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HERBERT GROSS: Hi.
00:00:35.720 --> 00:00:40.630
Our lecture today involves the
backbone of Calculus, namely
00:00:40.630 --> 00:00:43.440
the concept of limit, and
perhaps no place in the
00:00:43.440 --> 00:00:47.200
history of mathematics is there
a more subtle topic than
00:00:47.200 --> 00:00:48.950
this particular one.
00:00:48.950 --> 00:00:52.090
I'm reminded of the anecdote
of the professor who was
00:00:52.090 --> 00:00:55.710
staring quite intently,
philosophically, in his home,
00:00:55.710 --> 00:00:58.420
and his wife asked, what are
you contemplating, and he
00:00:58.420 --> 00:01:01.090
said, that it's amazing
how electricity works.
00:01:01.090 --> 00:01:03.350
And the wife said, well what's
so amazing about that?
00:01:03.350 --> 00:01:05.820
All you have to do is
flick the switch.
00:01:05.820 --> 00:01:08.890
And this somehow or other is
exactly the predicament that
00:01:08.890 --> 00:01:10.740
the limit concept falls into.
00:01:10.740 --> 00:01:13.910
From the point of view of
computation, it seems that
00:01:13.910 --> 00:01:18.370
what comes naturally 99% of the
time gives us the right
00:01:18.370 --> 00:01:23.480
answer, but unfortunately, the
1% of the time turns up almost
00:01:23.480 --> 00:01:26.410
all the time in differential
Calculus.
00:01:26.410 --> 00:01:29.380
For this reason, what we will do
is introduce the concept of
00:01:29.380 --> 00:01:32.540
limit through the eyes of
differential Calculus, and we
00:01:32.540 --> 00:01:37.370
shall call our lecture today
'Derivatives and Limits'.
00:01:37.370 --> 00:01:41.440
And what we shall do is go back
to our old friend who
00:01:41.440 --> 00:01:44.540
makes an appearance in almost
all of our lectures so far,
00:01:44.540 --> 00:01:47.080
the case of the freely
falling object.
00:01:47.080 --> 00:01:48.820
A ball is dropped.
00:01:48.820 --> 00:01:53.590
It falls freely in a vacuum, and
the distance, 's', that it
00:01:53.590 --> 00:01:57.380
falls in feet, at the end of 't'
seconds, is given by 's'
00:01:57.380 --> 00:01:59.640
equals '16t squared'.
00:01:59.640 --> 00:02:03.820
The question that we would like
to raise is, how fast is
00:02:03.820 --> 00:02:07.770
the ball falling when
't' equals one?
00:02:07.770 --> 00:02:10.940
In other words, notice that
the ball is covering a
00:02:10.940 --> 00:02:12.730
different distance
here as it falls.
00:02:12.730 --> 00:02:15.920
What we want to know is at the
instant that the time is one.
00:02:15.920 --> 00:02:18.280
And by the way, don't confuse
the speed with the
00:02:18.280 --> 00:02:19.060
displacement.
00:02:19.060 --> 00:02:22.617
We know that when the time is
one, the object has fallen a
00:02:22.617 --> 00:02:25.530
distance of 16 feet.
00:02:25.530 --> 00:02:28.290
When the time is 1, the object
has fallen 16 feet.
00:02:28.290 --> 00:02:31.560
What we want to know is how
fast is it falling at that
00:02:31.560 --> 00:02:32.960
particular point.
00:02:32.960 --> 00:02:36.810
Now keep in mind that
mathematics being a logical
00:02:36.810 --> 00:02:40.920
subject proceeds from the idea
of trying to study the
00:02:40.920 --> 00:02:43.160
unfamiliar in terms
of the familiar.
00:02:43.160 --> 00:02:46.860
What we are familiar with is the
concept known as average
00:02:46.860 --> 00:02:48.030
rate of speed.
00:02:48.030 --> 00:02:52.560
So what we will do is duck the
main question temporarily and
00:02:52.560 --> 00:02:54.500
proceed to a different
question.
00:02:54.500 --> 00:02:58.640
What we shall ask is, what is
the average speed of the ball
00:02:58.640 --> 00:03:03.330
during the time interval from t
equals one to t equals two?
00:03:03.330 --> 00:03:06.230
And we all know how to solve
this problem from our
00:03:06.230 --> 00:03:12.600
pre-calculus courses, namely,
we compute where the ball is
00:03:12.600 --> 00:03:14.430
when 't' is two.
00:03:14.430 --> 00:03:17.960
We compute where the ball
was when 't' is one.
00:03:17.960 --> 00:03:21.280
The difference between these two
distances is the distance
00:03:21.280 --> 00:03:24.070
that the ball has fallen during
this time interval, and
00:03:24.070 --> 00:03:26.740
we then divide by the
time interval.
00:03:26.740 --> 00:03:30.400
And the distance divided by
the time is precisely the
00:03:30.400 --> 00:03:32.890
definition of average
rate of speed.
00:03:32.890 --> 00:03:35.820
In other words, during the time
interval from 't' equals
00:03:35.820 --> 00:03:40.420
one to 't' equals two, the ball
falls at an average speed
00:03:40.420 --> 00:03:42.690
of 48 feet per second.
00:03:42.690 --> 00:03:47.230
Again, in terms of our diagram,
you see at 't' equals
00:03:47.230 --> 00:03:50.510
one, the object is over here.
00:03:50.510 --> 00:03:55.300
At 't' equals two, the object
is over here, it's fallen 64
00:03:55.300 --> 00:03:59.270
feet, and so we compute the
average speed by taking the
00:03:59.270 --> 00:04:03.150
total distance and dividing
by the time that it took.
00:04:03.150 --> 00:04:07.470
Now again, what we have done is
found the right answer, but
00:04:07.470 --> 00:04:08.690
to the wrong question.
00:04:08.690 --> 00:04:10.940
The question was not what was
the average rate of speed from
00:04:10.940 --> 00:04:14.190
't' equals one to 't' equals
two, it was, what is the speed
00:04:14.190 --> 00:04:16.360
at the instant 't' equals one?
00:04:16.360 --> 00:04:20.050
And we sense that between 't'
equals one and 't' equals two,
00:04:20.050 --> 00:04:23.170
an awful lot can happen, and
that therefore, our average
00:04:23.170 --> 00:04:26.450
speed need not be a good
approximation for the
00:04:26.450 --> 00:04:27.780
instantaneous speed.
00:04:27.780 --> 00:04:31.630
What we do next, you see, is we
say OK, let's do something
00:04:31.630 --> 00:04:32.850
a little bit differently.
00:04:32.850 --> 00:04:37.240
Let's now compute the average
speed, but not from 't' equals
00:04:37.240 --> 00:04:40.710
one to 't' equals two, but
rather from 't' equals one to
00:04:40.710 --> 00:04:42.970
't' equals 1.1.
00:04:42.970 --> 00:04:47.220
And computationally, we mimic
the same procedure as before.
00:04:47.220 --> 00:04:50.910
We compute 's' when 't'
is 1.1, we compute
00:04:50.910 --> 00:04:52.780
's' when 't' is one.
00:04:52.780 --> 00:04:54.510
By the way, the only difference
is that it's a
00:04:54.510 --> 00:04:58.850
little bit tougher to square
1.1 and multiply that by 16
00:04:58.850 --> 00:05:01.930
than it was to square two
and multiply that by 16.
00:05:01.930 --> 00:05:04.680
The arithmetic gets slightly
messier if you want to call it
00:05:04.680 --> 00:05:06.900
that, but the concept
stays the same.
00:05:06.900 --> 00:05:09.810
We find the total distance
traveled during the time
00:05:09.810 --> 00:05:13.980
interval, we divide by the
length of the time interval,
00:05:13.980 --> 00:05:17.560
and that quotient, by
definition, is the average
00:05:17.560 --> 00:05:18.560
rate of speed.
00:05:18.560 --> 00:05:23.050
Again, in terms of our diagram,
at t equals one, the
00:05:23.050 --> 00:05:25.640
object has fallen 16 feet.
00:05:25.640 --> 00:05:28.230
I'll distort this so that we can
see what's happening here.
00:05:28.230 --> 00:05:34.210
At 't' equals 1.1, the object
has fallen 19.36 feet, and now
00:05:34.210 --> 00:05:37.530
we compute the average speed
from here to here in the usual
00:05:37.530 --> 00:05:39.050
high school way.
00:05:39.050 --> 00:05:44.030
Now again, the same question
arises as before, namely, this
00:05:44.030 --> 00:05:45.390
is still an average speed.
00:05:45.390 --> 00:05:47.800
We wanted an instantaneous
speed.
00:05:47.800 --> 00:05:51.350
And our come back is to say look
at, we at least suspect
00:05:51.350 --> 00:05:56.940
intuitively that between one and
1.1, we will hopefully get
00:05:56.940 --> 00:06:00.430
a better idea as to what's
happening at exactly one than
00:06:00.430 --> 00:06:03.320
when we did on the bigger
interval from one to two.
