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HERBERT GROSS: Hi.

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Our lecture today involves the
backbone of Calculus, namely

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the concept of limit, and
perhaps no place in the

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history of mathematics is there
a more subtle topic than

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this particular one.

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I'm reminded of the anecdote
of the professor who was

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staring quite intently,
philosophically, in his home,

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and his wife asked, what are
you contemplating, and he

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said, that it's amazing
how electricity works.

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And the wife said, well what's
so amazing about that?

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All you have to do is
flick the switch.

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And this somehow or other is
exactly the predicament that

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the limit concept falls into.

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From the point of view of
computation, it seems that

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what comes naturally 99% of the
time gives us the right

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answer, but unfortunately, the
1% of the time turns up almost

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all the time in differential
Calculus.

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For this reason, what we will do
is introduce the concept of

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limit through the eyes of
differential Calculus, and we

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shall call our lecture today
'Derivatives and Limits'.

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And what we shall do is go back
to our old friend who

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makes an appearance in almost
all of our lectures so far,

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the case of the freely
falling object.

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A ball is dropped.

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It falls freely in a vacuum, and
the distance, 's', that it

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falls in feet, at the end of 't'
seconds, is given by 's'

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equals '16t squared'.

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00:01:59,640 --> 00:02:03,820
The question that we would like
to raise is, how fast is

38
00:02:03,820 --> 00:02:07,770
the ball falling when
't' equals one?

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00:02:07,770 --> 00:02:10,940
In other words, notice that
the ball is covering a

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different distance
here as it falls.

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00:02:12,730 --> 00:02:15,920
What we want to know is at the
instant that the time is one.

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And by the way, don't confuse
the speed with the

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displacement.

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We know that when the time is
one, the object has fallen a

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distance of 16 feet.

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When the time is 1, the object
has fallen 16 feet.

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00:02:28,290 --> 00:02:31,560
What we want to know is how
fast is it falling at that

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particular point.

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00:02:32,960 --> 00:02:36,810
Now keep in mind that
mathematics being a logical

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00:02:36,810 --> 00:02:40,920
subject proceeds from the idea
of trying to study the

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unfamiliar in terms
of the familiar.

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What we are familiar with is the
concept known as average

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rate of speed.

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So what we will do is duck the
main question temporarily and

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proceed to a different
question.

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What we shall ask is, what is
the average speed of the ball

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during the time interval from t
equals one to t equals two?

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And we all know how to solve
this problem from our

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pre-calculus courses, namely,
we compute where the ball is

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when 't' is two.

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We compute where the ball
was when 't' is one.

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The difference between these two
distances is the distance

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that the ball has fallen during
this time interval, and

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we then divide by the
time interval.

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00:03:26,740 --> 00:03:30,400
And the distance divided by
the time is precisely the

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definition of average
rate of speed.

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In other words, during the time
interval from 't' equals

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one to 't' equals two, the ball
falls at an average speed

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of 48 feet per second.

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00:03:42,690 --> 00:03:47,230
Again, in terms of our diagram,
you see at 't' equals

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00:03:47,230 --> 00:03:50,510
one, the object is over here.

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00:03:50,510 --> 00:03:55,300
At 't' equals two, the object
is over here, it's fallen 64

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feet, and so we compute the
average speed by taking the

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total distance and dividing
by the time that it took.

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00:04:03,150 --> 00:04:07,470
Now again, what we have done is
found the right answer, but

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to the wrong question.

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The question was not what was
the average rate of speed from

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't' equals one to 't' equals
two, it was, what is the speed

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at the instant 't' equals one?

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00:04:16,360 --> 00:04:20,050
And we sense that between 't'
equals one and 't' equals two,

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an awful lot can happen, and
that therefore, our average

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00:04:23,170 --> 00:04:26,450
speed need not be a good
approximation for the

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instantaneous speed.

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What we do next, you see, is we
say OK, let's do something

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a little bit differently.

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Let's now compute the average
speed, but not from 't' equals

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one to 't' equals two, but
rather from 't' equals one to

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't' equals 1.1.

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And computationally, we mimic
the same procedure as before.

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We compute 's' when 't'
is 1.1, we compute

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00:04:50,910 --> 00:04:52,780
's' when 't' is one.

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00:04:52,780 --> 00:04:54,510
By the way, the only difference
is that it's a

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00:04:54,510 --> 00:04:58,850
little bit tougher to square
1.1 and multiply that by 16

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00:04:58,850 --> 00:05:01,930
than it was to square two
and multiply that by 16.

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00:05:01,930 --> 00:05:04,680
The arithmetic gets slightly
messier if you want to call it

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00:05:04,680 --> 00:05:06,900
that, but the concept
stays the same.

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00:05:06,900 --> 00:05:09,810
We find the total distance
traveled during the time

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interval, we divide by the
length of the time interval,

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00:05:13,980 --> 00:05:17,560
and that quotient, by
definition, is the average

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00:05:17,560 --> 00:05:18,560
rate of speed.

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Again, in terms of our diagram,
at t equals one, the

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00:05:23,050 --> 00:05:25,640
object has fallen 16 feet.

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00:05:25,640 --> 00:05:28,230
I'll distort this so that we can
see what's happening here.

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00:05:28,230 --> 00:05:34,210
At 't' equals 1.1, the object
has fallen 19.36 feet, and now

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00:05:34,210 --> 00:05:37,530
we compute the average speed
from here to here in the usual

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00:05:37,530 --> 00:05:39,050
high school way.

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00:05:39,050 --> 00:05:44,030
Now again, the same question
arises as before, namely, this

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is still an average speed.

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We wanted an instantaneous
speed.

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And our come back is to say look
at, we at least suspect

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intuitively that between one and
1.1, we will hopefully get

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00:05:56,940 --> 00:06:00,430
a better idea as to what's
happening at exactly one than

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when we did on the bigger
interval from one to two.

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00:06:03,320 --> 00:06:06,810
In other words, whereas we don't
believe that 33.6 is the

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00:06:06,810 --> 00:06:10,350
answer to our original question,
we do believe that

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00:06:10,350 --> 00:06:15,600
33.6 is probably a better
approximation to the answer to

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00:06:15,600 --> 00:06:19,560
our original question than the
answer 48 feet per second that

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00:06:19,560 --> 00:06:21,860
was obtained over the
larger interval.

