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PROFESSOR: Hi.
Welcome, once again, to another
lecture on limits.
Actually, from a certain point
of view, today's lecture will
be the same as the last
lecture, only
from a different viewpoint.
Our lecture today is
called 'Limits: a
More Rigorous Approach'.
And what our objective for the
day is, aside from helping you
gain experience and facility
with using limit expressions
and absolute values and the
like, is to have you see how
we can use the power of
objective, well-defined
mathematical definitions to
find rather easy ways of
solving certain types
of problems.
Now to this end, let's very
briefly review our fundamental
definition of last time.
Namely, the limit of 'f of x'
as 'x' approaches 'a' equals
'l' means that for each epsilon
greater than 0, we can
find delta greater than 0,
such that whenever the
absolute value of 'x minus a' is
less than delta but greater
than 0, then the absolute value
of ''f of x' minus 'l''
will be less than epsilon.
To state that once again, but
in more intuitive terms, for
any tolerance limit epsilon at
all, we can suitably find
another tolerance limit, delta,
such that whenever we
make 'x' within delta of 'a', 'f
of x' will automatically be
within epsilon of 'l'.
And again, the very, very
important emphasis here, we do
not allow 'x' to equal 'a'.
Again, in terms of a diagram,
if this is the curve 'y'
equals 'f of x', this
is 'x' equals 'a'.
This is 'l'.
If we call this 'l' plus
epsilon, if we call this 'l'
minus epsilon-- in other words,
epsilon is this width
over here--
then the way we find delta is to
reflect back to the curve,
emphasizing, again, in
the neighborhood
of the point 'a'--
I can't stress that point
strongly enough, that if this
curve were not 1:1, there are
going to be other places where
this line meets the curve.
So if that happens, you must
make sure that you pick the
neighborhood of 'a', not some
other point over here.
We're interested in what
happens near 'a'.
But at any rate, again notice
that the fact that these two
intervals here were equal does
not guarantee that when you
project down here, they
will be equal.
In fact, the only time that
these two widths would be
equal is if the curve happened
to be a straight line.
And what it means, again, is
that the delta that we're
talking about, for example,
is the minimum
of these two widths.
In other words, as I've drawn
this diagram, delta would be
the distance from 'a'
to this end point.
And all we're saying is that
once 'x' is in this open
interval but not including 'a'
itself, 'f of x' will be in
this open interval.
And again, notice, as we were
emphasizing last time, once
this delta happens to work,
automatically any smaller
delta will also work.
In other words, if something
is true for everything in
here, it's certainly true
for everything in some
subinterval of this.
What you must be careful
about is not to
reverse this process.
Don't get outside and pick
bigger deltas, then you might
very well be in a little
bit of trouble.
At any rate, once we've reviewed
what the basic
definition is, it seems about
the only way to show what
mathematics is all about is to
actually do a few theorems,
that is, derive a few
inescapable consequences of
the definition that show how
our theorems coincide with
what we believe to be
intuitively true
in the first place.
And for obvious reasons, we
should start with what
hopefully would be the simplest
possible theorems and
then proceed to tougher
ones as we go along.
So as my first one, I've
chosen the following.
The limit of 'c' as 'x'
approaches 'a' is 'c', where
'c' is any constant.
Now again, that may look a
little bit strange to you.
Let's look at it this way.
What I'm saying is let
'f of x' equal 'c',
where 'c' is a constant.
Then what we're saying is for
this choice of 'f of x', the
limit of 'f of x' as 'x'
approaches 'a' is 'c' itself.
That's what this thing says.
How do we prove this?
Well, you see, we have
a criteria given.
Namely, what is our
basic definition?
Let's just juxtaposition our
basic definition with this
particular result.
To prove that the limit here
is 'c', notice that in this
particular problem, if we come
back and compare this with our
basic definition, notice that
we have the same basic
definition as before, only the
role of 'f of x' is played by
'c' and 'l' is also
played by 'c'.
In other words, in this
particular problem 'f of x' is
'c' and 'l' is also 'c'.
Now what must we do?
We must show that for each
epsilon greater than 0, or for
an arbitrary epsilon
greater than 0--
now what does it mean to say
an arbitrary epsilon?
In a way, think of it as being
a game, a battle of wits
between you and your worst
enemy, and you're out to win
and your worst enemy
is out to beat you.
So to make this as difficult a
game as possible, you allow
your worst enemy to choose the
epsilon, provided only that
it's positive.
For whichever one he picks, you
must be able to find the
delta that matches that epsilon,
such that what?
Whenever 'x' is within delta of
'a' but not equal to 'a',
'f of x' will be within
epsilon of 'l'.