00:06:03.320 --> 00:06:06.810
In other words, whereas we don't
believe that 33.6 is the
00:06:06.810 --> 00:06:10.350
answer to our original question,
we do believe that
00:06:10.350 --> 00:06:15.600
33.6 is probably a better
approximation to the answer to
00:06:15.600 --> 00:06:19.560
our original question than the
answer 48 feet per second that
00:06:19.560 --> 00:06:21.860
was obtained over the
larger interval.
00:06:21.860 --> 00:06:25.300
What the mathematician does next
is he says well, let's
00:06:25.300 --> 00:06:26.670
stop playing games.
00:06:26.670 --> 00:06:31.010
Let's not go between one and
1.1, or one and 1.01.
00:06:31.010 --> 00:06:37.530
Why don't we go between one and
'1+h', where 'h' is any
00:06:37.530 --> 00:06:38.630
non-zero amount?
00:06:38.630 --> 00:06:41.860
In other words, let's find the
average speed from 't' equals
00:06:41.860 --> 00:06:46.620
one to 't' equals '1+h', and
then what we'll do is we will
00:06:46.620 --> 00:06:48.930
investigate to see what
happens when h
00:06:48.930 --> 00:06:50.810
gets very, very small.
00:06:50.810 --> 00:06:54.510
Well again, we mimic exactly
our procedure before.
00:06:54.510 --> 00:06:59.090
We feed in 't' equals '1+h'
and compute 's', which of
00:06:59.090 --> 00:07:01.980
course is '16 times
'1+h' squared'.
00:07:01.980 --> 00:07:04.890
Then we subtract off the
distance that the particle had
00:07:04.890 --> 00:07:08.620
fallen when 't' is one, that's
16 feet, we divide by the
00:07:08.620 --> 00:07:09.960
length of the time interval.
00:07:09.960 --> 00:07:14.740
Well between one and '1+h', the
time interval is 'h', we
00:07:14.740 --> 00:07:18.650
simplify algebraically, and we
wind up with this particular
00:07:18.650 --> 00:07:23.680
expression, '32h plus 16h
squared' over 'h'.
00:07:23.680 --> 00:07:28.210
And now we come to the crux of
what limits, and derivatives,
00:07:28.210 --> 00:07:31.460
and instantaneous rate of
speeds are all about.
00:07:31.460 --> 00:07:34.630
If you recall, it's quite
tempting to look at this thing
00:07:34.630 --> 00:07:38.580
and say, ah-ha, I'll cancel an
'h' from both the numerator
00:07:38.580 --> 00:07:44.250
and the denominator, and that
will leave me '32 plus 16h'.
00:07:44.250 --> 00:07:47.260
And notice though, the very
important thing, that as we
00:07:47.260 --> 00:07:53.410
mentioned in previous lectures,
since you cannot
00:07:53.410 --> 00:07:57.960
divide by 0, it is crucial at
this stage that we emphasize
00:07:57.960 --> 00:08:00.090
that 'h' is not equal to 0.
00:08:00.090 --> 00:08:02.150
Now of course, this
is not a difficult
00:08:02.150 --> 00:08:03.770
rationalization to make.
00:08:03.770 --> 00:08:06.510
Namely, in terms of our
physical situation, we
00:08:06.510 --> 00:08:09.960
wouldn't want to compute an
average rate of speed over a
00:08:09.960 --> 00:08:12.810
time interval during which
no time transpired.
00:08:12.810 --> 00:08:16.630
And if we let 'h' be 0,
obviously, from one to '1+h',
00:08:16.630 --> 00:08:18.000
no time transpires.
00:08:18.000 --> 00:08:20.290
But the key statement is, and
we'll come back to exploit
00:08:20.290 --> 00:08:23.210
this not only for this lecture,
but for the next one
00:08:23.210 --> 00:08:27.310
too, is to exploit the idea that
in our entire discussion,
00:08:27.310 --> 00:08:31.170
it was crucial that h not
be allowed to equal 0.
00:08:31.170 --> 00:08:36.559
In other words, the average
speed between one and '1+h' is
00:08:36.559 --> 00:08:39.340
'32 plus 16h'.
00:08:39.340 --> 00:08:43.100
Now again, without trying to be
rigorous, observe that our
00:08:43.100 --> 00:08:47.740
senses tell us that as 'h' gets
us close to 0 as we want
00:08:47.740 --> 00:08:51.670
without ever being allowed to
get there, '16h' gets us close
00:08:51.670 --> 00:08:56.260
to 0 as we want also, and
hence, '32 plus 16h'
00:08:56.260 --> 00:09:01.490
apparently gets us close
to 32 as we wish.
00:09:01.490 --> 00:09:09.720
In other words, it appears that
the speed is 32 feet per
00:09:09.720 --> 00:09:12.180
second when t equals one.
00:09:12.180 --> 00:09:17.230
And to reuse the vernacular,
what we're saying is we
00:09:17.230 --> 00:09:20.550
compute the average
rate of speed--
00:09:20.550 --> 00:09:24.770
see 's' of '1+h' minus
's' of '1/h'--
00:09:24.770 --> 00:09:27.800
and see what happens to that
as 'h' approaches 0.
00:09:27.800 --> 00:09:29.740
And in our particular case,
notice that this
00:09:29.740 --> 00:09:30.700
simply means what?
00:09:30.700 --> 00:09:34.790
We came down to '32 plus 16h'
over here, and then as 'h' got
00:09:34.790 --> 00:09:38.580
arbitrarily small, we became
suspicious, and believed that
00:09:38.580 --> 00:09:41.230
as 'h' became arbitrarily
small, '32
00:09:41.230 --> 00:09:44.270
plus 16h' became 32.
00:09:44.270 --> 00:09:46.970
And I'm saying that in great
detail because this is not
00:09:46.970 --> 00:09:49.190
quite nearly as harmless
as it may seem.
00:09:49.190 --> 00:09:52.050
There is a great deal of
difficulty involved in just
00:09:52.050 --> 00:09:55.900
saying as 'h' gets close
to 0, '32 plus 16h'
00:09:55.900 --> 00:09:57.100
gets close to 32.
00:09:57.100 --> 00:09:59.910
Well there's no difficulty
in saying it.
00:09:59.910 --> 00:10:03.240
There's difficulty that comes up
in trying to show what this
00:10:03.240 --> 00:10:06.660
means precisely, as we shall
see in just a little while.
00:10:06.660 --> 00:10:10.280
By the way, it's rather easy
to generalize this result.
00:10:10.280 --> 00:10:13.220
Namely, you may have noticed in
our presentation there was
00:10:13.220 --> 00:10:16.730
nothing sacred about asking what
was the speed when 't'
00:10:16.730 --> 00:10:17.550
equals one.
00:10:17.550 --> 00:10:21.500
One could just as easily have
chosen any other time 't', say
00:10:21.500 --> 00:10:26.660
't sub one', and asked what was
the speed between 't1' and
00:10:26.660 --> 00:10:27.830
't1' plus 'h'?
00:10:27.830 --> 00:10:31.130
We could have done exactly
the same computation.
00:10:31.130 --> 00:10:34.670
In fact, notice down here, when
we do this computation,
00:10:34.670 --> 00:10:40.670
the answer '32t one' plus '16h'
becomes precisely '32
00:10:40.670 --> 00:10:44.500
plus 16h when 't' one
happens to be one.
00:10:44.500 --> 00:10:48.600
In other words, this result here
is the generalization of
00:10:48.600 --> 00:10:53.660
our previous result that
had us work with the
00:10:53.660 --> 00:10:57.170
limit of '32 plus 16h'.
00:10:57.170 --> 00:11:02.010
At any rate, by this approach,
it becomes clear then, if the
00:11:02.010 --> 00:11:05.090
equation of motion, the distance
versus the time, is
00:11:05.090 --> 00:11:08.840
given by 's' equals '16t
squared', then that time 't'
00:11:08.840 --> 00:11:12.670
equals 't one', the
instantaneous speed appears to
00:11:12.670 --> 00:11:16.020
be 32 times 't' one.
00:11:16.020 --> 00:11:21.270
And in fact, since 't' one
happens to be any arbitrary
00:11:21.270 --> 00:11:23.820
time that we choose, it is
customary to drop the
00:11:23.820 --> 00:11:25.880
subscript, and simply
say what?
00:11:25.880 --> 00:11:29.550
At any time, 't', our freely
falling body in this case,
00:11:29.550 --> 00:11:33.130
will have a speed
given by '32t'.
00:11:33.130 --> 00:11:37.400
And many of us will remember, as
this is a revisited course,
00:11:37.400 --> 00:11:40.740
that this is precisely the
recipe that turns out when one
00:11:40.740 --> 00:11:43.180
talks about bringing down
the exponent and
00:11:43.180 --> 00:11:45.250
replacing it by one less.
