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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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