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What the mathematician does next
is he says well, let's

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stop playing games.

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Let's not go between one and
1.1, or one and 1.01.

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Why don't we go between one and
'1+h', where 'h' is any

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non-zero amount?

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00:06:38,630 --> 00:06:41,860
In other words, let's find the
average speed from 't' equals

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00:06:41,860 --> 00:06:46,620
one to 't' equals '1+h', and
then what we'll do is we will

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investigate to see what
happens when h

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00:06:48,930 --> 00:06:50,810
gets very, very small.

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00:06:50,810 --> 00:06:54,510
Well again, we mimic exactly
our procedure before.

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00:06:54,510 --> 00:06:59,090
We feed in 't' equals '1+h'
and compute 's', which of

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course is '16 times
'1+h' squared'.

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00:07:01,980 --> 00:07:04,890
Then we subtract off the
distance that the particle had

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00:07:04,890 --> 00:07:08,620
fallen when 't' is one, that's
16 feet, we divide by the

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00:07:08,620 --> 00:07:09,960
length of the time interval.

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00:07:09,960 --> 00:07:14,740
Well between one and '1+h', the
time interval is 'h', we

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00:07:14,740 --> 00:07:18,650
simplify algebraically, and we
wind up with this particular

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00:07:18,650 --> 00:07:23,680
expression, '32h plus 16h
squared' over 'h'.

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00:07:23,680 --> 00:07:28,210
And now we come to the crux of
what limits, and derivatives,

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00:07:28,210 --> 00:07:31,460
and instantaneous rate of
speeds are all about.

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00:07:31,460 --> 00:07:34,630
If you recall, it's quite
tempting to look at this thing

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00:07:34,630 --> 00:07:38,580
and say, ah-ha, I'll cancel an
'h' from both the numerator

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00:07:38,580 --> 00:07:44,250
and the denominator, and that
will leave me '32 plus 16h'.

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And notice though, the very
important thing, that as we

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00:07:47,260 --> 00:07:53,410
mentioned in previous lectures,
since you cannot

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00:07:53,410 --> 00:07:57,960
divide by 0, it is crucial at
this stage that we emphasize

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00:07:57,960 --> 00:08:00,090
that 'h' is not equal to 0.

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00:08:00,090 --> 00:08:02,150
Now of course, this
is not a difficult

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00:08:02,150 --> 00:08:03,770
rationalization to make.

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00:08:03,770 --> 00:08:06,510
Namely, in terms of our
physical situation, we

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00:08:06,510 --> 00:08:09,960
wouldn't want to compute an
average rate of speed over a

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00:08:09,960 --> 00:08:12,810
time interval during which
no time transpired.

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00:08:12,810 --> 00:08:16,630
And if we let 'h' be 0,
obviously, from one to '1+h',

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00:08:16,630 --> 00:08:18,000
no time transpires.

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00:08:18,000 --> 00:08:20,290
But the key statement is, and
we'll come back to exploit

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00:08:20,290 --> 00:08:23,210
this not only for this lecture,
but for the next one

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00:08:23,210 --> 00:08:27,310
too, is to exploit the idea that
in our entire discussion,

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00:08:27,310 --> 00:08:31,170
it was crucial that h not
be allowed to equal 0.

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00:08:31,170 --> 00:08:36,559
In other words, the average
speed between one and '1+h' is

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00:08:36,559 --> 00:08:39,340
'32 plus 16h'.

159
00:08:39,340 --> 00:08:43,100
Now again, without trying to be
rigorous, observe that our

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00:08:43,100 --> 00:08:47,740
senses tell us that as 'h' gets
us close to 0 as we want

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00:08:47,740 --> 00:08:51,670
without ever being allowed to
get there, '16h' gets us close

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00:08:51,670 --> 00:08:56,260
to 0 as we want also, and
hence, '32 plus 16h'

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00:08:56,260 --> 00:09:01,490
apparently gets us close
to 32 as we wish.

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00:09:01,490 --> 00:09:09,720
In other words, it appears that
the speed is 32 feet per

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00:09:09,720 --> 00:09:12,180
second when t equals one.

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00:09:12,180 --> 00:09:17,230
And to reuse the vernacular,
what we're saying is we

167
00:09:17,230 --> 00:09:20,550
compute the average
rate of speed--

168
00:09:20,550 --> 00:09:24,770
see 's' of '1+h' minus
's' of '1/h'--

169
00:09:24,770 --> 00:09:27,800
and see what happens to that
as 'h' approaches 0.

170
00:09:27,800 --> 00:09:29,740
And in our particular case,
notice that this

171
00:09:29,740 --> 00:09:30,700
simply means what?

172
00:09:30,700 --> 00:09:34,790
We came down to '32 plus 16h'
over here, and then as 'h' got

173
00:09:34,790 --> 00:09:38,580
arbitrarily small, we became
suspicious, and believed that

174
00:09:38,580 --> 00:09:41,230
as 'h' became arbitrarily
small, '32

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00:09:41,230 --> 00:09:44,270
plus 16h' became 32.

176
00:09:44,270 --> 00:09:46,970
And I'm saying that in great
detail because this is not

177
00:09:46,970 --> 00:09:49,190
quite nearly as harmless
as it may seem.

178
00:09:49,190 --> 00:09:52,050
There is a great deal of
difficulty involved in just

179
00:09:52,050 --> 00:09:55,900
saying as 'h' gets close
to 0, '32 plus 16h'

180
00:09:55,900 --> 00:09:57,100
gets close to 32.

181
00:09:57,100 --> 00:09:59,910
Well there's no difficulty
in saying it.

182
00:09:59,910 --> 00:10:03,240
There's difficulty that comes up
in trying to show what this

183
00:10:03,240 --> 00:10:06,660
means precisely, as we shall
see in just a little while.

184
00:10:06,660 --> 00:10:10,280
By the way, it's rather easy
to generalize this result.

185
00:10:10,280 --> 00:10:13,220
Namely, you may have noticed in
our presentation there was

186
00:10:13,220 --> 00:10:16,730
nothing sacred about asking what
was the speed when 't'

187
00:10:16,730 --> 00:10:17,550
equals one.

188
00:10:17,550 --> 00:10:21,500
One could just as easily have
chosen any other time 't', say

189
00:10:21,500 --> 00:10:26,660
't sub one', and asked what was
the speed between 't1' and

190
00:10:26,660 --> 00:10:27,830
't1' plus 'h'?