00:11:45.250 --> 00:11:50.540
16 times two is 32, and 't' to
a power one less than two is
00:11:50.540 --> 00:11:53.150
just t to the first
power or 't'.
00:11:53.150 --> 00:11:56.200
But notice now, no recipes
involved here.
00:11:56.200 --> 00:12:01.060
Just a question of intuitively
defining instantaneous speed
00:12:01.060 --> 00:12:05.470
in terms of being a limit
of average speeds.
00:12:05.470 --> 00:12:08.340
Now in order that we don't
become hypocrites in what
00:12:08.340 --> 00:12:12.040
we're doing, recall also that
we have spent an entire
00:12:12.040 --> 00:12:16.610
lecture talking about the value
of analytic Geometry in
00:12:16.610 --> 00:12:18.070
the study of Calculus.
00:12:18.070 --> 00:12:21.470
The idea that a picture
is worth 1,000 words.
00:12:21.470 --> 00:12:25.360
So what we would like to do now
is to revisit our previous
00:12:25.360 --> 00:12:29.000
discussion in terms
of a graph.
00:12:29.000 --> 00:12:33.950
You see, suppose we have a curve
'y' equals 'f of x', and
00:12:33.950 --> 00:12:36.440
I've drawn the curve over
on this side over here.
00:12:36.440 --> 00:12:40.110
We'll make more reference to it
later as we go along, but
00:12:40.110 --> 00:12:44.120
what does it mean if I now
mechanically compute 'f of x1
00:12:44.120 --> 00:12:48.400
' plus 'delta x' minus 'f
of x1' over 'delta x'?
00:12:48.400 --> 00:12:53.240
And by the way, again, notice
just a question of symbolism.
00:12:53.240 --> 00:12:56.510
'Delta x' happens to be the
conventional symbol that one
00:12:56.510 --> 00:12:59.410
uses in the analytical Geometry
approach where we
00:12:59.410 --> 00:13:01.600
were using h before,
but this is simply
00:13:01.600 --> 00:13:03.490
a question of notation.
00:13:03.490 --> 00:13:12.410
Well the idea is this: observe
that if we go to the graph, 'f
00:13:12.410 --> 00:13:17.060
of x1' plus 'delta x' is
this particular height.
00:13:17.060 --> 00:13:21.790
On the other hand, this
height is 'f of x1' .
00:13:21.790 --> 00:13:29.000
Consequently, our numerator is
just this distance here, and
00:13:29.000 --> 00:13:31.960
our denominator, which is
'delta x', is just this
00:13:31.960 --> 00:13:33.460
distance here.
00:13:33.460 --> 00:13:37.690
And therefore, what we have done
in our quotient, is we
00:13:37.690 --> 00:13:41.650
have found the slope of the
straight line that joins 'p'
00:13:41.650 --> 00:13:45.870
to 'q.' And here then is why the
question of straight line
00:13:45.870 --> 00:13:48.430
becomes so important
in the study of
00:13:48.430 --> 00:13:49.900
differential Calculus.
00:13:49.900 --> 00:13:54.260
That what is average speed, in
terms of analysis, becomes the
00:13:54.260 --> 00:13:57.660
slope of a straight line
in terms of Geometry.
00:13:57.660 --> 00:14:01.030
And the idea, you see, is that
this gives me the average
00:14:01.030 --> 00:14:06.460
speed of the straight line that
joins the average rise of
00:14:06.460 --> 00:14:08.890
the straight line that
joins 'p' to 'q'.
00:14:08.890 --> 00:14:12.800
And by the way, this is quite
deceptive as any average is.
00:14:12.800 --> 00:14:14.990
It's like the man who sat with
his feet in the oven and an
00:14:14.990 --> 00:14:17.100
ice pack on his head, and when
somebody said how do you feel,
00:14:17.100 --> 00:14:18.850
he said, on the average,
pretty good.
00:14:18.850 --> 00:14:21.380
You see, the same thing is
happening over here.
00:14:21.380 --> 00:14:24.880
Notice if I had taken a
completely different curve, a
00:14:24.880 --> 00:14:28.560
curve which looked nothing at
all like my original curve,
00:14:28.560 --> 00:14:32.670
only that also passed through
the points 'p' and 'q', notice
00:14:32.670 --> 00:14:35.190
that even though these two
curves are quite different,
00:14:35.190 --> 00:14:38.760
the average rise from
'p' to 'q' is the
00:14:38.760 --> 00:14:40.220
same for both curves.
00:14:40.220 --> 00:14:43.260
In other words, when one deals
with the average, one is not
00:14:43.260 --> 00:14:46.690
too concerned with how one
got from the first
00:14:46.690 --> 00:14:47.870
point to the second.
00:14:47.870 --> 00:14:50.280
You see, this is the problem
with average rate of space.
00:14:50.280 --> 00:14:53.500
In fact, the bigger a time
interval we work over, the
00:14:53.500 --> 00:14:58.090
more danger there is that the
average rate of speed will not
00:14:58.090 --> 00:15:01.250
coincide intuitively with what
we would like to believe that
00:15:01.250 --> 00:15:03.450
the instantaneous rate
of speed is.
00:15:03.450 --> 00:15:06.940
And you see, that's exactly
what this idea here means.
00:15:06.940 --> 00:15:10.650
What we do to define the
derivative, the instantaneous
00:15:10.650 --> 00:15:15.820
rise at 'x' equals 'x1' , is
we take this quotient and
00:15:15.820 --> 00:15:20.050
investigate what happens to it
as 'delta x' is allowed to get
00:15:20.050 --> 00:15:22.080
arbitrarily close to 0.
00:15:22.080 --> 00:15:24.210
But let's emphasize
this again.
00:15:24.210 --> 00:15:28.160
'Delta x' is not allowed
to equal 0.
00:15:28.160 --> 00:15:31.020
You see, it's over a smaller and
smaller interval, and you
00:15:31.020 --> 00:15:33.120
see what's happening
here geometrically?
00:15:33.120 --> 00:15:36.580
To let 'delta x' get smaller and
smaller means, that for a
00:15:36.580 --> 00:15:41.810
given 'x1' , you are holding 'p'
fixed, and allowing 'q' to
00:15:41.810 --> 00:15:44.560
move in closer and
closer to 'p'.
00:15:44.560 --> 00:15:46.960
And what we are saying is, is
that when 'q' assumes this
00:15:46.960 --> 00:15:51.290
position, let's call it 'q sub
1', the slope of the line that
00:15:51.290 --> 00:15:55.080
joints 'p' to 'q' one looks a
lot more like the slope of the
00:15:55.080 --> 00:15:59.040
line that will be tangent to
the curve at the point 'p'.
00:15:59.040 --> 00:16:03.140
And for this reason, what is
instantaneous rate of speed,
00:16:03.140 --> 00:16:07.220
when we're talking about
functions, become slopes of
00:16:07.220 --> 00:16:11.980
curves at a given point when we
are talking about curves.
00:16:11.980 --> 00:16:15.270
You see that the tangent line to
a curve is viewed as being
00:16:15.270 --> 00:16:19.395
the limiting position of a cord
drawn from a point 'p' to
00:16:19.395 --> 00:16:21.150
a point 'q'.
00:16:21.150 --> 00:16:24.220
Now at any rate, this will be
emphasized in the text.
00:16:24.220 --> 00:16:26.990
It will be emphasized
in our exercises.
00:16:26.990 --> 00:16:31.180
The major issue now is what did
we really do when it came
00:16:31.180 --> 00:16:34.620
time to compute this
little gadget.
00:16:34.620 --> 00:16:38.050
In other words, when we talked
about the limit of 'f of x' as
00:16:38.050 --> 00:16:41.980
'x' approached 'a', how
did we compute this?
00:16:41.980 --> 00:16:46.220
The danger is, and this is why
I have underlined a question
00:16:46.220 --> 00:16:48.980
mark here, the intuitive
approach is to say something
00:16:48.980 --> 00:16:53.550
like, let's just replace
x by a in here.
00:16:53.550 --> 00:16:56.360
In other words, let the limit of
'f of x', as 'x' approaches
00:16:56.360 --> 00:16:58.310
'a', be 'f of a'.
00:16:58.310 --> 00:17:03.110
In fact, this is what we did
with our '32 plus 16h'.
00:17:03.110 --> 00:17:06.122
In a way, what we did was, with
'32 plus 16h', do you
00:17:06.122 --> 00:17:06.970
remember what we did?
00:17:06.970 --> 00:17:10.260
We said, let's let 'h' get
arbitrarily close to 0.
00:17:10.260 --> 00:17:12.450
And in our minds, we
let 'h' equal 0.
00:17:12.450 --> 00:17:17.730
'16h' was then 0, and then '32
plus 16h' was then 32.
00:17:17.730 --> 00:17:20.140
But what we had done is somehow
or other, if our
00:17:20.140 --> 00:17:24.530
intuition is correct, we had
arrived at the correct answer
00:17:24.530 --> 00:17:26.300
but for wrong reasons.
