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HERBERT GROSS: Hi.
Our lecture today involves the
backbone of Calculus, namely
the concept of limit, and
perhaps no place in the
history of mathematics is there
a more subtle topic than
this particular one.
I'm reminded of the anecdote
of the professor who was
staring quite intently,
philosophically, in his home,
and his wife asked, what are
you contemplating, and he
said, that it's amazing
how electricity works.
And the wife said, well what's
so amazing about that?
All you have to do is
flick the switch.
And this somehow or other is
exactly the predicament that
the limit concept falls into.
From the point of view of
computation, it seems that
what comes naturally 99% of the
time gives us the right
answer, but unfortunately, the
1% of the time turns up almost
all the time in differential
Calculus.
For this reason, what we will do
is introduce the concept of
limit through the eyes of
differential Calculus, and we
shall call our lecture today
'Derivatives and Limits'.
And what we shall do is go back
to our old friend who
makes an appearance in almost
all of our lectures so far,
the case of the freely
falling object.
A ball is dropped.
It falls freely in a vacuum, and
the distance, 's', that it
falls in feet, at the end of 't'
seconds, is given by 's'
equals '16t squared'.
The question that we would like
to raise is, how fast is
the ball falling when
't' equals one?
In other words, notice that
the ball is covering a
different distance
here as it falls.
What we want to know is at the
instant that the time is one.
And by the way, don't confuse
the speed with the
displacement.
We know that when the time is
one, the object has fallen a
distance of 16 feet.
When the time is 1, the object
has fallen 16 feet.
What we want to know is how
fast is it falling at that
particular point.
Now keep in mind that
mathematics being a logical
subject proceeds from the idea
of trying to study the
unfamiliar in terms
of the familiar.
What we are familiar with is the
concept known as average
rate of speed.
So what we will do is duck the
main question temporarily and
proceed to a different
question.
What we shall ask is, what is
the average speed of the ball
during the time interval from t
equals one to t equals two?
And we all know how to solve
this problem from our
pre-calculus courses, namely,
we compute where the ball is
when 't' is two.
We compute where the ball
was when 't' is one.
The difference between these two
distances is the distance
that the ball has fallen during
this time interval, and
we then divide by the
time interval.
And the distance divided by
the time is precisely the
definition of average
rate of speed.
In other words, during the time
interval from 't' equals
one to 't' equals two, the ball
falls at an average speed
of 48 feet per second.
Again, in terms of our diagram,
you see at 't' equals
one, the object is over here.
At 't' equals two, the object
is over here, it's fallen 64
feet, and so we compute the
average speed by taking the
total distance and dividing
by the time that it took.
Now again, what we have done is
found the right answer, but
to the wrong question.
The question was not what was
the average rate of speed from
't' equals one to 't' equals
two, it was, what is the speed
at the instant 't' equals one?
And we sense that between 't'
equals one and 't' equals two,
an awful lot can happen, and
that therefore, our average
speed need not be a good
approximation for the
instantaneous speed.
What we do next, you see, is we
say OK, let's do something
a little bit differently.
Let's now compute the average
speed, but not from 't' equals
one to 't' equals two, but
rather from 't' equals one to
't' equals 1.1.
And computationally, we mimic
the same procedure as before.
We compute 's' when 't'
is 1.1, we compute
's' when 't' is one.
By the way, the only difference
is that it's a
little bit tougher to square
1.1 and multiply that by 16
than it was to square two
and multiply that by 16.
The arithmetic gets slightly
messier if you want to call it
that, but the concept
stays the same.
We find the total distance
traveled during the time
interval, we divide by the
length of the time interval,
and that quotient, by
definition, is the average
rate of speed.
Again, in terms of our diagram,
at t equals one, the
object has fallen 16 feet.
I'll distort this so that we can
see what's happening here.
At 't' equals 1.1, the object
has fallen 19.36 feet, and now
we compute the average speed
from here to here in the usual
high school way.
Now again, the same question
arises as before, namely, this
is still an average speed.
We wanted an instantaneous
speed.
And our come back is to say look
at, we at least suspect
intuitively that between one and
1.1, we will hopefully get
a better idea as to what's
happening at exactly one than
when we did on the bigger
interval from one to two.