So let's write that
down over here.
Let's write what that says.
Given epsilon greater than 0,
we must find a delta greater
than 0, such that what?
Well, by definition.
Such that when the absolute
value of 'x minus a' is less
than delta but greater
than 0, the absolute
value of 'f of x'--
well, that of course, is
'c' in this case--
minus 'l', which is also 'c' in
this case, the 'l' stands
for the limit, has to be
less than epsilon.
And lo and behold, we find that
this is a rather simple
procedure because what
is 'c' minus 'c'?
'c' minus 'c' is 0, and
automatically that will be
smaller than any positive
epsilon.
In other words, what this says
is even your worst enemy can't
give you a tough time
with this problem.
Namely, no matter what epsilon
he prescribes, no matter how
sensitive, you say, well, for
delta I'll pick anything.
And it works.
And why does that work?
Well, here again we
can emphasize
the geometric approach.
Namely, if we plot the graph 'f
of x' equals 'c', observe
that that plot equals a straight
line, 'y' equals 'c'.
Now take 'x' equals
'a' over here.
What is 'f of a'?
'f of a' is 'c'.
Now pick an epsilon, which is
this half-width over here,
knock that off on either
side of 'c'.
And all you're saying in this
particular case is no matter
how far away 'x' is from 'a', 'f
of x' will be within these
tolerance limits of 'c'.
And why is that?
Because 'f' is defined in such
a way that the output for any
element in its domain
is 'c' itself.
In other words, every element
maps up here.
This is what you mean by saying
that the graph is the
straight line 'y' equals 'c'.
By the way, another
word of caution.
We must always make sure that
your solution does not depend
on the diagram.
You see, if a person were to
look at this diagram very
quickly, he would assume that
'c' had to be a positive
constant here, because look at
how I've drawn the line 'y'
equals 'c'.
It's above the x-axis.
Well, 'c' could just as easily
be a negative constant.
And if I drew the diagram
that way, 'c' would
be below the x-axis.
The important thing
to check is this.
When you draw a diagram, you
can't have a certain number
being positive and negative at
the same time, so you choose
it one way or the other.
Always make sure when you do
this that your formal proof,
your analytic proof, goes
through word for word if you
reverse the signature of the
sign of the number that you're
working with.
Make sure that your answer does
not depend on the picture
that you've drawn.
Make sure that your answer
follows, inescapably, from
your basic definition.
And notice that this is
what we did here.
We showed what?
That no matter what epsilon we
were given, we could find a
delta-- in fact, in this
case, any delta--
such that when 'x' was within
delta of a as long as 'x'
wasn't equal to 'a'.
Well in this case, even
if 'x' equaled 'a',
there was no harm done.
But the point is we want to keep
away from a 0 over 0 form
so we always impose
this condition.
In this case, once 'x' was
within delta of 'a', 'f of x'
was automatically within epsilon
of 'c', as long as
epsilon was positive,
because 'f of x' was
already equal to 'c'.
The difference was already 0.
Now again, notice what
happens here.
The more pragmatic student will
say, why did you use this
long math when it was obvious
from either the diagram or
from intuition that this
is the correct answer?
And the reason, is as we have
already seen and as we will
see many, many times during
our course, the intuitive
answer and the correct
answer will not
necessarily be the same.
The point is we always want to
make sure that when we prove
something, the proof follows
from the assumed definitions
in a logically rigorous way.
If it happens once you prove
this that you can intuitively
recognize the same result,
that's like a double reward
because now you won't have to
memorize what the result was,
you'll just use this thing
automatically.
But the beauty is that for
somebody who doesn't have the
same intuitive insights that you
have, if he says it's not
self-evident to me, explain
to me what's happening.
Then, you see, once he's
accepted the basic definition
and assuming that he knows the
rules of mathematics, he has
to come up with the same
answer that you did.
And this is what we mean by an
objective criterion for doing
mathematics, okay.
Well, this was kind
of an easy one.
Let's do another kind of
easy one that's harder.
Let's make some gradual
transitions, here.
Let's pick another theorem.
Here's another one that sounds
pretty self-evident.
The limit as 'x' approaches
'a' is 'a'.
In fact, what that seems to say
in a self-evident way is
that as 'x' gets arbitrarily
close to 'a', 'x' gets
arbitrarily close
to equal to 'a'.
And if anything is a truism,
I guess that's it.
So we certainly suspect that
this is a true statement.
All I'm trying to get you used
to is not to make a mountain
out of a mole hill, not to have
to think that mathematics
becomes a severe thing where
we try to find hard ways of
doing easy things, but rather
that we can find an
unambiguous, logically
constructed language from
which all of our results can be
proven without recourse to
intuition once our basic
definitions are chosen.