00:17:26.300 --> 00:17:30.000
Because you see, in this
expression, it is mandatory,
00:17:30.000 --> 00:17:32.620
as we've outlined this,
is that 'x' never be
00:17:32.620 --> 00:17:36.030
allowed to equal 'a'.
00:17:36.030 --> 00:17:37.960
Well you say, who cares
whether you
00:17:37.960 --> 00:17:38.850
allow it or you don't?
00:17:38.850 --> 00:17:40.750
Aren't we going to get
the same answer?
00:17:40.750 --> 00:17:43.000
And just to give you a quickie,
one that we talked
00:17:43.000 --> 00:17:45.910
about in our introductory
lecture, consider the function
00:17:45.910 --> 00:17:48.690
'f of x' to be 'x squared
minus nine'
00:17:48.690 --> 00:17:51.080
over 'x minus three'.
00:17:51.080 --> 00:17:53.970
And let's evaluate what
this limit is as
00:17:53.970 --> 00:17:56.700
'x' approaches three.
00:17:56.700 --> 00:18:00.690
If we were to blindly assume
that all we have to do is
00:18:00.690 --> 00:18:04.150
replace 'x' by 'a', that
would mean what here?
00:18:04.150 --> 00:18:06.510
Replace 'x' by three.
00:18:06.510 --> 00:18:08.700
Notice that what
we got is what?
00:18:08.700 --> 00:18:12.980
Three squared, which is nine,
minus nine, (which is 0) over
00:18:12.980 --> 00:18:15.730
three minus three,
which is also 0.
00:18:15.730 --> 00:18:19.010
In other words, if in the
expression 'x squared minus
00:18:19.010 --> 00:18:23.770
nine' over 'x minus three' you
replace 'x' by three, you do
00:18:23.770 --> 00:18:25.750
get 0 over 0.
00:18:25.750 --> 00:18:30.270
Unfortunately, or perhaps
fortunately, somehow or other,
00:18:30.270 --> 00:18:33.310
this definition is supposed to
incorporate the fact that you
00:18:33.310 --> 00:18:36.860
never allow 'x' to
equal three.
00:18:36.860 --> 00:18:40.240
See, the way we got 0 over 0
was we violated what the
00:18:40.240 --> 00:18:41.960
intuitive meaning
of limit was.
00:18:41.960 --> 00:18:46.630
We wound up with this 0 over 0
mess by allowing 'x' to equal
00:18:46.630 --> 00:18:50.130
the one value it is not
allowed to equal here.
00:18:50.130 --> 00:18:54.250
To see what happened more
anatomically in this
00:18:54.250 --> 00:18:57.750
particular problem, let's go
back to our old friend of
00:18:57.750 --> 00:18:59.140
cancellation again.
00:18:59.140 --> 00:19:03.520
We observe that 'x squared minus
nine' can be written as
00:19:03.520 --> 00:19:07.930
'x plus three' times 'x minus
three', and then we divide by
00:19:07.930 --> 00:19:12.020
'x minus three', and here's
where the main problem seems
00:19:12.020 --> 00:19:12.710
to come up.
00:19:12.710 --> 00:19:16.420
One automatically, somehow or
other, gets into the habit of
00:19:16.420 --> 00:19:18.600
canceling these things out.
00:19:18.600 --> 00:19:21.160
But the point is, since we have
already agreed that you
00:19:21.160 --> 00:19:25.490
cannot divide by 0, the idea
then becomes that the only
00:19:25.490 --> 00:19:30.850
time this is permissible is when
'x' is unequal to three.
00:19:30.850 --> 00:19:36.990
In fact, to have been more
precise, what we should have
00:19:36.990 --> 00:19:42.160
said was not that 'x squared
minus nine' over 'x minus
00:19:42.160 --> 00:19:46.310
three' equals 'x plus three',
but rather, 'x squared minus
00:19:46.310 --> 00:19:51.850
nine' over 'x minus three'
equals 'x plus three', unless
00:19:51.850 --> 00:19:53.820
'x' equals three.
00:19:53.820 --> 00:19:57.470
And by the way, I should
add here what?
00:19:57.470 --> 00:20:04.820
That if 'x' equals three, then
'x squared minus nine' over 'x
00:20:04.820 --> 00:20:08.290
minus three' is undefined.
00:20:08.290 --> 00:20:10.850
You see, that's what
we call 0 over 0.
00:20:10.850 --> 00:20:14.000
Undefined, indeterminate.
00:20:14.000 --> 00:20:16.570
But here's what bails us out.
00:20:16.570 --> 00:20:20.410
As soon as we say the limit as
'x' approaches three, what are
00:20:20.410 --> 00:20:21.490
we assuming?
00:20:21.490 --> 00:20:27.070
We're assuming that 'x' can be
any value at all except three,
00:20:27.070 --> 00:20:30.370
and the interesting point now
becomes, that as long as 'x'
00:20:30.370 --> 00:20:34.120
is anything but three, the only
time you could tell these
00:20:34.120 --> 00:20:39.000
two expressions apart was when
'x' happened to equal three.
00:20:39.000 --> 00:20:43.550
So if you now say we won't allow
'x' to equal three, then
00:20:43.550 --> 00:20:47.450
it is true, whereas the
bracketed expression by itself
00:20:47.450 --> 00:20:49.200
is not equal to the
parenthetical
00:20:49.200 --> 00:20:50.530
expression by itself.
00:20:50.530 --> 00:20:54.520
The limit as 'x' approaches
three of these two expressions
00:20:54.520 --> 00:20:58.160
are equal, and in fact, one
becomes tempted to say "what?"
00:20:58.160 --> 00:20:59.110
at this case.
00:20:59.110 --> 00:21:02.920
Oh, here it's quite easy to see,
that as 'x' approaches
00:21:02.920 --> 00:21:08.520
three, 'x plus three'
is equal to six.
00:21:08.520 --> 00:21:11.330
And let me put a question mark
here also, because it
00:21:11.330 --> 00:21:15.120
certainly is true that if you
replace 'x' by three, this
00:21:15.120 --> 00:21:17.310
will be six.
00:21:17.310 --> 00:21:20.600
But this says don't replace
'x' by three.
00:21:20.600 --> 00:21:24.030
Now before, when we replaced
'x' by three we got into
00:21:24.030 --> 00:21:25.860
trouble, so we said
we can't do that.
00:21:25.860 --> 00:21:29.280
Now we replace 'x' by three, we
get an answer that we like,
00:21:29.280 --> 00:21:32.210
and that is the danger that
we'll say OK, we won't allow
00:21:32.210 --> 00:21:35.860
this to be done, unless
we like the result.
00:21:35.860 --> 00:21:38.710
And of course this gives you
a highly subjective subject
00:21:38.710 --> 00:21:39.640
matter here.
00:21:39.640 --> 00:21:42.400
You see, the point is that
somehow or other we must come
00:21:42.400 --> 00:21:46.070
up with a more objective
criteria for distinguishing
00:21:46.070 --> 00:21:47.700
what limits are.
00:21:47.700 --> 00:21:51.200
And by the way, for the Polyanna
who has the intuitive
00:21:51.200 --> 00:21:55.370
feeling for saying things like
well look at, this may give
00:21:55.370 --> 00:21:58.390
you trouble if you wind
up with 0 over 0.
00:21:58.390 --> 00:22:02.240
But what is the likelihood that
given 'f of x' at random,
00:22:02.240 --> 00:22:05.270
and 'a' at random, that the
limit of 'f of x' as 'x'
00:22:05.270 --> 00:22:09.060
approaches 'a' is going to be
0 over 0 if you're careless?
00:22:09.060 --> 00:22:11.950
And the answer is that the
probability is quite small,
00:22:11.950 --> 00:22:16.540
except in Calculus, in which
case, every time, you take a
00:22:16.540 --> 00:22:17.460
derivative.
00:22:17.460 --> 00:22:21.470
Every time you are going to
wind up with 0 over 0.
00:22:21.470 --> 00:22:24.700
This is why we said earlier that
differential Calculus has
00:22:24.700 --> 00:22:28.360
been referred to as 'the
study of 0 over 0'.
00:22:28.360 --> 00:22:31.950
You see, the whole point is,
let's go back now to our basic
00:22:31.950 --> 00:22:36.190
definition of average rate of
change of 'f', 'f of x1' plus
00:22:36.190 --> 00:22:40.970
'delta x' minus 'f of
x1' over 'delta x'.
00:22:40.970 --> 00:22:43.430
I guess when you're writing in
black, you have to use white
00:22:43.430 --> 00:22:46.150
for emphasis, so
we'll use this.
00:22:46.150 --> 00:22:49.800
Notice that every time you let
'delta x' equal zero, you are
00:22:49.800 --> 00:22:51.230
going to wind up with what?
00:22:51.230 --> 00:22:55.790
'f of x1' minus 'f of x1' ,
which is 0, over 'delta x',
00:22:55.790 --> 00:22:56.720
which is 0.