In other words, whereas we don't
believe that 33.6 is the
answer to our original question,
we do believe that
33.6 is probably a better
approximation to the answer to
our original question than the
answer 48 feet per second that
was obtained over the
larger interval.
What the mathematician does next
is he says well, let's
stop playing games.
Let's not go between one and
1.1, or one and 1.01.
Why don't we go between one and
'1+h', where 'h' is any
non-zero amount?
In other words, let's find the
average speed from 't' equals
one to 't' equals '1+h', and
then what we'll do is we will
investigate to see what
happens when h
gets very, very small.
Well again, we mimic exactly
our procedure before.
We feed in 't' equals '1+h'
and compute 's', which of
course is '16 times
'1+h' squared'.
Then we subtract off the
distance that the particle had
fallen when 't' is one, that's
16 feet, we divide by the
length of the time interval.
Well between one and '1+h', the
time interval is 'h', we
simplify algebraically, and we
wind up with this particular
expression, '32h plus 16h
squared' over 'h'.
And now we come to the crux of
what limits, and derivatives,
and instantaneous rate of
speeds are all about.
If you recall, it's quite
tempting to look at this thing
and say, ah-ha, I'll cancel an
'h' from both the numerator
and the denominator, and that
will leave me '32 plus 16h'.
And notice though, the very
important thing, that as we
mentioned in previous lectures,
since you cannot
divide by 0, it is crucial at
this stage that we emphasize
that 'h' is not equal to 0.
Now of course, this
is not a difficult
rationalization to make.
Namely, in terms of our
physical situation, we
wouldn't want to compute an
average rate of speed over a
time interval during which
no time transpired.
And if we let 'h' be 0,
obviously, from one to '1+h',
no time transpires.
But the key statement is, and
we'll come back to exploit
this not only for this lecture,
but for the next one
too, is to exploit the idea that
in our entire discussion,
it was crucial that h not
be allowed to equal 0.
In other words, the average
speed between one and '1+h' is
'32 plus 16h'.
Now again, without trying to be
rigorous, observe that our
senses tell us that as 'h' gets
us close to 0 as we want
without ever being allowed to
get there, '16h' gets us close
to 0 as we want also, and
hence, '32 plus 16h'
apparently gets us close
to 32 as we wish.
In other words, it appears that
the speed is 32 feet per
second when t equals one.
And to reuse the vernacular,
what we're saying is we
compute the average
rate of speed--
see 's' of '1+h' minus
's' of '1/h'--
and see what happens to that
as 'h' approaches 0.
And in our particular case,
notice that this
simply means what?
We came down to '32 plus 16h'
over here, and then as 'h' got
arbitrarily small, we became
suspicious, and believed that
as 'h' became arbitrarily
small, '32
plus 16h' became 32.
And I'm saying that in great
detail because this is not
quite nearly as harmless
as it may seem.
There is a great deal of
difficulty involved in just
saying as 'h' gets close
to 0, '32 plus 16h'
gets close to 32.
Well there's no difficulty
in saying it.
There's difficulty that comes up
in trying to show what this
means precisely, as we shall
see in just a little while.
By the way, it's rather easy
to generalize this result.
Namely, you may have noticed in
our presentation there was
nothing sacred about asking what
was the speed when 't'
equals one.
One could just as easily have
chosen any other time 't', say
't sub one', and asked what was
the speed between 't1' and
't1' plus 'h'?
We could have done exactly
the same computation.
In fact, notice down here, when
we do this computation,
the answer '32t one' plus '16h'
becomes precisely '32
plus 16h when 't' one
happens to be one.
In other words, this result here
is the generalization of
our previous result that
had us work with the
limit of '32 plus 16h'.
At any rate, by this approach,
it becomes clear then, if the
equation of motion, the distance
versus the time, is
given by 's' equals '16t
squared', then that time 't'
equals 't one', the
instantaneous speed appears to
be 32 times 't' one.
And in fact, since 't' one
happens to be any arbitrary
time that we choose, it is
customary to drop the
subscript, and simply
say what?
At any time, 't', our freely
falling body in this case,
will have a speed
given by '32t'.
And many of us will remember, as
this is a revisited course,
that this is precisely the
recipe that turns out when one
talks about bringing down
the exponent and
replacing it by one less.