So let's see how this
would work.
Let's again go back to
our basic definition.
You see, again, what we're
saying here is that this is
just a special case now where
'f of x' equals 'x'.
You see the function
is 'f of x'.
In this case, it's 'x', and this
is a special case where
'a' is 'l'.
Now what are you saying here?
You're saying given epsilon
greater than 0, we must find a
delta greater than 0,
such that what?
Such that whenever 'x' is within
delta of 'a' but not
equal to 'a', then 'x' will
be within epsilon of 'a'.
And now you just look at this
thing, I hope, and you say
well, there's sort of a
similarity over here.
How close should 'x' be chosen
to 'a' if you want 'x' to be
within epsilon of 'a'?
And the answer, quite obviously
in this case
without, cause to define again,
is to simply say what
in this case?
For the given epsilon,
choose, for example,
delta to equal epsilon.
Because certainly, if the
absolute value of 'x minus a'
is less than epsilon but greater
than 0, then certainly
the absolute value of 'x minus
a' is less than epsilon.
Don't be upset that the proof
happened to be fairly
easy in this case.
More importantly, don't be upset
that there was a more
intuitive way of doing it.
Remember what we want to do is
to get these rigorous ways
down pat, interpret them in
terms of pictures wherever the
pictures are available.
And then, when we get to the
case where pictures aren't
available, to be able to extend
the analytic concepts.
Again, to see what happens here
pictorially, you notice
that if you start with the
function 'f of x' equals 'x,'
its graph is a straight
line 'y' equals 'x'.
And I'll risk some freehand
drawing here.
So in other words, we suspect
that this is 'a'.
It better be, if this says that
all points on this line
have the property that the
y-coordinate is equal to the
x-coordinate.
Now what do we want to do?
We pick an epsilon and
knock off a plus
epsilon and a minus epsilon.
And now what we want to find out
is how close must x be to
a on the x-axis in order that 'f
of x'-- namely 'y' in this
case, or 'x' itself--
be within this prescribed
tolerance limits of 'a'?
You see what happens here?
This is that one case where
it happens that what?
If you draw these lines over
and project down, but by
proportional parts--
and this is crucial here.
Even if this weren't the 45
degree line, the fact that
these two pieces here are equal
would guarantee that
these two pieces here are equal,
in spite of how I've
drawn this.
The fact that this is the 45
degree line not only says that
these two pieces here are
equal but it says what?
That each of these pieces is
equal to each of these pieces.
I wish I had drawn this better
for you, but in fact the worse
I draw it, the more you have
to rely on being able to
visualize abstractly what's
happening here.
What I'm saying is that this
point here would be labeled 'a
plus epsilon', this point here
would be labeled 'a minus
epsilon', and this is the
pictorial version of what it
means to say you could have
chosen delta to equal epsilon,
in this particular case.
Well, that's enough of the
easier ones, so let's pick one
that gets slightly tougher.
And this one gets tougher in one
sense, but insidiously--
meaning it's real sneaky--
simple in another sense.
In other words, it turns out
that one can reason falsely
and get the right answer
just by a quirk.
You see, what I want to prove
next is a very important
theorem that says that the limit
of a sum is equal to the
sum of the limits.
Written out more formally, if I
have two functions 'f of x'
and 'g of x' and formed the sum
''f of x' plus 'g of x''
and I want to take the limit of
that sum as 'x' approaches
'a', what this theorem says is
you can find the limit of each
of the functions separately
first, and
then add the two results.
Now at first glance, you might
be tempted to say, well, what
else would you expect
to happen?
The answer is, I don't know,
again, what else you would
expect to happen.
But this is a luxury.
You see, evidently what's
happened here is that we've
reversed the order
of operations.
You see, this says what?
First add these two, and
then take the limit.
What are we doing down here?
First we're taking the limits
and then we're adding.
Now, is it self-evident that
just by changing the order of
operations you don't
change anything?
We've seen many examples already
in the short time that
this course has been in
existence where changing the
order, changing the voice
inflection, what have you,
changes the answer.
And in fact, we're going
to see more drastic
examples later on.
I guess this is one of
the tragedies of
a course like this.
I guess it's typical of
problems every place.
If the place that the person is
going to get into trouble
comes far beyond the time at
which you're lecturing to him,
it's kind of empty to threaten
him with the trouble he's
going to get into.
So I'm not going to threaten
you with the trouble you're
going to get into until
we get into it.
All I'm going to say is be
careful when you say that it's
self-evident that we can first
add and then take the limit or
whether we first take the
limits and then add.