191
00:10:27,830 --> 00:10:31,130
We could have done exactly
the same computation.

192
00:10:31,130 --> 00:10:34,670
In fact, notice down here, when
we do this computation,

193
00:10:34,670 --> 00:10:40,670
the answer '32t one' plus '16h'
becomes precisely '32

194
00:10:40,670 --> 00:10:44,500
plus 16h when 't' one
happens to be one.

195
00:10:44,500 --> 00:10:48,600
In other words, this result here
is the generalization of

196
00:10:48,600 --> 00:10:53,660
our previous result that
had us work with the

197
00:10:53,660 --> 00:10:57,170
limit of '32 plus 16h'.

198
00:10:57,170 --> 00:11:02,010
At any rate, by this approach,
it becomes clear then, if the

199
00:11:02,010 --> 00:11:05,090
equation of motion, the distance
versus the time, is

200
00:11:05,090 --> 00:11:08,840
given by 's' equals '16t
squared', then that time 't'

201
00:11:08,840 --> 00:11:12,670
equals 't one', the
instantaneous speed appears to

202
00:11:12,670 --> 00:11:16,020
be 32 times 't' one.

203
00:11:16,020 --> 00:11:21,270
And in fact, since 't' one
happens to be any arbitrary

204
00:11:21,270 --> 00:11:23,820
time that we choose, it is
customary to drop the

205
00:11:23,820 --> 00:11:25,880
subscript, and simply
say what?

206
00:11:25,880 --> 00:11:29,550
At any time, 't', our freely
falling body in this case,

207
00:11:29,550 --> 00:11:33,130
will have a speed
given by '32t'.

208
00:11:33,130 --> 00:11:37,400
And many of us will remember, as
this is a revisited course,

209
00:11:37,400 --> 00:11:40,740
that this is precisely the
recipe that turns out when one

210
00:11:40,740 --> 00:11:43,180
talks about bringing down
the exponent and

211
00:11:43,180 --> 00:11:45,250
replacing it by one less.

212
00:11:45,250 --> 00:11:50,540
16 times two is 32, and 't' to
a power one less than two is

213
00:11:50,540 --> 00:11:53,150
just t to the first
power or 't'.

214
00:11:53,150 --> 00:11:56,200
But notice now, no recipes
involved here.

215
00:11:56,200 --> 00:12:01,060
Just a question of intuitively
defining instantaneous speed

216
00:12:01,060 --> 00:12:05,470
in terms of being a limit
of average speeds.

217
00:12:05,470 --> 00:12:08,340
Now in order that we don't
become hypocrites in what

218
00:12:08,340 --> 00:12:12,040
we're doing, recall also that
we have spent an entire

219
00:12:12,040 --> 00:12:16,610
lecture talking about the value
of analytic Geometry in

220
00:12:16,610 --> 00:12:18,070
the study of Calculus.

221
00:12:18,070 --> 00:12:21,470
The idea that a picture
is worth 1,000 words.

222
00:12:21,470 --> 00:12:25,360
So what we would like to do now
is to revisit our previous

223
00:12:25,360 --> 00:12:29,000
discussion in terms
of a graph.

224
00:12:29,000 --> 00:12:33,950
You see, suppose we have a curve
'y' equals 'f of x', and

225
00:12:33,950 --> 00:12:36,440
I've drawn the curve over
on this side over here.

226
00:12:36,440 --> 00:12:40,110
We'll make more reference to it
later as we go along, but

227
00:12:40,110 --> 00:12:44,120
what does it mean if I now
mechanically compute 'f of x1

228
00:12:44,120 --> 00:12:48,400
' plus 'delta x' minus 'f
of x1' over 'delta x'?

229
00:12:48,400 --> 00:12:53,240
And by the way, again, notice
just a question of symbolism.

230
00:12:53,240 --> 00:12:56,510
'Delta x' happens to be the
conventional symbol that one

231
00:12:56,510 --> 00:12:59,410
uses in the analytical Geometry
approach where we

232
00:12:59,410 --> 00:13:01,600
were using h before,
but this is simply

233
00:13:01,600 --> 00:13:03,490
a question of notation.

234
00:13:03,490 --> 00:13:12,410
Well the idea is this: observe
that if we go to the graph, 'f

235
00:13:12,410 --> 00:13:17,060
of x1' plus 'delta x' is
this particular height.

236
00:13:17,060 --> 00:13:21,790
On the other hand, this
height is 'f of x1' .

237
00:13:21,790 --> 00:13:29,000
Consequently, our numerator is
just this distance here, and

238
00:13:29,000 --> 00:13:31,960
our denominator, which is
'delta x', is just this

239
00:13:31,960 --> 00:13:33,460
distance here.

240
00:13:33,460 --> 00:13:37,690
And therefore, what we have done
in our quotient, is we

241
00:13:37,690 --> 00:13:41,650
have found the slope of the
straight line that joins 'p'

242
00:13:41,650 --> 00:13:45,870
to 'q.' And here then is why the
question of straight line

243
00:13:45,870 --> 00:13:48,430
becomes so important
in the study of

244
00:13:48,430 --> 00:13:49,900
differential Calculus.

245
00:13:49,900 --> 00:13:54,260
That what is average speed, in
terms of analysis, becomes the

246
00:13:54,260 --> 00:13:57,660
slope of a straight line
in terms of Geometry.

247
00:13:57,660 --> 00:14:01,030
And the idea, you see, is that
this gives me the average

248
00:14:01,030 --> 00:14:06,460
speed of the straight line that
joins the average rise of

249
00:14:06,460 --> 00:14:08,890
the straight line that
joins 'p' to 'q'.

250
00:14:08,890 --> 00:14:12,800
And by the way, this is quite
deceptive as any average is.

251
00:14:12,800 --> 00:14:14,990
It's like the man who sat with
his feet in the oven and an

252
00:14:14,990 --> 00:14:17,100
ice pack on his head, and when
somebody said how do you feel,

253
00:14:17,100 --> 00:14:18,850
he said, on the average,
pretty good.

254
00:14:18,850 --> 00:14:21,380
You see, the same thing is
happening over here.