16 times two is 32, and 't' to
a power one less than two is
just t to the first
power or 't'.
But notice now, no recipes
involved here.
Just a question of intuitively
defining instantaneous speed
in terms of being a limit
of average speeds.
Now in order that we don't
become hypocrites in what
we're doing, recall also that
we have spent an entire
lecture talking about the value
of analytic Geometry in
the study of Calculus.
The idea that a picture
is worth 1,000 words.
So what we would like to do now
is to revisit our previous
discussion in terms
of a graph.
You see, suppose we have a curve
'y' equals 'f of x', and
I've drawn the curve over
on this side over here.
We'll make more reference to it
later as we go along, but
what does it mean if I now
mechanically compute 'f of x1
' plus 'delta x' minus 'f
of x1' over 'delta x'?
And by the way, again, notice
just a question of symbolism.
'Delta x' happens to be the
conventional symbol that one
uses in the analytical Geometry
approach where we
were using h before,
but this is simply
a question of notation.
Well the idea is this: observe
that if we go to the graph, 'f
of x1' plus 'delta x' is
this particular height.
On the other hand, this
height is 'f of x1' .
Consequently, our numerator is
just this distance here, and
our denominator, which is
'delta x', is just this
distance here.
And therefore, what we have done
in our quotient, is we
have found the slope of the
straight line that joins 'p'
to 'q.' And here then is why the
question of straight line
becomes so important
in the study of
differential Calculus.
That what is average speed, in
terms of analysis, becomes the
slope of a straight line
in terms of Geometry.
And the idea, you see, is that
this gives me the average
speed of the straight line that
joins the average rise of
the straight line that
joins 'p' to 'q'.
And by the way, this is quite
deceptive as any average is.
It's like the man who sat with
his feet in the oven and an
ice pack on his head, and when
somebody said how do you feel,
he said, on the average,
pretty good.
You see, the same thing is
happening over here.
Notice if I had taken a
completely different curve, a
curve which looked nothing at
all like my original curve,
only that also passed through
the points 'p' and 'q', notice
that even though these two
curves are quite different,
the average rise from
'p' to 'q' is the
same for both curves.
In other words, when one deals
with the average, one is not
too concerned with how one
got from the first
point to the second.
You see, this is the problem
with average rate of space.
In fact, the bigger a time
interval we work over, the
more danger there is that the
average rate of speed will not
coincide intuitively with what
we would like to believe that
the instantaneous rate
of speed is.
And you see, that's exactly
what this idea here means.
What we do to define the
derivative, the instantaneous
rise at 'x' equals 'x1' , is
we take this quotient and
investigate what happens to it
as 'delta x' is allowed to get
arbitrarily close to 0.
But let's emphasize
this again.
'Delta x' is not allowed
to equal 0.
You see, it's over a smaller and
smaller interval, and you
see what's happening
here geometrically?
To let 'delta x' get smaller and
smaller means, that for a
given 'x1' , you are holding 'p'
fixed, and allowing 'q' to
move in closer and
closer to 'p'.
And what we are saying is, is
that when 'q' assumes this
position, let's call it 'q sub
1', the slope of the line that
joints 'p' to 'q' one looks a
lot more like the slope of the
line that will be tangent to
the curve at the point 'p'.
And for this reason, what is
instantaneous rate of speed,
when we're talking about
functions, become slopes of
curves at a given point when we
are talking about curves.
You see that the tangent line to
a curve is viewed as being
the limiting position of a cord
drawn from a point 'p' to
a point 'q'.
Now at any rate, this will be
emphasized in the text.
It will be emphasized
in our exercises.
The major issue now is what did
we really do when it came
time to compute this
little gadget.
In other words, when we talked
about the limit of 'f of x' as
'x' approached 'a', how
did we compute this?
The danger is, and this is why
I have underlined a question
mark here, the intuitive
approach is to say something
like, let's just replace
x by a in here.
In other words, let the limit of
'f of x', as 'x' approaches
'a', be 'f of a'.
In fact, this is what we did
with our '32 plus 16h'.
In a way, what we did was, with
'32 plus 16h', do you
remember what we did?
We said, let's let 'h' get
arbitrarily close to 0.
And in our minds, we
let 'h' equal 0.
'16h' was then 0, and then '32
plus 16h' was then 32.