00:22:56.720 --> 00:22:59.330
In other words, you are
going to get 0 over 0.
00:23:02.850 --> 00:23:07.260
If you believe that all this
means is to replace 'delta x'
00:23:07.260 --> 00:23:11.070
by 0, you see, and the point is,
this is not the definition
00:23:11.070 --> 00:23:14.130
of this, and this is why you
don't wind up with this
00:23:14.130 --> 00:23:15.436
particular thing.
00:23:15.436 --> 00:23:19.100
Well I've spent enough time,
I think, trying to give a
00:23:19.100 --> 00:23:22.950
picture as to what limits mean
intuitively, now what I'd like
00:23:22.950 --> 00:23:26.340
to do is to devote the remainder
of this lecture,
00:23:26.340 --> 00:23:30.320
plus our next one, to
investigating, more
00:23:30.320 --> 00:23:35.470
stringently, what limit means in
a way that is unambiguous.
00:23:35.470 --> 00:23:37.650
And you know, it's like
everything else in this world.
00:23:37.650 --> 00:23:41.400
When you want to get something
that is ironclad, you don't
00:23:41.400 --> 00:23:43.150
get something for nothing.
00:23:43.150 --> 00:23:48.560
To get an ironclad expression
that doesn't lend itself to
00:23:48.560 --> 00:23:52.020
misinterpretations usually
involves a rigorous enough
00:23:52.020 --> 00:23:55.050
language that scares off
the uninitiated.
00:23:55.050 --> 00:23:57.020
Sort of like a legal document.
00:23:57.020 --> 00:23:59.600
The chances are if you're not
a trained lawyer, and the
00:23:59.600 --> 00:24:02.450
document is simple enough for
you to understand, it must
00:24:02.450 --> 00:24:04.540
have thousands of
loopholes in it.
00:24:04.540 --> 00:24:07.740
But that same rigorous language
that makes it almost
00:24:07.740 --> 00:24:11.320
impossible for the layman to
understand what's happening,
00:24:11.320 --> 00:24:14.660
is precisely the language that
the lawyer needs to make the
00:24:14.660 --> 00:24:17.950
document well defined
to him, hopefully.
00:24:17.950 --> 00:24:21.410
Well with that as a background,
let's now try to
00:24:21.410 --> 00:24:24.560
give a more rigorous definition
of limit.
00:24:24.560 --> 00:24:27.380
And let's try to do it in such
a way that not only will the
00:24:27.380 --> 00:24:31.540
definition be rigorous, but it
will agree, to the best of our
00:24:31.540 --> 00:24:34.470
ability, with what we
intuitively believe a limit
00:24:34.470 --> 00:24:35.660
should mean.
00:24:35.660 --> 00:24:38.710
Now what I want to do here is
scare you a little bit, and
00:24:38.710 --> 00:24:40.980
the reason I want to scare you
a little bit is, I want to
00:24:40.980 --> 00:24:44.650
show you, once and for all,
how in many cases, what
00:24:44.650 --> 00:24:48.120
appears to be ominous
mathematical notation, is
00:24:48.120 --> 00:24:51.270
simply a rigorous way of saying
something which was
00:24:51.270 --> 00:24:53.840
quite intuitive pictorially
to do.
00:24:53.840 --> 00:24:56.630
The mathematical definition
of limit is the following.
00:24:56.630 --> 00:25:00.330
The limit of 'f of x' as 'x'
approaches 'a' equals 'l'
00:25:00.330 --> 00:25:04.660
means, and get this, given
epsilon greater than 0, we can
00:25:04.660 --> 00:25:08.070
find delta greater than 0, such
that when the absolute
00:25:08.070 --> 00:25:11.370
value of 'x minus a' is less
than delta, but greater than
00:25:11.370 --> 00:25:15.750
0, then the absolute value of
'f of x' minus 'l' is less
00:25:15.750 --> 00:25:17.210
than epsilon.
00:25:17.210 --> 00:25:19.850
Now doesn't that kind
of turn you on?
00:25:19.850 --> 00:25:21.460
See, this is the
whole problem.
00:25:21.460 --> 00:25:25.650
I will now cut you in sort of on
the trade secret that makes
00:25:25.650 --> 00:25:28.510
this thing very,
very readable.
00:25:28.510 --> 00:25:32.170
First of all, to say that you
want to get arbitrarily close
00:25:32.170 --> 00:25:36.210
in value, you want 'f of x' to
be arbitrarily close to 'l',
00:25:36.210 --> 00:25:38.020
another way of saying
that is what?
00:25:38.020 --> 00:25:41.460
You specify any tolerance limit
whatsoever, which we
00:25:41.460 --> 00:25:45.775
call epsilon, and I can make 'f
of x' get to within epsilon
00:25:45.775 --> 00:25:49.870
of 'l' just by choosing 'x'
sufficiently close to 'a'.
00:25:49.870 --> 00:25:53.640
And by the way again, in terms
of geometry, that's all the
00:25:53.640 --> 00:25:55.780
absolute value of a
difference means.
00:25:55.780 --> 00:25:59.330
The absolute value of 'x minus
a' simply means what?
00:25:59.330 --> 00:26:02.030
The distance between
'x' and 'a'.
00:26:02.030 --> 00:26:03.750
All this says is what?
00:26:03.750 --> 00:26:09.980
I can make 'f of x' fall within
epsilon of 'l' by
00:26:09.980 --> 00:26:15.050
finding a delta such that 'x' is
within delta of 'a', and by
00:26:15.050 --> 00:26:20.090
the way, this little tidbit over
here is simply a fancy
00:26:20.090 --> 00:26:23.600
way of saying that you're not
letting 'x' equal 'a'.
00:26:23.600 --> 00:26:25.860
You see, the only way the
absolute value of a number can
00:26:25.860 --> 00:26:28.770
be 0 is if the number
itself is 0.
00:26:28.770 --> 00:26:32.270
The only way this could be 0
is if 'x' equals 'a', so by
00:26:32.270 --> 00:26:35.360
saying this little tidbit over
here, that this is greater
00:26:35.360 --> 00:26:38.800
than 0, that's our fancy way
of saying that we are not
00:26:38.800 --> 00:26:42.050
going to allow 'x'
to equal to 'a'.
00:26:42.050 --> 00:26:45.430
Now again, whereas this may
sound a little bit simpler
00:26:45.430 --> 00:26:48.740
than the original definition,
let's see how much easier the
00:26:48.740 --> 00:26:50.060
picture makes it.
00:26:50.060 --> 00:26:52.970
Let's go to a graph of
'y' equals 'f of x'.
00:26:56.860 --> 00:27:00.390
And in my diagram here, here is
'a', and here is 'l', and
00:27:00.390 --> 00:27:04.070
what I'm saying is, I
intuitively suspect.
00:27:04.070 --> 00:27:05.530
Let me write down
what I suspect.
00:27:05.530 --> 00:27:09.620
I suspect that the limit of 'f
of x', as 'x' approaches 'a',
00:27:09.620 --> 00:27:11.580
is 'l' in this case.
00:27:11.580 --> 00:27:14.900
Namely I suspect that as 'x'
gets closer and closer to 'a',
00:27:14.900 --> 00:27:18.950
no matter which side you come
in on, that's why we use
00:27:18.950 --> 00:27:19.790
absolute value.
00:27:19.790 --> 00:27:22.550
It makes no difference whether
it's positive or negative.
00:27:22.550 --> 00:27:25.950
As 'x' gets in tighter and
tighter on 'a', I suspect that
00:27:25.950 --> 00:27:30.290
I can make 'f of x' come
arbitrarily equal to 'l'.
00:27:30.290 --> 00:27:33.260
In other words, if 'x' is close
enough to 'a', 'f of x',
00:27:33.260 --> 00:27:35.310
it seems, will be
close to 'l'.
00:27:35.310 --> 00:27:38.310
And by the way, to complete this
problem geometrically,
00:27:38.310 --> 00:27:39.990
pick your epsilon.
00:27:39.990 --> 00:27:43.220
Let's say that epsilon is this
'y', and since epsilon is
00:27:43.220 --> 00:27:45.960
positive, this will be called
'l plus epsilon'.
00:27:45.960 --> 00:27:49.920
This will be called
'l minus epsilon'.
00:27:49.920 --> 00:27:53.010
Now what we do is we come
over to the curve here.
00:27:56.460 --> 00:27:59.190
All right, hopefully these
will look like horizontal
00:27:59.190 --> 00:28:00.295
lines to you.
00:28:00.295 --> 00:28:03.330
When I get down to here, I then
project down vertical
00:28:03.330 --> 00:28:04.580
lines to meet the x-axis.
00:28:07.230 --> 00:28:10.140
And now all I'm saying, and I
think this is fairly intuitive
00:28:10.140 --> 00:28:12.790
clear, is what?
00:28:12.790 --> 00:28:18.810
That if 'x' is this close to
'a', 'f of x' will be within
00:28:18.810 --> 00:28:20.140
these tolerance limits of 'l'.
