. I am trying to get all this example up here at once. We need to rewrite
that x-5 expression into the form of x plus 4 again to see the maximum value, to find
the maximum, value that will take on. So we add each of these three parts of our compound
inequality by 9, getting this x plus 4 that we need. While we can see that it is greater
than 8, we are tying to find the maximum value that x plus 4 can take on and that is going
to be 10. So, from that we have the absolute value of x plus 4 is going to be less than
10. So, from this I can come over here and write one more line. The absolute value of
x-5 is less than epsilon over 10. So now from here it would seem that we could let delta
equal epsilon over 10. But for the rest of this proof is it really acceptable to say
that delta is equal to epsilon over 10? This not as simple of a direct relationship as
it was with the linear function of course. I had to work on a limitation on delta to
get that direct relationship. Because of that, we are going to say that delta is going to
be equal to minimum of one and epsilon over 10. Again, with this extra condition now put
on this what I am going to say is my direct relationship between delta and epsilon, is
it going to be acceptable to replace any time I see a delta with epsilon over 10. Well,
if delta is equal to epsilon over 10, then based on this condition that means that epsilon
over ten must be less than one. Since delta is the minimum of these two values we can
go ahead and say that we have got with the idea starting off with the absolute value
of x-5 has to be...again whenever that is less than epsilon...then the absolute value
of x-5 is going to be less than epsilon over 10. Close up that absolute value symbol. Ok,
let's keep that in the back of our minds as we work through our proof. What if we have
the condition where delta is equal to one? That epsilon is given is such a manor that
has this fraction be larger than one. Again, we only came up with that expression, that
direct relationship delta is epsilon over 10 with the condition that delta can be no
more than one. So, if delta comes out...if we have to set delta equal to one, well then
that means that one had to have been less than the value of epsilon over ten. Ok, so
again with this idea that starting off with find delta greater than zero such that the
absolute value of f(x) minus L or in this case 14 is less than epsilon whenever the
absolute value of x-5 is less then delta. If we have the absolute value of x-5 is less
then delta which we are saying is one, we are saying that it is one because the epsilon
was such that this fraction came out to be bigger than one, well I would still like to
be able to know that when I see an absolute value of x-5 that I can relate it to this
epsilon over 10 idea. Well, if I picked one then one was less than epsilon over 10. If
the absolute value of x-5 is less than one, because we had to set delta equal to one because
it was smaller than e/10, well then it is also less than epsilon over 10. I just said
e. That means that we can clean this up and say that if the absolute value of x-5 is less
than one which is less than epsilon over 10, then it is just simply the absolute value
of x-5 is less than epsilon over 10. So, I can still write that statement that letting
delta be equal to epsilon over 10, that direct relationship that we got by setting delta
to be a maximum of one...has to be less than or equal to one, either way whether the delta
is going to be set equal to epsilon over 10 or if delta is equal to one, with that condition
delta is the minimum of 1 or e/10 it is still acceptable to write the same statement that
the absolute value of x-5 is less than e/10. Now we are ready to start our proof. So we
are officially starting the proof now that all of this scratch work is done. For every
epsilon greater than zero, let delta equal the minimum of one or epsilon over 10 and
let zero be less than the absolute value of x-5 which is less than delta. Now with these
conditions we are trying to prove that the absolute value of f(x)-L is less than epsilon.
We have the absolute value of again f(x) minus the limit L is equal to the function again
is x squared minus x minus 6. Our limit that we found when we plugged in 5 came out to
be 14. Can I manipulate this expression and show that ultimately this is less than epsilon.
So this is going to be, just like the before, the absolute value of x squared minus x minus
20. This is equal to the absolute value of x-5 times the absolute value of x+4. Now we
start working with these relationships that we just came up with. We were able to show
or actually not show but we have the initial assumption that the absolute value of x-5
has to be less than delta. You know, we are trying to prove that the absolute value of
f(x) minus L is less than epsilon whenever this is true. This now, this is going to be
come...I don't have room to right it over there, less than delta and the... I don't
know why I am using parenthesis... absolute value of x minus four. We have from our scratch
work over here that the absolute value of x-4 is less then ten. Ok, well we are trying
to prove that this is less than epsilon so we need to take the delta out and replace
it with an expression that has an epsilon in it. We just happen to have it right there
and we were able to validate with our check or our work here. Working with this extra
condition that we created which was that delta can be no more than one, that every time that
we see an absolute value of x-5 yes that is less than delta but we also showed that it
is less than epsilon over ten which we set to be equal to delta. Now we can see that
the 10's are going to cancel out. Bringing this down now, we have the absolute value
of f(x) minus L is indeed shown with these conditions that it is less than epsilon. So,
by definition so therefore, or box, or whatever your book is using by definition
the limit as x approaches five of x squared minus x minus 6 is indeed equal to 14. That
is the end our second example. nanananana... Now before we move on to our last example
which is going to be an absolute value function. I wanted to put some concrete arithmetic,
some examples that go along this proof. Not validating the proof is true because you can't
do a proof with a just bunch of examples. But, show you some examples on how they relate
to what we said in this proof. That if delta is the minimum of 1 and epsilon over 10, and
we are looking for x approaching 5 or letting the absolute value of x-5 be less than delta,
we showed that the absolute value of f(x) minus L is indeed less than epsilon. But,
that idea of you give me an arbitrarily small epsilon and from that I will develop a delta
such that if x is within delta of c, that when you plug that x into the original function
you get a difference between the functions value and the limit which is less than epsilon.
So let's let epsilon be equal to 20. If that is the case, then delta is the minimum between
and 1 and epsilon over 10. If our vertical distance from the limit is set to be 20, then
20 over 10 is going to be equal to 2. So, in that case your epsilon over 10 expression
is indeed bigger than 1...it is 2. But we are setting delta to be the minimum of 1 and
e/10. So we are going to let delta be equal to 1. In that case, delta being the horizontal
distance from c or a however your book is denoting that, and we are finding the limit
as x approaches 5 then c is 5 and c minus delta is 4 and c plus delta is 6. So, picking
an x that is within delta of c let's let x be 4.5. If we let x be 4.5, f(x)...f(4.5)
is 9.75. The absolute of f(x)-L is it indeed less than epsilon? Well, is the absolute value
of 9.5 minus 14 less than 20? Yes it is. Now, if we pick an epsilon value, again that vertical
distance from our limit, that is 1...that is smaller. If we let epsilon be 1, then we
have again delta is the minimum between 1 and epsilon over 10. Well, if I say epsilon
is equal to delta is the minimum of 1 and 1/10. Since delta is the minimum of one of
these two values, delta...I keep saying d all of a sudden...delta is going to be set
equal to 1/10. Again we are finding the limit as x approaches 5 and moving to the left and
the right of five by a delta of 1/10, we get c minus delta is 4.9 and c plus delta is 5.1.
We need to pick an x that is within delta of c or that c is within delta units of x,
so we are going to let x be 5.05... somewhere within this interval. Well if you take 5.05
and you plug into your function you get a value of 14.4525. Indeed, or with this example,
is the absolute value of f(x) minus L less than epsilon. Well f(x) again, just said it
from the calculator, 14.4525. The limit is equal to 14. The absolute value of this comes
out to be .4525 and that is less than our epsilon, our given epsilon, or our arbitrary
epsilon of 1. So, there is some arithmetic examples go along with what exactly these
proofs are telling us and how they work. BAM!!!