255
00:14:21,380 --> 00:14:24,880
Notice if I had taken a
completely different curve, a

256
00:14:24,880 --> 00:14:28,560
curve which looked nothing at
all like my original curve,

257
00:14:28,560 --> 00:14:32,670
only that also passed through
the points 'p' and 'q', notice

258
00:14:32,670 --> 00:14:35,190
that even though these two
curves are quite different,

259
00:14:35,190 --> 00:14:38,760
the average rise from
'p' to 'q' is the

260
00:14:38,760 --> 00:14:40,220
same for both curves.

261
00:14:40,220 --> 00:14:43,260
In other words, when one deals
with the average, one is not

262
00:14:43,260 --> 00:14:46,690
too concerned with how one
got from the first

263
00:14:46,690 --> 00:14:47,870
point to the second.

264
00:14:47,870 --> 00:14:50,280
You see, this is the problem
with average rate of space.

265
00:14:50,280 --> 00:14:53,500
In fact, the bigger a time
interval we work over, the

266
00:14:53,500 --> 00:14:58,090
more danger there is that the
average rate of speed will not

267
00:14:58,090 --> 00:15:01,250
coincide intuitively with what
we would like to believe that

268
00:15:01,250 --> 00:15:03,450
the instantaneous rate
of speed is.

269
00:15:03,450 --> 00:15:06,940
And you see, that's exactly
what this idea here means.

270
00:15:06,940 --> 00:15:10,650
What we do to define the
derivative, the instantaneous

271
00:15:10,650 --> 00:15:15,820
rise at 'x' equals 'x1' , is
we take this quotient and

272
00:15:15,820 --> 00:15:20,050
investigate what happens to it
as 'delta x' is allowed to get

273
00:15:20,050 --> 00:15:22,080
arbitrarily close to 0.

274
00:15:22,080 --> 00:15:24,210
But let's emphasize
this again.

275
00:15:24,210 --> 00:15:28,160
'Delta x' is not allowed
to equal 0.

276
00:15:28,160 --> 00:15:31,020
You see, it's over a smaller and
smaller interval, and you

277
00:15:31,020 --> 00:15:33,120
see what's happening
here geometrically?

278
00:15:33,120 --> 00:15:36,580
To let 'delta x' get smaller and
smaller means, that for a

279
00:15:36,580 --> 00:15:41,810
given 'x1' , you are holding 'p'
fixed, and allowing 'q' to

280
00:15:41,810 --> 00:15:44,560
move in closer and
closer to 'p'.

281
00:15:44,560 --> 00:15:46,960
And what we are saying is, is
that when 'q' assumes this

282
00:15:46,960 --> 00:15:51,290
position, let's call it 'q sub
1', the slope of the line that

283
00:15:51,290 --> 00:15:55,080
joints 'p' to 'q' one looks a
lot more like the slope of the

284
00:15:55,080 --> 00:15:59,040
line that will be tangent to
the curve at the point 'p'.

285
00:15:59,040 --> 00:16:03,140
And for this reason, what is
instantaneous rate of speed,

286
00:16:03,140 --> 00:16:07,220
when we're talking about
functions, become slopes of

287
00:16:07,220 --> 00:16:11,980
curves at a given point when we
are talking about curves.

288
00:16:11,980 --> 00:16:15,270
You see that the tangent line to
a curve is viewed as being

289
00:16:15,270 --> 00:16:19,395
the limiting position of a cord
drawn from a point 'p' to

290
00:16:19,395 --> 00:16:21,150
a point 'q'.

291
00:16:21,150 --> 00:16:24,220
Now at any rate, this will be
emphasized in the text.

292
00:16:24,220 --> 00:16:26,990
It will be emphasized
in our exercises.

293
00:16:26,990 --> 00:16:31,180
The major issue now is what did
we really do when it came

294
00:16:31,180 --> 00:16:34,620
time to compute this
little gadget.

295
00:16:34,620 --> 00:16:38,050
In other words, when we talked
about the limit of 'f of x' as

296
00:16:38,050 --> 00:16:41,980
'x' approached 'a', how
did we compute this?

297
00:16:41,980 --> 00:16:46,220
The danger is, and this is why
I have underlined a question

298
00:16:46,220 --> 00:16:48,980
mark here, the intuitive
approach is to say something

299
00:16:48,980 --> 00:16:53,550
like, let's just replace
x by a in here.

300
00:16:53,550 --> 00:16:56,360
In other words, let the limit of
'f of x', as 'x' approaches

301
00:16:56,360 --> 00:16:58,310
'a', be 'f of a'.

302
00:16:58,310 --> 00:17:03,110
In fact, this is what we did
with our '32 plus 16h'.

303
00:17:03,110 --> 00:17:06,122
In a way, what we did was, with
'32 plus 16h', do you

304
00:17:06,122 --> 00:17:06,970
remember what we did?

305
00:17:06,970 --> 00:17:10,260
We said, let's let 'h' get
arbitrarily close to 0.

306
00:17:10,260 --> 00:17:12,450
And in our minds, we
let 'h' equal 0.

307
00:17:12,450 --> 00:17:17,730
'16h' was then 0, and then '32
plus 16h' was then 32.

308
00:17:17,730 --> 00:17:20,140
But what we had done is somehow
or other, if our

309
00:17:20,140 --> 00:17:24,530
intuition is correct, we had
arrived at the correct answer

310
00:17:24,530 --> 00:17:26,300
but for wrong reasons.

311
00:17:26,300 --> 00:17:30,000
Because you see, in this
expression, it is mandatory,

312
00:17:30,000 --> 00:17:32,620
as we've outlined this,
is that 'x' never be

313
00:17:32,620 --> 00:17:36,030
allowed to equal 'a'.

314
00:17:36,030 --> 00:17:37,960
Well you say, who cares
whether you

315
00:17:37,960 --> 00:17:38,850
allow it or you don't?

316
00:17:38,850 --> 00:17:40,750
Aren't we going to get
the same answer?

317
00:17:40,750 --> 00:17:43,000
And just to give you a quickie,
one that we talked

318
00:17:43,000 --> 00:17:45,910
about in our introductory
lecture, consider the function

319
00:17:45,910 --> 00:17:48,690
'f of x' to be 'x squared
minus nine'

320
00:17:48,690 --> 00:17:51,080
over 'x minus three'.

321
00:17:51,080 --> 00:17:53,970
And let's evaluate what
this limit is as

322
00:17:53,970 --> 00:17:56,700
'x' approaches three.