00:28:20.140 --> 00:28:23.750
In other words, for any value of
'x' in here, see, for this
00:28:23.750 --> 00:28:27.960
neighborhood of 'a', every
input has its output come
00:28:27.960 --> 00:28:30.610
within epsilon of 'l'.
00:28:30.610 --> 00:28:33.500
By the way, just one brief
note here, something that
00:28:33.500 --> 00:28:36.410
we'll have to come back to in
more detail-- in fact, I'll do
00:28:36.410 --> 00:28:38.290
that in our next example--
00:28:38.290 --> 00:28:47.560
notice here how important it
was that our curve was not
00:28:47.560 --> 00:28:49.160
single value, but one to one.
00:28:49.160 --> 00:28:52.180
In other words, suppose our
curve had doubled back, say
00:28:52.180 --> 00:28:53.780
something like this.
00:28:53.780 --> 00:28:57.710
Notice then, when you try to
project over to the curve from
00:28:57.710 --> 00:29:01.090
'l plus epsilon', you don't know
whether to stop at this
00:29:01.090 --> 00:29:02.810
point or at this point.
00:29:02.810 --> 00:29:05.260
You see, in other words, notice
that there is another
00:29:05.260 --> 00:29:09.690
point down here which has the
same 'l plus epsilon' value.
00:29:09.690 --> 00:29:12.560
In other words, whenever our
function is not one to one,
00:29:12.560 --> 00:29:14.490
there is going to be a
problem, when we work
00:29:14.490 --> 00:29:18.470
analytically, of being able to
solve algebraic equations, and
00:29:18.470 --> 00:29:21.620
trying to distinguish this point
of intersection from
00:29:21.620 --> 00:29:22.440
this point.
00:29:22.440 --> 00:29:26.650
You see, if we made a mistake
and bypass this point and came
00:29:26.650 --> 00:29:29.920
all the way over to here, and
thought that this was our
00:29:29.920 --> 00:29:32.840
tolerance limit, we would be in
a great deal of difficulty.
00:29:32.840 --> 00:29:37.680
Because you see, for example,
for this value of 'x', its 'f
00:29:37.680 --> 00:29:42.720
of x' value projects up here,
which is outside of the
00:29:42.720 --> 00:29:45.120
tolerance limits
that are given.
00:29:45.120 --> 00:29:47.950
You see, many things which are
self evident from the picture
00:29:47.950 --> 00:29:52.180
are not nearly so self evident
analytically, to which we then
00:29:52.180 --> 00:29:55.850
raise the question, why
use analysis at all?
00:29:55.850 --> 00:29:57.560
Let's just use pictures.
00:29:57.560 --> 00:30:00.030
And again the answer is, in
this case, we can get away
00:30:00.030 --> 00:30:02.640
with it, but in more complicated
situations where
00:30:02.640 --> 00:30:04.760
either we can't draw the diagram
because it's too
00:30:04.760 --> 00:30:08.300
complicated, or because we have
several variables, and we
00:30:08.300 --> 00:30:11.400
get into a dimension problem,
the point is that sometimes we
00:30:11.400 --> 00:30:14.490
have no recourse other
than the analysis.
00:30:14.490 --> 00:30:18.820
Let me give you an exercise
that we'll do in terms of
00:30:18.820 --> 00:30:22.990
putting the geometry and the
analysis side by side.
00:30:22.990 --> 00:30:27.750
Let's look at the expression the
limit of 'x squared minus
00:30:27.750 --> 00:30:30.590
2x' as 'x' approaches three.
00:30:30.590 --> 00:30:33.900
Now of course, we don't want 'x'
to equal three here, but
00:30:33.900 --> 00:30:37.840
let's cheat again, and see
what this thing means.
00:30:37.840 --> 00:30:41.950
Our intuition tells us that when
'x' is three, 'x squared'
00:30:41.950 --> 00:30:44.940
is nine, '2x' is six.
00:30:44.940 --> 00:30:48.810
So nine minus six is three, and
we would suspect that the
00:30:48.810 --> 00:30:51.600
limit here should be three.
00:30:51.600 --> 00:30:56.990
Now to see what this means,
let's draw a little diagram.
00:30:56.990 --> 00:31:00.610
The graph 'y' equals 'x squared
minus 2x' crosses the
00:31:00.610 --> 00:31:03.040
x-axis at 0 and two.
00:31:03.040 --> 00:31:07.140
It has its low point at the
0.1 comma minus one.
00:31:07.140 --> 00:31:10.620
We can sketch these curves,
it's a parabola.
00:31:10.620 --> 00:31:21.520
And now what we're saying is, is
it true that when 'x' gets
00:31:21.520 --> 00:31:24.890
very close to three,
that 'f of x' gets
00:31:24.890 --> 00:31:26.790
very close to three?
00:31:26.790 --> 00:31:29.640
And the answer is, from the
diagram, it appears very
00:31:29.640 --> 00:31:31.550
obvious that this is the case.
00:31:31.550 --> 00:31:34.840
Fact, here's that example of
non-single value again.
00:31:34.840 --> 00:31:37.860
If you call this thing 'l',
and you now pick tolerance
00:31:37.860 --> 00:31:41.310
limits which we'll call 'l plus
epsilon' and 'l minus
00:31:41.310 --> 00:31:44.740
epsilon', and now you want to
see what interval you need on
00:31:44.740 --> 00:31:49.800
the x-axis, notice that in this
particular diagram, you
00:31:49.800 --> 00:31:54.600
will get two intervals, one of
which surrounds the value 'x'
00:31:54.600 --> 00:31:57.350
equals minus one, and the other
of which surrounds the
00:31:57.350 --> 00:31:59.620
'x' value 'x' equals three.
00:31:59.620 --> 00:32:03.400
In other words, both when 'x' is
minus one and 'x' is three,
00:32:03.400 --> 00:32:07.690
'x squared minus 2x'
will equal three.
00:32:07.690 --> 00:32:11.350
At any rate, leaving the diagram
as an aid, let's see
00:32:11.350 --> 00:32:14.650
what our epsilon delta
definition says, and how we
00:32:14.650 --> 00:32:17.380
handle the stuff algebraically,
and how we
00:32:17.380 --> 00:32:20.370
correlate the algebra
with the geometry.
00:32:22.870 --> 00:32:25.970
Given epsilon greater than
0, what must I do?
00:32:25.970 --> 00:32:29.860
I must find a delta greater
than 0 such that, and I'm
00:32:29.860 --> 00:32:31.770
going to read this colloquially,
I'll write it
00:32:31.770 --> 00:32:34.600
formally, but read it
colloquially, such that
00:32:34.600 --> 00:32:39.950
whenever 'x' is within delta
of three, but not equal to
00:32:39.950 --> 00:32:44.340
three, then 'x squared
minus 2x' will be
00:32:44.340 --> 00:32:47.080
within epsilon of three.
00:32:47.080 --> 00:32:48.350
Now here's the whole point.
00:32:48.350 --> 00:32:52.360
We know, algebraically, how
to handle absolute values.
00:32:52.360 --> 00:32:56.700
Namely, the absolute value of
'x squared minus 2x minus
00:32:56.700 --> 00:32:59.730
three' being less than epsilon,
is the same as saying
00:32:59.730 --> 00:33:04.180
that 'x squared minus 2x minus
three' itself must be between
00:33:04.180 --> 00:33:07.100
epsilon and minus epsilon.
00:33:07.100 --> 00:33:10.300
Now again, we rewrite things
in fancy ways if we wish.
00:33:10.300 --> 00:33:11.850
There other ways
of doing this.
00:33:11.850 --> 00:33:13.760
I call this completing
the square.
00:33:13.760 --> 00:33:16.215
Well, people for 2,000 years
have called this completing
00:33:16.215 --> 00:33:17.280
the square.
00:33:17.280 --> 00:33:21.630
Namely, I rewrite minus three as
one minus four, so that 'x
00:33:21.630 --> 00:33:24.580
squared minus 2x plus one'
can be factored as
00:33:24.580 --> 00:33:26.500
'x minus one squared'.
00:33:26.500 --> 00:33:28.750
I'm now down to this
form here.
00:33:28.750 --> 00:33:32.360
Then I add four to all sides
of my inequality, and have
00:33:32.360 --> 00:33:36.210
that 'x minus one squared' is
between four plus epsilon and
00:33:36.210 --> 00:33:37.460
four minus epsilon.
00:33:37.460 --> 00:33:41.330
Some fairly elementary Algebra
of inequalities here.
00:33:41.330 --> 00:33:42.660
Now here's the key point.
00:33:42.660 --> 00:33:46.060
Remembering my diagram, I said
to you, how do we know whether
00:33:46.060 --> 00:33:48.300
we're near three or
near minus one?
00:33:48.300 --> 00:33:51.300
How do we distinguish, when we
draw the lines to the curve,
00:33:51.300 --> 00:33:52.890
what neighborhood we want?