But what we had done is somehow
or other, if our
intuition is correct, we had
arrived at the correct answer
but for wrong reasons.
Because you see, in this
expression, it is mandatory,
as we've outlined this,
is that 'x' never be
allowed to equal 'a'.
Well you say, who cares
whether you
allow it or you don't?
Aren't we going to get
the same answer?
And just to give you a quickie,
one that we talked
about in our introductory
lecture, consider the function
'f of x' to be 'x squared
minus nine'
over 'x minus three'.
And let's evaluate what
this limit is as
'x' approaches three.
If we were to blindly assume
that all we have to do is
replace 'x' by 'a', that
would mean what here?
Replace 'x' by three.
Notice that what
we got is what?
Three squared, which is nine,
minus nine, (which is 0) over
three minus three,
which is also 0.
In other words, if in the
expression 'x squared minus
nine' over 'x minus three' you
replace 'x' by three, you do
get 0 over 0.
Unfortunately, or perhaps
fortunately, somehow or other,
this definition is supposed to
incorporate the fact that you
never allow 'x' to
equal three.
See, the way we got 0 over 0
was we violated what the
intuitive meaning
of limit was.
We wound up with this 0 over 0
mess by allowing 'x' to equal
the one value it is not
allowed to equal here.
To see what happened more
anatomically in this
particular problem, let's go
back to our old friend of
cancellation again.
We observe that 'x squared minus
nine' can be written as
'x plus three' times 'x minus
three', and then we divide by
'x minus three', and here's
where the main problem seems
to come up.
One automatically, somehow or
other, gets into the habit of
canceling these things out.
But the point is, since we have
already agreed that you
cannot divide by 0, the idea
then becomes that the only
time this is permissible is when
'x' is unequal to three.
In fact, to have been more
precise, what we should have
said was not that 'x squared
minus nine' over 'x minus
three' equals 'x plus three',
but rather, 'x squared minus
nine' over 'x minus three'
equals 'x plus three', unless
'x' equals three.
And by the way, I should
add here what?
That if 'x' equals three, then
'x squared minus nine' over 'x
minus three' is undefined.
You see, that's what
we call 0 over 0.
Undefined, indeterminate.
But here's what bails us out.
As soon as we say the limit as
'x' approaches three, what are
we assuming?
We're assuming that 'x' can be
any value at all except three,
and the interesting point now
becomes, that as long as 'x'
is anything but three, the only
time you could tell these
two expressions apart was when
'x' happened to equal three.
So if you now say we won't allow
'x' to equal three, then
it is true, whereas the
bracketed expression by itself
is not equal to the
parenthetical
expression by itself.
The limit as 'x' approaches
three of these two expressions
are equal, and in fact, one
becomes tempted to say "what?"
at this case.
Oh, here it's quite easy to see,
that as 'x' approaches
three, 'x plus three'
is equal to six.
And let me put a question mark
here also, because it
certainly is true that if you
replace 'x' by three, this
will be six.
But this says don't replace
'x' by three.
Now before, when we replaced
'x' by three we got into
trouble, so we said
we can't do that.
Now we replace 'x' by three, we
get an answer that we like,
and that is the danger that
we'll say OK, we won't allow
this to be done, unless
we like the result.
And of course this gives you
a highly subjective subject
matter here.
You see, the point is that
somehow or other we must come
up with a more objective
criteria for distinguishing
what limits are.
And by the way, for the Polyanna
who has the intuitive
feeling for saying things like
well look at, this may give
you trouble if you wind
up with 0 over 0.
But what is the likelihood that
given 'f of x' at random,
and 'a' at random, that the
limit of 'f of x' as 'x'
approaches 'a' is going to be
0 over 0 if you're careless?
And the answer is that the
probability is quite small,
except in Calculus, in which
case, every time, you take a
derivative.
Every time you are going to
wind up with 0 over 0.
This is why we said earlier that
differential Calculus has
been referred to as 'the
study of 0 over 0'.
You see, the whole point is,
let's go back now to our basic
definition of average rate of
change of 'f', 'f of x1' plus
'delta x' minus 'f of
x1' over 'delta x'.
I guess when you're writing in
black, you have to use white
for emphasis, so
we'll use this.