In general, it does make a
difference in which order you
perform operations.
Well, let's take this just a
little bit more formally to
see what this thing says.
First of all, when we write
something like this, we assume
that the limit of 'f of x' as
'x' approaches 'a' exists,
otherwise we wouldn't
write this thing.
So let's call that limit 'l1'.
In other words, let the limit of
'f of x' as 'x' approaches
'a' equal 'l1'.
Let the limit of 'g of x' as 'x'
approaches 'a' equal 'l2'.
Now define a new function 'h of
x' to be equal to ''f of x'
plus 'g of x''.
And by the way, this is
something I didn't say
strongly enough in one of our
early lectures, and I want to
make sure that it's very clear
that this is understood.
And that is, notice that when
you define 'h' to be the sum
of 'f' and 'g', you had better
make sure that 'f' and 'g'
have the same domain.
You see, if some number 'x'
belongs to the domain of 'f'
but it doesn't belong to the
domain of 'g', then how can
you form ''f of x' plus
'g of x''? 'g'
doesn't operate on 'x'.
By the way, this is not quite
as serious a problem as it
seems if you understand the
language of our new
mathematics and sets
and the like.
Namely, if the domain of 'f'
happens to look like this and
the domain of 'g' happens to
look like this, what we do is
we restrict the sum to the
intersection of the two domains.
In other words, referring back
to our function 'h', we define
the domain of 'h' to
be the intersection
of these two domains.
And that way, for any 'x', which
is in the domain of 'h',
it automatically belongs to both
the domain of 'f' and the
domain of 'g'.
And so this becomes
well-defined.
Another way of looking at this
is what you're really saying
is that 'f' and 'g' must
include in their domain
intervals surrounding
'x' equals 'a'.
And since you're only interested
in what's happening
near 'x' equals 'a', you don't
really care whether these have
the same domains or not,
provided they have what?
An intersection that can serve
as a common domain.
But that, I think, is more
clear from the context.
It's a very, very important
fine point.
It's a tragedy to try to add
two numbers, one of which
doesn't exist.
I don't know if it's a tragedy,
it's certainly futile.
At any rate, though, let's see
what this thing then says.
If we now define 'h' to be 'f
plus g', what we want to prove
is that the limit of 'h of x'
as 'x' approaches 'a' equals
'l1 plus l2'.
Now here's the point again.
What does this mean
by definition?
It means that given epsilon
greater than 0, we must be
able to find a delta such that
when 'x' is within delta of
'a' but not equal to 'a',
'h of x' is within
epsilon of 'l1 plus l2'.
That's probably kind of hard
to keep track of, so I've
taken the liberty of writing
this down for
you right over here.
See, given epsilon greater than
0-- so that's given, we
have no control over that--
what we must do is find delta
greater than 0, such that 0
less than the absolute value
of 'x minus a' less than
delta, implies--
now, what is 'h of x'?
It's ''f of x' plus 'g of x'',
and our limit that we're
looking for in this case
is 'l1 plus l2'.
So mathematically, we replace 'h
of x' by ''f of x' plus 'g
of x'', the limit
is 'l1 plus l2'.
So what must we show that the
absolute value of the quantity
of ''f of x' plus 'g of x''
minus the quantity 'l1 plus
l2' is less than epsilon.
And now we start to play
detective again.
This is the expression
that we want to
make less than epsilon.
So what we do is we look at this
particular expression and
we try to see what kind of cute
things we can do with it.
Now, what do I mean
by a cute thing?
Well, we're assuming that the
limit of 'f of x' as 'x'
approaches 'a' equals 'l1'.
Let me write this as
an aside over here.
Among other things, what this
tells us is that we have a
hold on expressions like this.
In other words, the fact that
the limit of 'f of x' as 'x'
approaches 'a' is equal to 'l1'
tells us that we can make
this as small as we want.
I'll use a subscript over here
for epsilon 1 because it
doesn't have to be the same
epsilon that was given here.
For any positive number, say
epsilon 1, the point is I can
make 'f of x' within epsilon 1
of 'l1' just by choosing 'x'
sufficiently close to
'a' by definition
of what limit means.
So in other words, I like
expressions of this form.
And similarly, I like
expressions of this form.
And again, the reason is that
since the limit of 'g of x' as
'x' approaches 'a' is 'l sub 2',
it gives me a hold on the
difference between 'g
of x' and 'l2'.
So again, using the old adage
that hindsight is better than
foresight by a darn site,
knowing exactly what it is I
have to do, I come back here
and try to doctor things up
for myself.
The first thing I observe is
that this can be rewritten.