323
00:17:56,700 --> 00:18:00,690
If we were to blindly assume
that all we have to do is

324
00:18:00,690 --> 00:18:04,150
replace 'x' by 'a', that
would mean what here?

325
00:18:04,150 --> 00:18:06,510
Replace 'x' by three.

326
00:18:06,510 --> 00:18:08,700
Notice that what
we got is what?

327
00:18:08,700 --> 00:18:12,980
Three squared, which is nine,
minus nine, (which is 0) over

328
00:18:12,980 --> 00:18:15,730
three minus three,
which is also 0.

329
00:18:15,730 --> 00:18:19,010
In other words, if in the
expression 'x squared minus

330
00:18:19,010 --> 00:18:23,770
nine' over 'x minus three' you
replace 'x' by three, you do

331
00:18:23,770 --> 00:18:25,750
get 0 over 0.

332
00:18:25,750 --> 00:18:30,270
Unfortunately, or perhaps
fortunately, somehow or other,

333
00:18:30,270 --> 00:18:33,310
this definition is supposed to
incorporate the fact that you

334
00:18:33,310 --> 00:18:36,860
never allow 'x' to
equal three.

335
00:18:36,860 --> 00:18:40,240
See, the way we got 0 over 0
was we violated what the

336
00:18:40,240 --> 00:18:41,960
intuitive meaning
of limit was.

337
00:18:41,960 --> 00:18:46,630
We wound up with this 0 over 0
mess by allowing 'x' to equal

338
00:18:46,630 --> 00:18:50,130
the one value it is not
allowed to equal here.

339
00:18:50,130 --> 00:18:54,250
To see what happened more
anatomically in this

340
00:18:54,250 --> 00:18:57,750
particular problem, let's go
back to our old friend of

341
00:18:57,750 --> 00:18:59,140
cancellation again.

342
00:18:59,140 --> 00:19:03,520
We observe that 'x squared minus
nine' can be written as

343
00:19:03,520 --> 00:19:07,930
'x plus three' times 'x minus
three', and then we divide by

344
00:19:07,930 --> 00:19:12,020
'x minus three', and here's
where the main problem seems

345
00:19:12,020 --> 00:19:12,710
to come up.

346
00:19:12,710 --> 00:19:16,420
One automatically, somehow or
other, gets into the habit of

347
00:19:16,420 --> 00:19:18,600
canceling these things out.

348
00:19:18,600 --> 00:19:21,160
But the point is, since we have
already agreed that you

349
00:19:21,160 --> 00:19:25,490
cannot divide by 0, the idea
then becomes that the only

350
00:19:25,490 --> 00:19:30,850
time this is permissible is when
'x' is unequal to three.

351
00:19:30,850 --> 00:19:36,990
In fact, to have been more
precise, what we should have

352
00:19:36,990 --> 00:19:42,160
said was not that 'x squared
minus nine' over 'x minus

353
00:19:42,160 --> 00:19:46,310
three' equals 'x plus three',
but rather, 'x squared minus

354
00:19:46,310 --> 00:19:51,850
nine' over 'x minus three'
equals 'x plus three', unless

355
00:19:51,850 --> 00:19:53,820
'x' equals three.

356
00:19:53,820 --> 00:19:57,470
And by the way, I should
add here what?

357
00:19:57,470 --> 00:20:04,820
That if 'x' equals three, then
'x squared minus nine' over 'x

358
00:20:04,820 --> 00:20:08,290
minus three' is undefined.

359
00:20:08,290 --> 00:20:10,850
You see, that's what
we call 0 over 0.

360
00:20:10,850 --> 00:20:14,000
Undefined, indeterminate.

361
00:20:14,000 --> 00:20:16,570
But here's what bails us out.

362
00:20:16,570 --> 00:20:20,410
As soon as we say the limit as
'x' approaches three, what are

363
00:20:20,410 --> 00:20:21,490
we assuming?

364
00:20:21,490 --> 00:20:27,070
We're assuming that 'x' can be
any value at all except three,

365
00:20:27,070 --> 00:20:30,370
and the interesting point now
becomes, that as long as 'x'

366
00:20:30,370 --> 00:20:34,120
is anything but three, the only
time you could tell these

367
00:20:34,120 --> 00:20:39,000
two expressions apart was when
'x' happened to equal three.

368
00:20:39,000 --> 00:20:43,550
So if you now say we won't allow
'x' to equal three, then

369
00:20:43,550 --> 00:20:47,450
it is true, whereas the
bracketed expression by itself

370
00:20:47,450 --> 00:20:49,200
is not equal to the
parenthetical

371
00:20:49,200 --> 00:20:50,530
expression by itself.

372
00:20:50,530 --> 00:20:54,520
The limit as 'x' approaches
three of these two expressions

373
00:20:54,520 --> 00:20:58,160
are equal, and in fact, one
becomes tempted to say "what?"

374
00:20:58,160 --> 00:20:59,110
at this case.

375
00:20:59,110 --> 00:21:02,920
Oh, here it's quite easy to see,
that as 'x' approaches

376
00:21:02,920 --> 00:21:08,520
three, 'x plus three'
is equal to six.

377
00:21:08,520 --> 00:21:11,330
And let me put a question mark
here also, because it

378
00:21:11,330 --> 00:21:15,120
certainly is true that if you
replace 'x' by three, this

379
00:21:15,120 --> 00:21:17,310
will be six.

380
00:21:17,310 --> 00:21:20,600
But this says don't replace
'x' by three.

381
00:21:20,600 --> 00:21:24,030
Now before, when we replaced
'x' by three we got into

382
00:21:24,030 --> 00:21:25,860
trouble, so we said
we can't do that.

383
00:21:25,860 --> 00:21:29,280
Now we replace 'x' by three, we
get an answer that we like,

384
00:21:29,280 --> 00:21:32,210
and that is the danger that
we'll say OK, we won't allow

385
00:21:32,210 --> 00:21:35,860
this to be done, unless
we like the result.

386
00:21:35,860 --> 00:21:38,710
And of course this gives you
a highly subjective subject

387
00:21:38,710 --> 00:21:39,640
matter here.

388
00:21:39,640 --> 00:21:42,400
You see, the point is that
somehow or other we must come

389
00:21:42,400 --> 00:21:46,070
up with a more objective
criteria for distinguishing

390
00:21:46,070 --> 00:21:47,700
what limits are.