Notice that every time you let
'delta x' equal zero, you are
going to wind up with what?
'f of x1' minus 'f of x1' ,
which is 0, over 'delta x',
which is 0.
In other words, you are
going to get 0 over 0.
If you believe that all this
means is to replace 'delta x'
by 0, you see, and the point is,
this is not the definition
of this, and this is why you
don't wind up with this
particular thing.
Well I've spent enough time,
I think, trying to give a
picture as to what limits mean
intuitively, now what I'd like
to do is to devote the remainder
of this lecture,
plus our next one, to
investigating, more
stringently, what limit means in
a way that is unambiguous.
And you know, it's like
everything else in this world.
When you want to get something
that is ironclad, you don't
get something for nothing.
To get an ironclad expression
that doesn't lend itself to
misinterpretations usually
involves a rigorous enough
language that scares off
the uninitiated.
Sort of like a legal document.
The chances are if you're not
a trained lawyer, and the
document is simple enough for
you to understand, it must
have thousands of
loopholes in it.
But that same rigorous language
that makes it almost
impossible for the layman to
understand what's happening,
is precisely the language that
the lawyer needs to make the
document well defined
to him, hopefully.
Well with that as a background,
let's now try to
give a more rigorous definition
of limit.
And let's try to do it in such
a way that not only will the
definition be rigorous, but it
will agree, to the best of our
ability, with what we
intuitively believe a limit
should mean.
Now what I want to do here is
scare you a little bit, and
the reason I want to scare you
a little bit is, I want to
show you, once and for all,
how in many cases, what
appears to be ominous
mathematical notation, is
simply a rigorous way of saying
something which was
quite intuitive pictorially
to do.
The mathematical definition
of limit is the following.
The limit of 'f of x' as 'x'
approaches 'a' equals 'l'
means, and get this, given
epsilon greater than 0, we can
find delta greater than 0, such
that when 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' is less
than epsilon.
Now doesn't that kind
of turn you on?
See, this is the
whole problem.
I will now cut you in sort of on
the trade secret that makes
this thing very,
very readable.
First of all, to say that you
want to get arbitrarily close
in value, you want 'f of x' to
be arbitrarily close to 'l',
another way of saying
that is what?
You specify any tolerance limit
whatsoever, which we
call epsilon, and I can make 'f
of x' get to within epsilon
of 'l' just by choosing 'x'
sufficiently close to 'a'.
And by the way again, in terms
of geometry, that's all the
absolute value of a
difference means.
The absolute value of 'x minus
a' simply means what?
The distance between
'x' and 'a'.
All this says is what?
I can make 'f of x' fall within
epsilon of 'l' by
finding a delta such that 'x' is
within delta of 'a', and by
the way, this little tidbit over
here is simply a fancy
way of saying that you're not
letting 'x' equal 'a'.
You see, the only way the
absolute value of a number can
be 0 is if the number
itself is 0.
The only way this could be 0
is if 'x' equals 'a', so by
saying this little tidbit over
here, that this is greater
than 0, that's our fancy way
of saying that we are not
going to allow 'x'
to equal to 'a'.
Now again, whereas this may
sound a little bit simpler
than the original definition,
let's see how much easier the
picture makes it.
Let's go to a graph of
'y' equals 'f of x'.
And in my diagram here, here is
'a', and here is 'l', and
what I'm saying is, I
intuitively suspect.
Let me write down
what I suspect.
I suspect that the limit of 'f
of x', as 'x' approaches 'a',
is 'l' in this case.
Namely I suspect that as 'x'
gets closer and closer to 'a',
no matter which side you come
in on, that's why we use
absolute value.
It makes no difference whether
it's positive or negative.
As 'x' gets in tighter and
tighter on 'a', I suspect that
I can make 'f of x' come
arbitrarily equal to 'l'.
In other words, if 'x' is close
enough to 'a', 'f of x',
it seems, will be
close to 'l'.
And by the way, to complete this
problem geometrically,
pick your epsilon.
Let's say that epsilon is this
'y', and since epsilon is
positive, this will be called
'l plus epsilon'.
This will be called
'l minus epsilon'.
Now what we do is we come
over to the curve here.
All right, hopefully these
will look like horizontal
lines to you.
When I get down to here, I then
project down vertical
lines to meet the x-axis.