There's no calculus
in this, notice.
Just plain ordinary algebra,
arithmetic.
This can be rewritten so that
I can group the 'f sub 'f of
x' and 'l1'' together and 'g
of x' and 'l2' together.
In other words, this is indeed
nothing more than 1.
The absolute value of the
quantity of ''f of x' minus
l1' plus the absolute value
of the quantity
''g of x' minus l2'.
Now, the point is since we
already know that the absolute
value of a sum is less than
or equal to the sum of the
absolute values, that
tells me that--
treating this is one number and
this is another number,
the absolute value of a sum is
less than or equal to the sum
of the absolute values.
What this now tells me is I can
say that this, that I'm
trying to get a hold on,
is less than this.
But look at this expression.
This expression is the
absolute value of
''f of x' minus l1'.
And this expression is
the absolute value of
''g of x' minus l2'.
In other words, then, since
I can make 'f of x' as
arbitrarily nearly equal to 'l1'
as I want and 'g of x' as
close to 'l2' as I want just by
choosing 'x' sufficiently
close to 'a', why don't I choose
'x' close enough to 'a'
so each of these will be less
than epsilon over 2?
Now again, this calls
for a little aside.
When one talks about epsilon,
that is 'a',
what shall we say?
A generic name for any number
which exceeds 0.
Or I could have written that
less mystically by just saying
any positive number.
Well, if epsilon is positive,
what can you say
about half of epsilon?
It's also positive.
In other words, if I had chosen
a different epsilon,
say 'epsilon sub 1', equal to
the original epsilon divided
by 2, then I'm guaranteed
what?
That I can get 'f of x' with an
epsilon 2 of 'l1', 'g of x'
with an epsilon 2 of 'l2'.
And now adding these two
together, if this term is less
than epsilon over 2 and this
term is less than epsilon over
2, the whole sum is
less than epsilon.
And it seems now
semi-intuitively--
what do I mean by
semi-intuitively?
Well, this is far from an
intuitive job over here.
It's quite mathematical.
It's rigorous in that sense.
It's intuitive in the sense that
I'm not playing around
with the deltas here.
All I'm saying is look, I can
make this as small is I want,
I could make this as small as
I want, therefore I can make
the sum as small as I want.
And the fancy way of saying
that is I can make it less
than any given epsilon, and
therefore it appears that this
will be the limit.
Using our old adage again of
being able, knowing what we
want, to be able to doctor
things up rigorously, once
we've gone through this it's now
relatively easy to clean
up the details.
In other words, for those
of us who are
mathematically-oriented enough
to say, the way you've proven
this last result is the same
sloppiness that I was used to
seeing in certain types of
engineering proofs where
people were more interested in
the result than with the
answer, I still don't see how
you used the epsilons and the
deltas here.
Let me show you what a simple
step it is to now go from the
semi-rigorous approach to the
completely rigorous approach.
All we do is reword what
we've done before.
In fact, this is true
in most mathematics.
You take a geometry book and
there's a theorem that says
something like if 'a', 'b', 'c',
and 'd' are true, then
'e' is true.
And you learn this proof
quite mechanically.
You sort of memorize it.
Well you know, the man who
proved that theorem didn't, in
general, start out by saying, I
wonder what happens if 'a',
'b', 'c', and 'd' are true.
In general, what he tries to
do is to prove that some
result, like 'e', is true.
And as he's proving it,
he hits pitfalls.
And he says, you know, if I
could only be sure 'a' was
true, I could get over
this pitfall.
And if I could be sure 'b' was
true, I could get over the
second pitfall, et cetera.
And when he makes enough
assumptions to get over all
the pitfalls and he has his
answer, he then we writes down
the answer in the reverse
order from
which he invented it.
Namely, he says what?
Suppose 'a', 'b', 'c',
and 'd' are true.
Let's prove that 'e' is true.
And the student is then robbed
of any attempt to see
intuitively how this whole
thing came about.
As a case in point, let
me show you what
I'm driving at here.
My first exposure to formal
limit proofs was
something like this.
When somebody said prove the
limit of a sum equals the sum
of the limits, something
like this would happen.
The first statement in the book
would say, given epsilon
greater than 0, let epsilon
1 equal epsilon over 2.
Now, I respected my teacher,
I respected the
author of the book.
If he says let epsilon
1 equal epsilon over
2, all right, fine.
We can do that.
And more to the point, it turned
out that the problem
worked if you did that.
The part that bothered me is why
did he say epsilon over 2?
Why not 2 epsilon over 3?
Or epsilon over 5?
Or epsilon over 6,872?
Why epsilon over 2?