00:33:52.890 --> 00:33:55.510
And notice that all we're saying
is that if 'x' is near
00:33:55.510 --> 00:33:58.560
three, that means what?
'x' is close to three.
00:33:58.560 --> 00:34:01.910
If 'x' is reasonably close to
three, then 'x minus one' is
00:34:01.910 --> 00:34:06.190
going to be reasonably close to
two, and hence be positive.
00:34:06.190 --> 00:34:08.920
Point is, that as long as you're
dealing with positive
00:34:08.920 --> 00:34:11.940
numbers over here, see if
epsilon is sufficiently small,
00:34:11.940 --> 00:34:14.070
these will then all be
positive numbers.
00:34:14.070 --> 00:34:18.100
For positive numbers, if the
squares obey a certain
00:34:18.100 --> 00:34:22.730
inequality, the square roots
will obey the same inequality,
00:34:22.730 --> 00:34:26.090
as we have emphasized in
one of our exercises.
00:34:26.090 --> 00:34:31.139
In other words, from here, we
can now say this, and from
00:34:31.139 --> 00:34:34.409
this, we can now conclude, that
if you want 'x squared
00:34:34.409 --> 00:34:39.710
minus 2x' to be within epsilon
of three, 'x' itself is going
00:34:39.710 --> 00:34:42.850
to have to be between 'one plus
the square root of 'four
00:34:42.850 --> 00:34:45.699
plus epsilon'', and 'one
plus the square root
00:34:45.699 --> 00:34:47.190
of 'four minus epsilon''.
00:34:47.190 --> 00:34:50.030
And by the way, this is quite
rigorous, but I don't think it
00:34:50.030 --> 00:34:50.639
turns you on.
00:34:50.639 --> 00:34:52.989
I don't think there's any
picture that you associate
00:34:52.989 --> 00:34:56.159
with this, and so what I thought
I'd like to do for our
00:34:56.159 --> 00:35:01.960
next little thing, is to come
back to our diagram here, and
00:35:01.960 --> 00:35:04.800
show you what this
really means.
00:35:04.800 --> 00:35:09.850
Namely, notice that if epsilon
is a small positive number,
00:35:09.850 --> 00:35:11.060
let's take a look
at this again.
00:35:11.060 --> 00:35:14.260
If epsilon is a small positive
number, this is
00:35:14.260 --> 00:35:16.420
slightly less than four.
00:35:16.420 --> 00:35:19.280
Therefore, the square root is
slightly less than two,
00:35:19.280 --> 00:35:22.560
therefore, this number will be
slightly less than three.
00:35:22.560 --> 00:35:25.300
On the other hand, 'four plus
epsilon' is slightly more than
00:35:25.300 --> 00:35:29.080
four, so its square root will be
slightly more than two, and
00:35:29.080 --> 00:35:32.320
therefore, one plus that square
root will be slightly
00:35:32.320 --> 00:35:33.840
more than three.
00:35:33.840 --> 00:35:36.890
And what these two numbers, as
abstract as they look like,
00:35:36.890 --> 00:35:39.660
correspond to, is nothing
more than this.
00:35:39.660 --> 00:35:43.870
Coming back to our diagram and
choosing an epsilon, and
00:35:43.870 --> 00:35:47.980
coming over to the curve like
this, and projecting down like
00:35:47.980 --> 00:35:51.990
this, all we're saying is,
see, what are we saying
00:35:51.990 --> 00:35:52.970
pictorially?
00:35:52.970 --> 00:35:58.130
That for 'f of x' to be within
epsilon of 'l', 'x' itself has
00:35:58.130 --> 00:36:00.260
to be in this range here.
00:36:00.260 --> 00:36:05.950
And all this says is, this very
simple point to compute,
00:36:05.950 --> 00:36:09.740
just by projecting down like
this, its rigorous name would
00:36:09.740 --> 00:36:13.560
be 'one plus the square root
of 'four minus epsilon''.
00:36:13.560 --> 00:36:16.420
That's this number over here.
00:36:16.420 --> 00:36:20.460
And the furthest point, namely
this point here, which again,
00:36:20.460 --> 00:36:24.490
in terms of the picture is very
easy to find, that point
00:36:24.490 --> 00:36:28.220
corresponds to one plus
the square root
00:36:28.220 --> 00:36:31.660
of 'four plus epsilon'.
00:36:31.660 --> 00:36:35.390
And again notice, in terms
of the Algebra, I need no
00:36:35.390 --> 00:36:36.870
recourse to a picture.
00:36:36.870 --> 00:36:40.460
But in terms of the picture, I
get a heck of a good feeling
00:36:40.460 --> 00:36:43.960
as to what these abstract
symbols are telling me.
00:36:43.960 --> 00:36:48.830
Well let's continue on with the
solution of this exercise.
00:36:48.830 --> 00:36:51.760
You see, we didn't want to find
where 'x' was, we wanted
00:36:51.760 --> 00:36:54.360
to see where 'x minus
three' had to be.
00:36:54.360 --> 00:36:56.400
In other words, we're trying
to find the delta.
00:36:56.400 --> 00:37:00.290
We know that 'x minus three'
is between these
00:37:00.290 --> 00:37:01.830
two extremes here.
00:37:01.830 --> 00:37:04.460
Consequently, and here's a place
we have to be a little
00:37:04.460 --> 00:37:07.840
bit careful here, this of
course, is a positive number,
00:37:07.840 --> 00:37:10.450
because this is more than two.
00:37:10.450 --> 00:37:13.200
This, on the other hand, is a
negative number, because you
00:37:13.200 --> 00:37:16.640
see 'four minus epsilon' is less
than four, so its square
00:37:16.640 --> 00:37:18.940
root is less than two.
00:37:18.940 --> 00:37:21.490
So if I subtract two from--
it's a negative number--
00:37:21.490 --> 00:37:24.580
and since delta has to be
positive, if this thing is
00:37:24.580 --> 00:37:28.880
negative, two minus the square
root of 'four minus epsilon'
00:37:28.880 --> 00:37:30.730
will be positive.
00:37:30.730 --> 00:37:35.040
So what I do is, to solve this
problem, is I simply choose
00:37:35.040 --> 00:37:39.530
delta to be the minimum
of these two numbers.
00:37:39.530 --> 00:37:42.330
And then by definition,
what do I have?
00:37:42.330 --> 00:37:45.280
That when the absolute value
of 'x minus three' is less
00:37:45.280 --> 00:37:48.920
than delta, and at the same
time, this is crucial,
00:37:48.920 --> 00:37:49.770
greater than 0.
00:37:49.770 --> 00:37:52.600
In other words, we never
let 'x' equal three.
00:37:52.600 --> 00:37:56.570
For this choice of delta, by how
we worked backwards, this
00:37:56.570 --> 00:37:59.920
will come out to be
less than epsilon.
00:37:59.920 --> 00:38:03.200
And that is exactly what you
mean by our formal definition
00:38:03.200 --> 00:38:06.140
of saying that the limit of 'x
squared minus 2x' as 'x'
00:38:06.140 --> 00:38:08.980
approaches three, in this
case, equals three.
00:38:08.980 --> 00:38:12.000
For someone who wants more
concrete evidence, I think all
00:38:12.000 --> 00:38:17.860
one has to do is, for example,
take a number like 'one plus
00:38:17.860 --> 00:38:20.480
the square root of 'four plus
epsilon'', that is the widest
00:38:20.480 --> 00:38:24.240
point on our interval, and
actually plug that in for 'x',
00:38:24.240 --> 00:38:27.830
and go through this computation
of squaring 'one
00:38:27.830 --> 00:38:30.190
plus the square root of
'four plus epsilon''.
00:38:30.190 --> 00:38:32.890
Subtract twice, 'one plus
the square root
00:38:32.890 --> 00:38:34.390
of 'four plus epsilon''.
00:38:34.390 --> 00:38:37.720
Carry out that computation
and you get what?
00:38:37.720 --> 00:38:39.750
'Three plus epsilon'.
00:38:39.750 --> 00:38:43.150
That using the upper extreme,
you wind up with what?
00:38:43.150 --> 00:38:48.570
In excess of three by epsilon,
which by the way is exactly
00:38:48.570 --> 00:38:49.740
what's supposed to happen.
00:38:49.740 --> 00:38:52.560
In fact, I hope this doesn't
give you an eye sore, I will
00:38:52.560 --> 00:38:56.180
pull down the top board again
to show you also what these
00:38:56.180 --> 00:38:59.180
two distances mean, and
what this means.
00:38:59.180 --> 00:39:02.390
You see, all these two numbers
mean is the following, and
00:39:02.390 --> 00:39:04.910
I'll pull this down just
far enough to see it.
00:39:04.910 --> 00:39:10.140
That these two numbers were the
widths from this end point
00:39:10.140 --> 00:39:14.550
to three on the one extreme,
and from this end point to
00:39:14.550 --> 00:39:17.270
three on the other extreme.