And now all I'm saying, and I
think this is fairly intuitive
clear, is what?
That if 'x' is this close to
'a', 'f of x' will be within
these tolerance limits of 'l'.
In other words, for any value of
'x' in here, see, for this
neighborhood of 'a', every
input has its output come
within epsilon of 'l'.
By the way, just one brief
note here, something that
we'll have to come back to in
more detail-- in fact, I'll do
that in our next example--
notice here how important it
was that our curve was not
single value, but one to one.
In other words, suppose our
curve had doubled back, say
something like this.
Notice then, when you try to
project over to the curve from
'l plus epsilon', you don't know
whether to stop at this
point or at this point.
You see, in other words, notice
that there is another
point down here which has the
same 'l plus epsilon' value.
In other words, whenever our
function is not one to one,
there is going to be a
problem, when we work
analytically, of being able to
solve algebraic equations, and
trying to distinguish this point
of intersection from
this point.
You see, if we made a mistake
and bypass this point and came
all the way over to here, and
thought that this was our
tolerance limit, we would be in
a great deal of difficulty.
Because you see, for example,
for this value of 'x', its 'f
of x' value projects up here,
which is outside of the
tolerance limits
that are given.
You see, many things which are
self evident from the picture
are not nearly so self evident
analytically, to which we then
raise the question, why
use analysis at all?
Let's just use pictures.
And again the answer is, in
this case, we can get away
with it, but in more complicated
situations where
either we can't draw the diagram
because it's too
complicated, or because we have
several variables, and we
get into a dimension problem,
the point is that sometimes we
have no recourse other
than the analysis.
Let me give you an exercise
that we'll do in terms of
putting the geometry and the
analysis side by side.
Let's look at the expression the
limit of 'x squared minus
2x' as 'x' approaches three.
Now of course, we don't want 'x'
to equal three here, but
let's cheat again, and see
what this thing means.
Our intuition tells us that when
'x' is three, 'x squared'
is nine, '2x' is six.
So nine minus six is three, and
we would suspect that the
limit here should be three.
Now to see what this means,
let's draw a little diagram.
The graph 'y' equals 'x squared
minus 2x' crosses the
x-axis at 0 and two.
It has its low point at the
0.1 comma minus one.
We can sketch these curves,
it's a parabola.
And now what we're saying is, is
it true that when 'x' gets
very close to three,
that 'f of x' gets
very close to three?
And the answer is, from the
diagram, it appears very
obvious that this is the case.
Fact, here's that example of
non-single value again.
If you call this thing 'l',
and you now pick tolerance
limits which we'll call 'l plus
epsilon' and 'l minus
epsilon', and now you want to
see what interval you need on
the x-axis, notice that in this
particular diagram, you
will get two intervals, one of
which surrounds the value 'x'
equals minus one, and the other
of which surrounds the
'x' value 'x' equals three.
In other words, both when 'x' is
minus one and 'x' is three,
'x squared minus 2x'
will equal three.
At any rate, leaving the diagram
as an aid, let's see
what our epsilon delta
definition says, and how we
handle the stuff algebraically,
and how we
correlate the algebra
with the geometry.
Given epsilon greater than
0, what must I do?
I must find a delta greater
than 0 such that, and I'm
going to read this colloquially,
I'll write it
formally, but read it
colloquially, such that
whenever 'x' is within delta
of three, but not equal to
three, then 'x squared
minus 2x' will be
within epsilon of three.
Now here's the whole point.
We know, algebraically, how
to handle absolute values.
Namely, the absolute value of
'x squared minus 2x minus
three' being less than epsilon,
is the same as saying
that 'x squared minus 2x minus
three' itself must be between
epsilon and minus epsilon.
Now again, we rewrite things
in fancy ways if we wish.
There other ways
of doing this.
I call this completing
the square.
Well, people for 2,000 years
have called this completing
the square.
Namely, I rewrite minus three as
one minus four, so that 'x
squared minus 2x plus one'
can be factored as
'x minus one squared'.
I'm now down to this
form here.
Then I add four to all sides
of my inequality, and have
that 'x minus one squared' is
between four plus epsilon and
four minus epsilon.
Some fairly elementary Algebra
of inequalities here.
Now here's the key point.