391
00:21:47,700 --> 00:21:51,200
And by the way, for the Polyanna
who has the intuitive

392
00:21:51,200 --> 00:21:55,370
feeling for saying things like
well look at, this may give

393
00:21:55,370 --> 00:21:58,390
you trouble if you wind
up with 0 over 0.

394
00:21:58,390 --> 00:22:02,240
But what is the likelihood that
given 'f of x' at random,

395
00:22:02,240 --> 00:22:05,270
and 'a' at random, that the
limit of 'f of x' as 'x'

396
00:22:05,270 --> 00:22:09,060
approaches 'a' is going to be
0 over 0 if you're careless?

397
00:22:09,060 --> 00:22:11,950
And the answer is that the
probability is quite small,

398
00:22:11,950 --> 00:22:16,540
except in Calculus, in which
case, every time, you take a

399
00:22:16,540 --> 00:22:17,460
derivative.

400
00:22:17,460 --> 00:22:21,470
Every time you are going to
wind up with 0 over 0.

401
00:22:21,470 --> 00:22:24,700
This is why we said earlier that
differential Calculus has

402
00:22:24,700 --> 00:22:28,360
been referred to as 'the
study of 0 over 0'.

403
00:22:28,360 --> 00:22:31,950
You see, the whole point is,
let's go back now to our basic

404
00:22:31,950 --> 00:22:36,190
definition of average rate of
change of 'f', 'f of x1' plus

405
00:22:36,190 --> 00:22:40,970
'delta x' minus 'f of
x1' over 'delta x'.

406
00:22:40,970 --> 00:22:43,430
I guess when you're writing in
black, you have to use white

407
00:22:43,430 --> 00:22:46,150
for emphasis, so
we'll use this.

408
00:22:46,150 --> 00:22:49,800
Notice that every time you let
'delta x' equal zero, you are

409
00:22:49,800 --> 00:22:51,230
going to wind up with what?

410
00:22:51,230 --> 00:22:55,790
'f of x1' minus 'f of x1' ,
which is 0, over 'delta x',

411
00:22:55,790 --> 00:22:56,720
which is 0.

412
00:22:56,720 --> 00:22:59,330
In other words, you are
going to get 0 over 0.

413
00:22:59,330 --> 00:23:02,850

414
00:23:02,850 --> 00:23:07,260
If you believe that all this
means is to replace 'delta x'

415
00:23:07,260 --> 00:23:11,070
by 0, you see, and the point is,
this is not the definition

416
00:23:11,070 --> 00:23:14,130
of this, and this is why you
don't wind up with this

417
00:23:14,130 --> 00:23:15,436
particular thing.

418
00:23:15,436 --> 00:23:19,100
Well I've spent enough time,
I think, trying to give a

419
00:23:19,100 --> 00:23:22,950
picture as to what limits mean
intuitively, now what I'd like

420
00:23:22,950 --> 00:23:26,340
to do is to devote the remainder
of this lecture,

421
00:23:26,340 --> 00:23:30,320
plus our next one, to
investigating, more

422
00:23:30,320 --> 00:23:35,470
stringently, what limit means in
a way that is unambiguous.

423
00:23:35,470 --> 00:23:37,650
And you know, it's like
everything else in this world.

424
00:23:37,650 --> 00:23:41,400
When you want to get something
that is ironclad, you don't

425
00:23:41,400 --> 00:23:43,150
get something for nothing.

426
00:23:43,150 --> 00:23:48,560
To get an ironclad expression
that doesn't lend itself to

427
00:23:48,560 --> 00:23:52,020
misinterpretations usually
involves a rigorous enough

428
00:23:52,020 --> 00:23:55,050
language that scares off
the uninitiated.

429
00:23:55,050 --> 00:23:57,020
Sort of like a legal document.

430
00:23:57,020 --> 00:23:59,600
The chances are if you're not
a trained lawyer, and the

431
00:23:59,600 --> 00:24:02,450
document is simple enough for
you to understand, it must

432
00:24:02,450 --> 00:24:04,540
have thousands of
loopholes in it.

433
00:24:04,540 --> 00:24:07,740
But that same rigorous language
that makes it almost

434
00:24:07,740 --> 00:24:11,320
impossible for the layman to
understand what's happening,

435
00:24:11,320 --> 00:24:14,660
is precisely the language that
the lawyer needs to make the

436
00:24:14,660 --> 00:24:17,950
document well defined
to him, hopefully.

437
00:24:17,950 --> 00:24:21,410
Well with that as a background,
let's now try to

438
00:24:21,410 --> 00:24:24,560
give a more rigorous definition
of limit.

439
00:24:24,560 --> 00:24:27,380
And let's try to do it in such
a way that not only will the

440
00:24:27,380 --> 00:24:31,540
definition be rigorous, but it
will agree, to the best of our

441
00:24:31,540 --> 00:24:34,470
ability, with what we
intuitively believe a limit

442
00:24:34,470 --> 00:24:35,660
should mean.

443
00:24:35,660 --> 00:24:38,710
Now what I want to do here is
scare you a little bit, and

444
00:24:38,710 --> 00:24:40,980
the reason I want to scare you
a little bit is, I want to

445
00:24:40,980 --> 00:24:44,650
show you, once and for all,
how in many cases, what

446
00:24:44,650 --> 00:24:48,120
appears to be ominous
mathematical notation, is

447
00:24:48,120 --> 00:24:51,270
simply a rigorous way of saying
something which was

448
00:24:51,270 --> 00:24:53,840
quite intuitive pictorially
to do.

449
00:24:53,840 --> 00:24:56,630
The mathematical definition
of limit is the following.

450
00:24:56,630 --> 00:25:00,330
The limit of 'f of x' as 'x'
approaches 'a' equals 'l'

451
00:25:00,330 --> 00:25:04,660
means, and get this, given
epsilon greater than 0, we can

452
00:25:04,660 --> 00:25:08,070
find delta greater than 0, such
that when the absolute

453
00:25:08,070 --> 00:25:11,370
value of 'x minus a' is less
than delta, but greater than

454
00:25:11,370 --> 00:25:15,750
0, then the absolute value of
'f of x' minus 'l' is less

455
00:25:15,750 --> 00:25:17,210
than epsilon.

