nananana...
Last example.
Find the limit, then use the Epsilon Delta definition to prove the limit is actually
L.
This time we have an absolute function.
We are going to find the limit as x approaches negative seven of the absolute value of 2x-14.
So where is my chalk at?
Here we go, I have lots of chalk but the one in the holder.
We are going to let x approach negative seven and that means that we are looking at the
absolute value of 2 times negative seven minus fourteen.
That is going to be the absolute value of negative 28 which is equal to 28 because the
absolute value function tells you how far a number is away from zero on the number line...even
though we just always say it gets rid of the negative.
Now, one thing to take a note of as we do our proof is to pay attention to the idea
that when we let x approach negative seven, the absolute value function had to change
the sign of our answer because...well 2 times negative 7 minus 14 came out to be negative
28 so we had to change the sign to get the limit.
We see the graph is here, here is negative seven, and we are approaching that y value...that
limit of 28.
So, again we are going to say given that epsilon is greater than zero, that the absolute value
of f(x)-L is less than epsilon.
Again, assuming that to be true as we work out the relationship between delta and epsilon,
so one last time this is scratch work this is not the beginning of the proof.
So, the absolute value, our function, is again the absolute value of 2x-14 minus the limit
that we found which was 28 is less than epsilon.
Ok, well we have an absolute value inside an absolute value and how are we going to
get this expression to look like...oh, yes.
Let's change these notes.
I really meant to write x minus c there, but I was going to change it into terms of this
problem any way.
Find delta great than zero such that the absolute value of f(x) minus 28, the limit that we
found it to be and we are tying to prove that it is, is less than epsilon whenever 0 is
less than the absolute value of x minus negative seven, or effectively just x+7 is less than
delta.
How do we get this expression with an absolute value symbol in it to look like simply x+7?
Well this is just right there already where we need to take into account the idea of what
happened, what did the absolute value symbol do when we tried to find the limit as x approached
negative seven.
It changed the sign and made negative 28 be basically positive 28.
What math notation do we use, or what math operation can we use to change the sign other
than just...
Well, it is not the same as the absolute value function of course, but if you want to make
a negative number into a positive you multiply it by negative one.
So, understanding that as x approaches negative seven that this is going to be a negative
answer, that 2x-14 is going to be less than zero, that absolute value is going to change
the sign of that answer.
We can replace this set of absolute value symbols with a set of negative parenthesis.
Not all the time, just for this particular problem with x approaching negative seven.
We have the absolute value of the negative parenthesis 2x minus 14 minus 28 is less than
epsilon.
Distribute the negative through and we get the absolute value of negative 2x, that is
going to be negative times negative is positive...I am not sure why I am pausing there...
plus 14 minus 28 is less than epsilon.
Now we have the absolute value of negative 2x minus 14 is less than epsilon.
This is going to be the absolute value of negative 2 times the absolute value of x +7,
because negative two divided by negative two is one and negative fourteen divided by negative
two is seven, is less than epsilon.
This is going to be the absolute value of negative two is two.
Bring that over to the other side, we have...I want to make sure that I don't run out of
room here... the absolute value of x+7 is less than epsilon over 2.
So we have basically x plus seven is less than delta, we have x plus seven is less than
epsilon/2, really understanding how to deal with that absolute value symbol.
Now if we are finding the limit as x approached...I don't know...five or six or seven, or some
where to the right of that sharp bend where the absolute value function did not have to
change the sign of that value, then we would just drop those absolute values all together
but in this case we needed those negative parenthesis.
Dealing with that we get right off the bat a nice direct relationship between delta and
epsilon.
This is a bit easier that it was in our previous example dealing with the quadratics.
We are going to let delta, not d, but delta equal epsilon over two.
Now we are ready to start our proof.
So, the proof is for every, you know what I know I am going to have to cram this in
here, for every epsilon greater than zero let delta equal epsilon over two and zero
is less than the absolute value of x - (-7) which is less than delta.
I have this statement up here, but I really don't want to have this start to flow and
I hate proofs that go all over the place.
So let me step off, erase all of this and clean this up, and put this proof statement
at the top of the board.
That is better.
I wanted to have all of this on the board so you could pause it at the end and see everything
together.
Again our proof, for every epsilon greater than zero let delta equal epsilon over two,
and zero be less than the absolute value of x+7 which is then less than delta.
With those conditions again we are trying to whenever these conditions are met, that
the absolute value...we are trying to prove the absolute value of f(x)-L is less than
epsilon.
Let's start off with that one last time absolute value of f(x) minus L. That is equal to absolute
value of the absolute value of 2x-14 minus our limit that we found to be positive 28.
That is going to be equal to just looking very similar to this work over here, or over
there, the absolute value of negative one time 2x minus 14 minus 28.
That is equal to the absolute value of negative 2x, distributing the negative again positive
14 minus 28 is negative 14.
Factoring out that negative two and writing the absolute value of negative two is not
just completely factoring it out but separating this one absolute value statement into two
absolute value statements.
The absolute value of negative two times the absolute value of x+7, or x minus negative
seven.
Now the absolute value of negative two is equal to positive two and uh of course the
absolute value of x+7, we showed that the absolute value of x+7 was...where is it at...oh
right here... the absolute value of x+7 is less than delta.
So, oh...less than excuse me.
This expression, this absolute of f(x)-L is equal to this, and this is less than the absolute
value if you want to leave that in there for a second, the absolute value of negative two
times delta because |x+7| is less than delta.
So when I replace that absolute of x+7 with delta, well that is less than delta so this
entire expression is less than the absolute value of negative two times delta.
Ok, so now at any rate, the absolute value of negative two is of course just two, and
delta we found that direct relationship between delta and epsilon to be delta is equal to
epsilon over two.
Like before thankfully these values of ten or in this case of two, or whatever it was
in the previous example, we are canceling everything out until we get just an epsilon
and we are right back to thankfully again finishing our proof and validating the statement
that we found that the limit as x approaches negative seven of the absolute value of 2x-14
is 28 because with these conditions in this proof we have that the absolute value of f(x)
minus L is indeed less than epsilon when we have the zero is less than the absolute value
of x plus seven which is less than delta and than delta has this relationship with epsilon
which is it is going to be half of whatever epsilon is.
By definition, we have the limit...I am just going to write what I said a second ago...as
x approaches negative seven of the absolute value of 2x-14 is indeed equal to 28.
That is the end of my last third example.
I am Mr. Tarrou.
BAM!!!
Go Do Your Homework:)