Remembering my diagram, I said
to you, how do we know whether
we're near three or
near minus one?
How do we distinguish, when we
draw the lines to the curve,
what neighborhood we want?
And notice that all we're saying
is that if 'x' is near
three, that means what?
'x' is close to three.
If 'x' is reasonably close to
three, then 'x minus one' is
going to be reasonably close to
two, and hence be positive.
Point is, that as long as you're
dealing with positive
numbers over here, see if
epsilon is sufficiently small,
these will then all be
positive numbers.
For positive numbers, if the
squares obey a certain
inequality, the square roots
will obey the same inequality,
as we have emphasized in
one of our exercises.
In other words, from here, we
can now say this, and from
this, we can now conclude, that
if you want 'x squared
minus 2x' to be within epsilon
of three, 'x' itself is going
to have to be between 'one plus
the square root of 'four
plus epsilon'', and 'one
plus the square root
of 'four minus epsilon''.
And by the way, this is quite
rigorous, but I don't think it
turns you on.
I don't think there's any
picture that you associate
with this, and so what I thought
I'd like to do for our
next little thing, is to come
back to our diagram here, and
show you what this
really means.
Namely, notice that if epsilon
is a small positive number,
let's take a look
at this again.
If epsilon is a small positive
number, this is
slightly less than four.
Therefore, the square root is
slightly less than two,
therefore, this number will be
slightly less than three.
On the other hand, 'four plus
epsilon' is slightly more than
four, so its square root will be
slightly more than two, and
therefore, one plus that square
root will be slightly
more than three.
And what these two numbers, as
abstract as they look like,
correspond to, is nothing
more than this.
Coming back to our diagram and
choosing an epsilon, and
coming over to the curve like
this, and projecting down like
this, all we're saying is,
see, what are we saying
pictorially?
That for 'f of x' to be within
epsilon of 'l', 'x' itself has
to be in this range here.
And all this says is, this very
simple point to compute,
just by projecting down like
this, its rigorous name would
be 'one plus the square root
of 'four minus epsilon''.
That's this number over here.
And the furthest point, namely
this point here, which again,
in terms of the picture is very
easy to find, that point
corresponds to one plus
the square root
of 'four plus epsilon'.
And again notice, in terms
of the Algebra, I need no
recourse to a picture.
But in terms of the picture, I
get a heck of a good feeling
as to what these abstract
symbols are telling me.
Well let's continue on with the
solution of this exercise.
You see, we didn't want to find
where 'x' was, we wanted
to see where 'x minus
three' had to be.
In other words, we're trying
to find the delta.
We know that 'x minus three'
is between these
two extremes here.
Consequently, and here's a place
we have to be a little
bit careful here, this of
course, is a positive number,
because this is more than two.
This, on the other hand, is a
negative number, because you
see 'four minus epsilon' is less
than four, so its square
root is less than two.
So if I subtract two from--
it's a negative number--
and since delta has to be
positive, if this thing is
negative, two minus the square
root of 'four minus epsilon'
will be positive.
So what I do is, to solve this
problem, is I simply choose
delta to be the minimum
of these two numbers.
And then by definition,
what do I have?
That when the absolute value
of 'x minus three' is less
than delta, and at the same
time, this is crucial,
greater than 0.
In other words, we never
let 'x' equal three.
For this choice of delta, by how
we worked backwards, this
will come out to be
less than epsilon.
And that is exactly what you
mean by our formal definition
of saying that the limit of 'x
squared minus 2x' as 'x'
approaches three, in this
case, equals three.
For someone who wants more
concrete evidence, I think all
one has to do is, for example,
take a number like 'one plus
the square root of 'four plus
epsilon'', that is the widest
point on our interval, and
actually plug that in for 'x',
and go through this computation
of squaring 'one
plus the square root of
'four plus epsilon''.
Subtract twice, 'one plus
the square root
of 'four plus epsilon''.
Carry out that computation
and you get what?
'Three plus epsilon'.
That using the upper extreme,
you wind up with what?
In excess of three by epsilon,
which by the way is exactly
what's supposed to happen.
In fact, I hope this doesn't
give you an eye sore, I will
pull down the top board again
to show you also what these
two distances mean, and
what this means.
You see, all these two numbers
mean is the following, and
I'll pull this down just
far enough to see it.