456
00:25:17,210 --> 00:25:19,850
Now doesn't that kind
of turn you on?

457
00:25:19,850 --> 00:25:21,460
See, this is the
whole problem.

458
00:25:21,460 --> 00:25:25,650
I will now cut you in sort of on
the trade secret that makes

459
00:25:25,650 --> 00:25:28,510
this thing very,
very readable.

460
00:25:28,510 --> 00:25:32,170
First of all, to say that you
want to get arbitrarily close

461
00:25:32,170 --> 00:25:36,210
in value, you want 'f of x' to
be arbitrarily close to 'l',

462
00:25:36,210 --> 00:25:38,020
another way of saying
that is what?

463
00:25:38,020 --> 00:25:41,460
You specify any tolerance limit
whatsoever, which we

464
00:25:41,460 --> 00:25:45,775
call epsilon, and I can make 'f
of x' get to within epsilon

465
00:25:45,775 --> 00:25:49,870
of 'l' just by choosing 'x'
sufficiently close to 'a'.

466
00:25:49,870 --> 00:25:53,640
And by the way again, in terms
of geometry, that's all the

467
00:25:53,640 --> 00:25:55,780
absolute value of a
difference means.

468
00:25:55,780 --> 00:25:59,330
The absolute value of 'x minus
a' simply means what?

469
00:25:59,330 --> 00:26:02,030
The distance between
'x' and 'a'.

470
00:26:02,030 --> 00:26:03,750
All this says is what?

471
00:26:03,750 --> 00:26:09,980
I can make 'f of x' fall within
epsilon of 'l' by

472
00:26:09,980 --> 00:26:15,050
finding a delta such that 'x' is
within delta of 'a', and by

473
00:26:15,050 --> 00:26:20,090
the way, this little tidbit over
here is simply a fancy

474
00:26:20,090 --> 00:26:23,600
way of saying that you're not
letting 'x' equal 'a'.

475
00:26:23,600 --> 00:26:25,860
You see, the only way the
absolute value of a number can

476
00:26:25,860 --> 00:26:28,770
be 0 is if the number
itself is 0.

477
00:26:28,770 --> 00:26:32,270
The only way this could be 0
is if 'x' equals 'a', so by

478
00:26:32,270 --> 00:26:35,360
saying this little tidbit over
here, that this is greater

479
00:26:35,360 --> 00:26:38,800
than 0, that's our fancy way
of saying that we are not

480
00:26:38,800 --> 00:26:42,050
going to allow 'x'
to equal to 'a'.

481
00:26:42,050 --> 00:26:45,430
Now again, whereas this may
sound a little bit simpler

482
00:26:45,430 --> 00:26:48,740
than the original definition,
let's see how much easier the

483
00:26:48,740 --> 00:26:50,060
picture makes it.

484
00:26:50,060 --> 00:26:52,970
Let's go to a graph of
'y' equals 'f of x'.

485
00:26:52,970 --> 00:26:56,860

486
00:26:56,860 --> 00:27:00,390
And in my diagram here, here is
'a', and here is 'l', and

487
00:27:00,390 --> 00:27:04,070
what I'm saying is, I
intuitively suspect.

488
00:27:04,070 --> 00:27:05,530
Let me write down
what I suspect.

489
00:27:05,530 --> 00:27:09,620
I suspect that the limit of 'f
of x', as 'x' approaches 'a',

490
00:27:09,620 --> 00:27:11,580
is 'l' in this case.

491
00:27:11,580 --> 00:27:14,900
Namely I suspect that as 'x'
gets closer and closer to 'a',

492
00:27:14,900 --> 00:27:18,950
no matter which side you come
in on, that's why we use

493
00:27:18,950 --> 00:27:19,790
absolute value.

494
00:27:19,790 --> 00:27:22,550
It makes no difference whether
it's positive or negative.

495
00:27:22,550 --> 00:27:25,950
As 'x' gets in tighter and
tighter on 'a', I suspect that

496
00:27:25,950 --> 00:27:30,290
I can make 'f of x' come
arbitrarily equal to 'l'.

497
00:27:30,290 --> 00:27:33,260
In other words, if 'x' is close
enough to 'a', 'f of x',

498
00:27:33,260 --> 00:27:35,310
it seems, will be
close to 'l'.

499
00:27:35,310 --> 00:27:38,310
And by the way, to complete this
problem geometrically,

500
00:27:38,310 --> 00:27:39,990
pick your epsilon.

501
00:27:39,990 --> 00:27:43,220
Let's say that epsilon is this
'y', and since epsilon is

502
00:27:43,220 --> 00:27:45,960
positive, this will be called
'l plus epsilon'.

503
00:27:45,960 --> 00:27:49,920
This will be called
'l minus epsilon'.

504
00:27:49,920 --> 00:27:53,010
Now what we do is we come
over to the curve here.

505
00:27:53,010 --> 00:27:56,460

506
00:27:56,460 --> 00:27:59,190
All right, hopefully these
will look like horizontal

507
00:27:59,190 --> 00:28:00,295
lines to you.

508
00:28:00,295 --> 00:28:03,330
When I get down to here, I then
project down vertical

509
00:28:03,330 --> 00:28:04,580
lines to meet the x-axis.

510
00:28:04,580 --> 00:28:07,230

511
00:28:07,230 --> 00:28:10,140
And now all I'm saying, and I
think this is fairly intuitive

512
00:28:10,140 --> 00:28:12,790
clear, is what?

513
00:28:12,790 --> 00:28:18,810
That if 'x' is this close to
'a', 'f of x' will be within

514
00:28:18,810 --> 00:28:20,140
these tolerance limits of 'l'.

515
00:28:20,140 --> 00:28:23,750
In other words, for any value of
'x' in here, see, for this

516
00:28:23,750 --> 00:28:27,960
neighborhood of 'a', every
input has its output come

517
00:28:27,960 --> 00:28:30,610
within epsilon of 'l'.

518
00:28:30,610 --> 00:28:33,500
By the way, just one brief
note here, something that

519
00:28:33,500 --> 00:28:36,410
we'll have to come back to in
more detail-- in fact, I'll do

520
00:28:36,410 --> 00:28:38,290
that in our next example--

521
00:28:38,290 --> 00:28:47,560
notice here how important it
was that our curve was not

522
00:28:47,560 --> 00:28:49,160
single value, but one to one.