00:39:17.270 --> 00:39:21.170
In other words, those two
numbers that we took the
00:39:21.170 --> 00:39:25.250
minimum of were the half-width
intervals
00:39:25.250 --> 00:39:26.880
that surrounded three.
00:39:26.880 --> 00:39:29.550
You see the problem, as we
mentioned before is, is that
00:39:29.550 --> 00:39:33.330
even though epsilon and minus
epsilon are symmetric with
00:39:33.330 --> 00:39:37.340
respect to 'l', when you
translate over to the curve,
00:39:37.340 --> 00:39:40.260
because this curve is not a
straight line when you project
00:39:40.260 --> 00:39:43.610
down, these two widths
will not be equal.
00:39:46.750 --> 00:39:49.030
By picking the minimum of
these two widths, you
00:39:49.030 --> 00:39:52.740
guarantee that you have
locked yourself inside
00:39:52.740 --> 00:39:54.000
the required area.
00:39:54.000 --> 00:39:57.710
You see, that's all this
particular thing means.
00:39:57.710 --> 00:40:01.550
Now the trouble is that one
might now get the idea that
00:40:01.550 --> 00:40:05.120
there are no shorter ways of
doing the same problem.
00:40:05.120 --> 00:40:07.580
You see, I showed you
a rigorous way.
00:40:07.580 --> 00:40:10.770
There was no law that said we
had to find the biggest delta.
00:40:10.770 --> 00:40:14.485
In other words, if delta equals,
a half will work, in
00:40:14.485 --> 00:40:16.670
other words, if you're within
a half of three, something's
00:40:16.670 --> 00:40:19.060
going to happen, then certainly
any smaller number
00:40:19.060 --> 00:40:21.410
will work just as well.
00:40:21.410 --> 00:40:23.020
In other words, I can
get estimates.
00:40:23.020 --> 00:40:24.790
Let me show you what
I mean by that.
00:40:24.790 --> 00:40:28.660
Another way of tackling how to
make 'x squared minus 2x minus
00:40:28.660 --> 00:40:32.580
three' smaller than epsilon is
to use our properties of
00:40:32.580 --> 00:40:35.180
absolute values, and
to factor this.
00:40:35.180 --> 00:40:37.880
In other words, we know that
'x squared minus 2x minus
00:40:37.880 --> 00:40:41.040
three' is 'x minus three'
times 'x plus one'.
00:40:41.040 --> 00:40:44.560
We know that the absolute
value of a product is a
00:40:44.560 --> 00:40:47.360
product of the absolute values,
so these two things
00:40:47.360 --> 00:40:54.050
here are synonyms. Then we
know that near 'x' equals
00:40:54.050 --> 00:40:56.410
three, which we're interested
in here, 'x plus
00:40:56.410 --> 00:40:59.550
one' is near four.
00:40:59.550 --> 00:41:02.700
Well what we mean more
rigorously is this, choose an
00:41:02.700 --> 00:41:06.430
epsilon, and if the absolute
value of 'x minus three' is
00:41:06.430 --> 00:41:09.560
less than epsilon, in other
words, if 'x minus three' is
00:41:09.560 --> 00:41:13.040
less than epsilon but greater
than minus epsilon, then by
00:41:13.040 --> 00:41:16.400
adding on four to all three
sides of the inequality here,
00:41:16.400 --> 00:41:20.390
we see that 'x plus one' is less
than 'four plus epsilon',
00:41:20.390 --> 00:41:22.140
but greater than 'four
minus epsilon'.
00:41:31.650 --> 00:41:34.550
If the size of epsilon is less
than one, then certainly this
00:41:34.550 --> 00:41:38.490
is less than five, and this
is greater than three.
00:41:38.490 --> 00:41:41.200
Now of course, somebody may say
to us, but what if epsilon
00:41:41.200 --> 00:41:42.070
is more than one?
00:41:42.070 --> 00:41:47.770
Well obviously, if a guy says
make this thing within 10, and
00:41:47.770 --> 00:41:50.190
I can make it within one,
certainly within
00:41:50.190 --> 00:41:51.600
10 is within one.
00:41:51.600 --> 00:41:53.870
I can always pick the
smaller number.
00:41:53.870 --> 00:41:57.140
It's like that nonsense of the
fellow asking for an ice cream
00:41:57.140 --> 00:41:59.240
cone, and the waitress said
what flavor, and he said
00:41:59.240 --> 00:42:00.460
anything except chocolate.
00:42:00.460 --> 00:42:01.880
And she said, I'm all out of
chocolate, will you take
00:42:01.880 --> 00:42:03.540
anything except vanilla?
00:42:03.540 --> 00:42:05.600
The idea here is that if
somebody says make this less
00:42:05.600 --> 00:42:09.240
than 15, if you've made it less
than 10, in particular,
00:42:09.240 --> 00:42:10.840
you've made it less than 15.
00:42:10.840 --> 00:42:13.140
And so all we do over
here, you see, is
00:42:13.140 --> 00:42:13.890
something like this.
00:42:13.890 --> 00:42:17.780
We say look it, if we want to
make this number here very
00:42:17.780 --> 00:42:22.170
small, we know that this number
here is less than five,
00:42:22.170 --> 00:42:25.710
we know that 'epsilon over five'
is a positive number if
00:42:25.710 --> 00:42:29.640
epsilon is small, so why don't
we just pick the absolute
00:42:29.640 --> 00:42:34.100
value of 'x minus three' to be
less than 'epsilon over five?'
00:42:34.100 --> 00:42:36.260
In fact, that's what this delta
will be in that case.
00:42:36.260 --> 00:42:38.590
In other words, if the absolute
value of x minus
00:42:38.590 --> 00:42:48.820
three is within 'epsilon over
five', if 'x' is within
00:42:48.820 --> 00:42:51.620
'epsilon over five' of
three, notice what
00:42:51.620 --> 00:42:53.040
this product becomes.
00:42:53.040 --> 00:42:56.160
This is less than 'epsilon over
five', this we know from
00:42:56.160 --> 00:43:00.640
before is less than five, and
therefore, this product is
00:43:00.640 --> 00:43:02.080
less than epsilon.
00:43:02.080 --> 00:43:05.640
In other words, we have
exhibited the delta such that
00:43:05.640 --> 00:43:09.040
when this happens, the absolute
value of what we want
00:43:09.040 --> 00:43:10.860
to make less than
epsilon indeed
00:43:10.860 --> 00:43:13.090
becomes less than epsilon.
00:43:13.090 --> 00:43:18.210
Now because I recognize that
this is a hard topic, you'll
00:43:18.210 --> 00:43:21.290
notice in the reading
assignments that these
00:43:21.290 --> 00:43:24.120
problems are covered
in great detail.
00:43:24.120 --> 00:43:28.220
Everything that I've said in the
lesson so far is repeated
00:43:28.220 --> 00:43:31.920
in great computational depth
in our learning exercises.
00:43:31.920 --> 00:43:35.350
The point is that even if I
can make the epsilons and
00:43:35.350 --> 00:43:39.230
deltas seem a little bit more
meaningful for you than by the
00:43:39.230 --> 00:43:43.400
formal definition, notice that
it's simple only in comparison
00:43:43.400 --> 00:43:44.870
to what we had before.
00:43:44.870 --> 00:43:46.940
But it's still very,
very difficult.
00:43:46.940 --> 00:43:51.370
The beauty of Calculus is that
in many cases, we do not have
00:43:51.370 --> 00:43:54.500
to know what delta looks like
for a given epsilon.
00:43:54.500 --> 00:43:57.400
What we shall do there
therefore, in our next
00:43:57.400 --> 00:44:01.430
lecture, is to develop recipes
that will allow us to get the
00:44:01.430 --> 00:44:04.970
answers to these limit problems
without having to go
00:44:04.970 --> 00:44:09.580
through this genuinely difficult
problem of finding a
00:44:09.580 --> 00:44:12.550
given delta to match
a given epsilon.
00:44:12.550 --> 00:44:15.940
Though we must admit, in real
life, in many cases where
00:44:15.940 --> 00:44:19.460
you're making approximations, we
will want to know what the
00:44:19.460 --> 00:44:20.810
tolerance limits are.
00:44:20.810 --> 00:44:24.140
I am not belittling the epsilon
delta approach.
00:44:24.140 --> 00:44:27.790
All I'm saying is that in
certain problems, you do not
00:44:27.790 --> 00:44:31.360
need the epsilon delta approach
to get nice results,
00:44:31.360 --> 00:44:34.560
and that will be the topic
of our next lecture.
00:44:34.560 --> 00:44:36.090
So until next time, goodbye.
00:44:39.210 --> 00:44:41.740
GUEST SPEAKER: Funding for the
publication of this video was
00:44:41.740 --> 00:44:46.460
provided by the Gabriella and
Paul Rosenbaum Foundation.
00:44:46.460 --> 00:44:50.630
Help OCW continue to provide
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00:44:50.630 --> 00:44:54.840
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