And the point was that
he had cheated.
He had already done the problem
that we had over here,
and knowing what he needed, then
came back here and said
let epsilon 1 equal
epsilon over 2.
And notice how this
is going to mimic
everything we said before.
For this choice of epsilon 1, we
can find a delta 1 greater
than 0 such that if the absolute
value of 'x minus a'
is less than delta 1 but greater
than 0, then the
absolute value of ''f of
x' minus l1' is less
than epsilon 1.
Why do we know that?
That's the definition of what it
means to say that the limit
of 'f of x' as 'x' approaches
'a' equals 'l1'.
In a similar way, he says we can
find delta 2 greater than
0, such that whenever the
absolute value of 'x minus a'
is greater than 0 but less than
delta over 2, we can make
the absolute value of ''g of
x' minus 'l sub 2'' and be
less than epsilon 1.
And now comes the
beautiful step.
He says, now that these delta
1 and delta 2 exist
separately, pick delta to be
the minimum of these two.
In other words, if I let delta
equal the minimum of these
two, what does that
guarantee me?
If delta is the minimum of these
two, it guarantees me
that both of these conditions
are met at the same time.
What does that tell me?
It tells me that as soon as the
absolute value of 'x minus
a' is less than delta but
greater than 0, automatically
these two conditions hold.
And that, in turn,
tells me what?
That the absolute value of ''f
of x' minus l1' is less than
epsilon 1 and the absolute value
of ''g of x' minus l2'
is less than epsilon 1.
And now by adding unequals to
equals, that tells me what?
That this plus this is less
than 2 epsilon 1.
But see, using my hindsight, I
picked epsilon to be what?
Epsilon over 2.
So 2 epsilon 1 is just another
way of saying epsilon.
In other words, what this now
implies is that the absolute
value of ''f of x' minus l1'
plus the absolute value of ''g
of x' minus l2' is less
than 2 epsilon 1,
which equals epsilon.
But you see, this
in turn is what?
This is greater than the
absolute value of ''f of x'
minus l1' plus ''g
of x' minus l2'.
And so this was the thing
we wanted to make
smaller than epsilon.
Since this is smaller than
this and this is already
smaller than epsilon, then
this must be smaller than
epsilon too.
Again, notice, this
is rigorous.
But if you understand what's
happening here piece by piece,
you never really have
to memorize a thing.
Let's just take a look back
here to reinforce what I'm
saying here.
Notice what we did.
We started with the answer that
we wanted to show, worked
around to get ahold of things
that we wanted.
We were lucky enough--
and this is true in any
game, for example.
One can plot masterful strategy
and still lose,
there's no guarantee we're
going to win with our
masterful strategy.
But we usually do in this
course, usually.
What we do is we masterfully
come back to this, see what
has to be done.
Knowing what the right answer
has got to be, we come back
here and then formalize it.
Essentially, what we've done
is reverse the steps here.
I guess there is one thing that
bothers many people that
I should make an aside about.
Why do you need a different
delta 1 and delta 2 for the
same epsilon 1?
See, if you're memorizing,
there's a danger that you
won't realize this and
why this happens.
Let me show you in terms of
a picture what this means.
Let this, for the sake of
argument, be the curve 'y'
equals 'f of x'.
And let this be the curve
'y' equals 'g of x'.
All we're saying is this.
That when you prescribe an
epsilon, the same epsilon that
surrounds 'l2', if you have
that surround 'l1'.
Because the curves may have
different slopes.
Look what happens over
here, even as badly
as I've drawn this.
Notice that in epsilon,
neighborhood of 'l2', projects
down into this size neighborhood
around 'a'.
On the other hand, an epsilon
neighborhood of 'l1'.
You get what you pay
for, I guess.
An epsilon, neighborhood of
'l1', projects into a much
smaller region.
And by the way, that's exactly
what we meant.
This is a delta 1,
for example.
This is delta 2.
And when we said pick delta to
be the minimum of delta 1 and
delta 2, all we were saying was
listen, if we guarantee
that 'x' stays in here, then
certainly both of these two
things will be true
at the same time.
Well again, we have many
exercises and reading material
to reinforce these points.
All I want to do with the
lecture is to give you an
overview as to what's happening
so that you see
these things.
And I'm afraid I might cure
you with details if I just
keep hammering home these
rigorous little points.
As I say, I hope you
get the main idea
from what we're doing.
Let me just, for the sake of
argument, try to work with
just one more idea and we'll see
how this works out also.
Let's, for example, try to play
around with the idea that
a companion to the limit of a
sum equals the sum of the
limits would be what?
The limit of a product equals
the product of the limits.