That these two numbers were the
widths from this end point
to three on the one extreme,
and from this end point to
three on the other extreme.
In other words, those two
numbers that we took the
minimum of were the half-width
intervals
that surrounded three.
You see the problem, as we
mentioned before is, is that
even though epsilon and minus
epsilon are symmetric with
respect to 'l', when you
translate over to the curve,
because this curve is not a
straight line when you project
down, these two widths
will not be equal.
By picking the minimum of
these two widths, you
guarantee that you have
locked yourself inside
the required area.
You see, that's all this
particular thing means.
Now the trouble is that one
might now get the idea that
there are no shorter ways of
doing the same problem.
You see, I showed you
a rigorous way.
There was no law that said we
had to find the biggest delta.
In other words, if delta equals,
a half will work, in
other words, if you're within
a half of three, something's
going to happen, then certainly
any smaller number
will work just as well.
In other words, I can
get estimates.
Let me show you what
I mean by that.
Another way of tackling how to
make 'x squared minus 2x minus
three' smaller than epsilon is
to use our properties of
absolute values, and
to factor this.
In other words, we know that
'x squared minus 2x minus
three' is 'x minus three'
times 'x plus one'.
We know that the absolute
value of a product is a
product of the absolute values,
so these two things
here are synonyms. Then we
know that near 'x' equals
three, which we're interested
in here, 'x plus
one' is near four.
Well what we mean more
rigorously is this, choose an
epsilon, and if the absolute
value of 'x minus three' is
less than epsilon, in other
words, if 'x minus three' is
less than epsilon but greater
than minus epsilon, then by
adding on four to all three
sides of the inequality here,
we see that 'x plus one' is less
than 'four plus epsilon',
but greater than 'four
minus epsilon'.
If the size of epsilon is less
than one, then certainly this
is less than five, and this
is greater than three.
Now of course, somebody may say
to us, but what if epsilon
is more than one?
Well obviously, if a guy says
make this thing within 10, and
I can make it within one,
certainly within
10 is within one.
I can always pick the
smaller number.
It's like that nonsense of the
fellow asking for an ice cream
cone, and the waitress said
what flavor, and he said
anything except chocolate.
And she said, I'm all out of
chocolate, will you take
anything except vanilla?
The idea here is that if
somebody says make this less
than 15, if you've made it less
than 10, in particular,
you've made it less than 15.
And so all we do over
here, you see, is
something like this.
We say look it, if we want to
make this number here very
small, we know that this number
here is less than five,
we know that 'epsilon over five'
is a positive number if
epsilon is small, so why don't
we just pick the absolute
value of 'x minus three' to be
less than 'epsilon over five?'
In fact, that's what this delta
will be in that case.
In other words, if the absolute
value of x minus
three is within 'epsilon over
five', if 'x' is within
'epsilon over five' of
three, notice what
this product becomes.
This is less than 'epsilon over
five', this we know from
before is less than five, and
therefore, this product is
less than epsilon.
In other words, we have
exhibited the delta such that
when this happens, the absolute
value of what we want
to make less than
epsilon indeed
becomes less than epsilon.
Now because I recognize that
this is a hard topic, you'll
notice in the reading
assignments that these
problems are covered
in great detail.
Everything that I've said in the
lesson so far is repeated
in great computational depth
in our learning exercises.
The point is that even if I
can make the epsilons and
deltas seem a little bit more
meaningful for you than by the
formal definition, notice that
it's simple only in comparison
to what we had before.
But it's still very,
very difficult.
The beauty of Calculus is that
in many cases, we do not have
to know what delta looks like
for a given epsilon.
What we shall do there
therefore, in our next
lecture, is to develop recipes
that will allow us to get the
answers to these limit problems
without having to go
through this genuinely difficult
problem of finding a
given delta to match
a given epsilon.
Though we must admit, in real
life, in many cases where
you're making approximations, we
will want to know what the
tolerance limits are.
I am not belittling the epsilon
delta approach.
All I'm saying is that in
certain problems, you do not
need the epsilon delta approach
to get nice results,
and that will be the topic
of our next lecture.
So until next time, goodbye.
GUEST SPEAKER: Funding for the
publication of this video was
provided by the Gabriella and
Paul Rosenbaum Foundation.
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