In other words, as before, if
the limit of 'f of x' as 'x'
approaches 'a' is 'l1' and the
limit of 'g of x' as 'x'
approaches 'a' is 'l2', let's
form a new function, which we
could call 'k of x' which
is equal to the
product of 'f' and 'g'.
Again, noticing that the 'f' and
'g' have to have a common
domain here.
You want to show that the limit
of 'f of x' times 'g of
x' is 'l1' times 'l2'.
And this is the hard part of the
course, this is the part
of the course that I don't think
anybody in the world can
really teach.
All one can do is try to expose
the student to ideas
and hope that the student has
the knack of putting these
things together to form
his own repertoire.
The idea is something
like this.
I'll show you alternative
methods and the like.
One way is we want to
get ahold of 'f of
x' times 'g of x'.
Now what do we have a hold on?
It's very clever what
we do here.
We have a hold of ''f of x'
minus l1' and we have a hold
of ''g of x' minus l2'.
So as we so often do in
mathematics, we simply add and
subtract the same thing.
We frequently add on zeroes
in this cute way.
We add and subtract
the same thing.
Notice what we did over here.
We wrote 'f of x' as 'l1' plus
''f of x' minus l1'.
Again, why did we do that?
Because from our definition of
the limit of 'f of x' as 'x'
approaches 'a' equaling 'l1',
we know that we can
control this side.
And the same thing is
true over here.
The fact that the limit of 'g
of x' as 'x' approaches 'a'
equals 'l2' means we have
some control over this.
Now let's just multiply
everything out.
This is what? 'f of
x' times 'g of x'.
I'm going to save myself some
space and keep the board
somewhat symmetric.
When I multiply these terms out,
I'm going to get what?
An 'l1' times 'l2' term over
here as one of my four terms?
Let me already transpose
that one so I kill two
birds with one stone.
One is I keep a little bit
of symmetry in what
I'm going to write.
And secondly, notice that later
on, this is what I want
to get a hold on.
In other words, notice that the
proof of the limit of 'f
of x' times 'g of x' as 'x'
approaches 'a' is 'l1' times
'l2', this is precisely the
expression I must make small.
Our show can be made as small is
I wish just by picking 'x'
sufficiently close to 'a'.
And again, I will not go through
all the details here,
I will simply outline
what we do here.
Namely, let's multiply out
the rest of this thing.
We have what here?
This times this,
which is what?
'l1' times ''g of
x' minus l2'.
Then we have this times this,
which is 'l2' times
''f of x' minus l1'.
And now we have what?
This times this.
That's ''f of x' minus l1' times
''g of x' minus l2'.
Now, the point is this is the
thing that we'd like to make
very small in absolute value.
Well again, the absolute value
of this is equal to the
absolute value of this, which is
less than or equal to-- and
I'm going to go through these
details rather rapidly, and
allow you to fill these
in for yourself-- all
I'm using is what?
That the absolute value of a sum
is less than or equal to
the sum of the absolute
values.
The absolute value of a product
is equal to the
product of the absolute values,
so this is going to be
less than or equal to.
Let's break these things up.
And now, you see what
the key idea is?
'l1' and 'l2' are certain fixed
numbers, fixed numbers.
Notice that because 'g of x' can
be made as close to 'l2'
as I want and 'f of x' can be
made as close to 'l1' as I
want, notice that no matter what
epsilon I'm given, I can
certainly make this as small as
I wish, just by picking 'x'
close enough to 'a'.
For example, for a given
epsilon, how many
terms do I have here?
One, two, three.
To make this whole sum less than
epsilon, it's sufficient
to make each of these three
factors less than
epsilon over 3.
I can make these two less than
epsilon over 3 pretty easy.
How do I make this less
than epsilon over 3?
And the answer is, if you want
this times this to be less
than epsilon over 3 where these
are positive numbers,
make each of these less than
the square root of
epsilon over 3.
Then when you multiply these two
together, if this is less
than this and this is less than
this, this times this
will be less than
epsilon over 3.
And now what you see what
we do is very simple.
To finish this proof off, all we
do now is say, let epsilon
greater than 0 be given.
Choose epsilon 1 to equal
epsilon over 3.
Choose epsilon 2 to equal
epsilon over 3.
Choose epsilon sub 3 and epsilon
sub 4 to each be the
square root of epsilon over 3.
Then we can find delta
1, delta 2, delta
3, delta 4, et cetera.
Clean up all of these, you see,
and pick delta to be the
minimum of the four
deltas involved.
Again, this is done much more
explicitly both in the text
and in the exercises.
I just wanted you to get
an overview here.
And by the way, while we're
speaking of this, there's
always the danger that some
of you may respect the
professor too much.
So that danger is one I don't
mind living with.
The problem is this.
You may get the idea out of
respect for me that I have
invented a unique proof here.
Let me tell you this.
One, I did not invent
this proof.
Two, it is not unique.
There are many different ways
of proving the same result.
For example, a person being
told to work on this--
and I'm not going to carry
the details out here--
but a person being told to work
on something like this
might have decided that instead
of doing the clever
thing that we did before, he
was going to do the clever
thing of adding and subtracting
'l1' times 'g of x'.
Because you see, if he did
that, what would happen?
He could now factor out a 'g
of x' from here and rewrite
this as 'g of x' times
''f of x' minus l1'.
And these two terms could be
combined together, we factor
out an 'l1' times what?
''g of x' minus 'l2'.
And even though the details
would have been considerably
different, the intuitive
approach would have been--
well, look.
This is pretty close to 'l2'
when 'x' is near 'a'.
This I can make as
small as I want.
Similarly, I can do the same
things over here, and pretty
soon you've got the idea that
you can make this sum as small
as you want just by choosing
these sufficiently small.
And there's no unique
way of doing this.
Now, here's a main point.
Once all this work has been
done, for a wide variety of
problems we never again have to
use an epsilon or a delta.
Let me illustrate with
one problem.
Let's suppose we were given
the problem limit of 'x
squared plus 7x' as 'x
approaches 3', and we wanted
to find that limit.
Our intuitive thing would
be to do what?
Let 'x' equal 3, in which case
we get 9 plus 21 is 30.
But we know by now that this
instruction says that 'x'
can't equal 3.
This is the problem.
We get what appears to be a nice
answer that we believe
in, but by an illegal method.
Can we use our legal methods
to gain the same result?
The answer is yes.
Because, you see, what is
'x squared plus 7x'?
It's the sum of two functions,
and we've just proven that the
limit of the sum is the
sum of the limits.
For example, what I
can say is this.
I don't know if this is true.
What I do know is true is that
the limit of 'x squared plus
7x' as 'x' approaches 3 is a
limit of 'x squared' as 'x'
approaches 3 plus the limit of
'7x' as 'x' approaches 3.
How do I know that?
I've proven that the
limit of a sum is
the sum of the limits.
Now what is this?
This is really a product.
This is really the limit
of 'x' times 'x'.
But we already know that the
limit of a product is the
product of the limits.
See, we proved that theorem.
Well, we almost proved it,
certainly close enough so I
think that we can
say that we did.
And this is a product also.
So you see, we can get from here
to here to here just by
our theorem.
Now didn't we also prove as one
of our theorems earlier
that the limit of 'x' as 'x'
approaches 'a' is 'a' itself?
Sure, we did.
In particular then, the limit of
'x' as 'x' approaches 3 has
already been proven to be 3.
So this is 3, this is 3.
What is the limit of 7
as 'x' approaches 3?
Well, 7 is a constant, and we
already proved that the limit
of a constant as 'x'
approaches 'a'
is that same constant.
So what is the constant here?
It's 7.
And what's the limit of 'x'
as 'x' approaches 3?
That's 3.
So by using our theorems,
we get from what?
From here to here
to here to here.
And now hopefully from a theorem
that comes from some
place around the third grade,
3 times 3 is 9, 7 times 3 is
21, the sum is 30.
And now you see this is no
longer a conjecture.
This follows, inescapably, from
the rules of our game,
from the rules and our
basic definition.
By the way, you may have a
tendency to feel when you see
something like this that, why
did we need the epsilons and
deltas in the first place?
Wasn't it a terrible waste?
Well, two things.
First of all, we couldn't prove
our theorems without the
epsilons and deltas.
And secondly, and don't lose
sight of this, in many real
life situations you may very
well be faced with the type of
a problem that doesn't ask you
to prove that the limit of 'x
squared plus 7x' is 30 as 'x'
approaches 3, but rather might
say how close must 'x' be chosen
to 3 if we want 'x
squared plus 7x' to be
less than 30.023.
And then, you see, if that's
the kind of a problem you
have, these new theorems
will not solve
that problem for you.
So we're not making
a choice here.
All we're saying is that the
epsilons and deltas are the
backbone of limits, but that
fortunately through
mathematical theorems, we can
get simpler ways of getting
important results.
And that was our main
purpose of today's
lecture, these two things.
That completes our lecture for
today, and so on until next
time, good bye.
NARRATOR: Funding for the
publication of this video was
provided by the Gabriella and
Paul Rosenbaum foundation.
Help OCW continue to provide
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