523
00:28:49,160 --> 00:28:52,180
In other words, suppose our
curve had doubled back, say

524
00:28:52,180 --> 00:28:53,780
something like this.

525
00:28:53,780 --> 00:28:57,710
Notice then, when you try to
project over to the curve from

526
00:28:57,710 --> 00:29:01,090
'l plus epsilon', you don't know
whether to stop at this

527
00:29:01,090 --> 00:29:02,810
point or at this point.

528
00:29:02,810 --> 00:29:05,260
You see, in other words, notice
that there is another

529
00:29:05,260 --> 00:29:09,690
point down here which has the
same 'l plus epsilon' value.

530
00:29:09,690 --> 00:29:12,560
In other words, whenever our
function is not one to one,

531
00:29:12,560 --> 00:29:14,490
there is going to be a
problem, when we work

532
00:29:14,490 --> 00:29:18,470
analytically, of being able to
solve algebraic equations, and

533
00:29:18,470 --> 00:29:21,620
trying to distinguish this point
of intersection from

534
00:29:21,620 --> 00:29:22,440
this point.

535
00:29:22,440 --> 00:29:26,650
You see, if we made a mistake
and bypass this point and came

536
00:29:26,650 --> 00:29:29,920
all the way over to here, and
thought that this was our

537
00:29:29,920 --> 00:29:32,840
tolerance limit, we would be in
a great deal of difficulty.

538
00:29:32,840 --> 00:29:37,680
Because you see, for example,
for this value of 'x', its 'f

539
00:29:37,680 --> 00:29:42,720
of x' value projects up here,
which is outside of the

540
00:29:42,720 --> 00:29:45,120
tolerance limits
that are given.

541
00:29:45,120 --> 00:29:47,950
You see, many things which are
self evident from the picture

542
00:29:47,950 --> 00:29:52,180
are not nearly so self evident
analytically, to which we then

543
00:29:52,180 --> 00:29:55,850
raise the question, why
use analysis at all?

544
00:29:55,850 --> 00:29:57,560
Let's just use pictures.

545
00:29:57,560 --> 00:30:00,030
And again the answer is, in
this case, we can get away

546
00:30:00,030 --> 00:30:02,640
with it, but in more complicated
situations where

547
00:30:02,640 --> 00:30:04,760
either we can't draw the diagram
because it's too

548
00:30:04,760 --> 00:30:08,300
complicated, or because we have
several variables, and we

549
00:30:08,300 --> 00:30:11,400
get into a dimension problem,
the point is that sometimes we

550
00:30:11,400 --> 00:30:14,490
have no recourse other
than the analysis.

551
00:30:14,490 --> 00:30:18,820
Let me give you an exercise
that we'll do in terms of

552
00:30:18,820 --> 00:30:22,990
putting the geometry and the
analysis side by side.

553
00:30:22,990 --> 00:30:27,750
Let's look at the expression the
limit of 'x squared minus

554
00:30:27,750 --> 00:30:30,590
2x' as 'x' approaches three.

555
00:30:30,590 --> 00:30:33,900
Now of course, we don't want 'x'
to equal three here, but

556
00:30:33,900 --> 00:30:37,840
let's cheat again, and see
what this thing means.

557
00:30:37,840 --> 00:30:41,950
Our intuition tells us that when
'x' is three, 'x squared'

558
00:30:41,950 --> 00:30:44,940
is nine, '2x' is six.

559
00:30:44,940 --> 00:30:48,810
So nine minus six is three, and
we would suspect that the

560
00:30:48,810 --> 00:30:51,600
limit here should be three.

561
00:30:51,600 --> 00:30:56,990
Now to see what this means,
let's draw a little diagram.

562
00:30:56,990 --> 00:31:00,610
The graph 'y' equals 'x squared
minus 2x' crosses the

563
00:31:00,610 --> 00:31:03,040
x-axis at 0 and two.

564
00:31:03,040 --> 00:31:07,140
It has its low point at the
0.1 comma minus one.

565
00:31:07,140 --> 00:31:10,620
We can sketch these curves,
it's a parabola.

566
00:31:10,620 --> 00:31:21,520
And now what we're saying is, is
it true that when 'x' gets

567
00:31:21,520 --> 00:31:24,890
very close to three,
that 'f of x' gets

568
00:31:24,890 --> 00:31:26,790
very close to three?

569
00:31:26,790 --> 00:31:29,640
And the answer is, from the
diagram, it appears very

570
00:31:29,640 --> 00:31:31,550
obvious that this is the case.

571
00:31:31,550 --> 00:31:34,840
Fact, here's that example of
non-single value again.

572
00:31:34,840 --> 00:31:37,860
If you call this thing 'l',
and you now pick tolerance

573
00:31:37,860 --> 00:31:41,310
limits which we'll call 'l plus
epsilon' and 'l minus

574
00:31:41,310 --> 00:31:44,740
epsilon', and now you want to
see what interval you need on

575
00:31:44,740 --> 00:31:49,800
the x-axis, notice that in this
particular diagram, you

576
00:31:49,800 --> 00:31:54,600
will get two intervals, one of
which surrounds the value 'x'

577
00:31:54,600 --> 00:31:57,350
equals minus one, and the other
of which surrounds the

578
00:31:57,350 --> 00:31:59,620
'x' value 'x' equals three.

579
00:31:59,620 --> 00:32:03,400
In other words, both when 'x' is
minus one and 'x' is three,

580
00:32:03,400 --> 00:32:07,690
'x squared minus 2x'
will equal three.

581
00:32:07,690 --> 00:32:11,350
At any rate, leaving the diagram
as an aid, let's see

582
00:32:11,350 --> 00:32:14,650
what our epsilon delta
definition says, and how we

583
00:32:14,650 --> 00:32:17,380
handle the stuff algebraically,
and how we

584
00:32:17,380 --> 00:32:20,370
correlate the algebra
with the geometry.

585
00:32:20,370 --> 00:32:22,870

586
00:32:22,870 --> 00:32:25,970
Given epsilon greater than
0, what must I do?

587
00:32:25,970 --> 00:32:29,860
I must find a delta greater
than 0 such that, and I'm

588
00:32:29,860 --> 00:32:31,770
going to read this colloquially,
I'll write it

589
00:32:31,770 --> 00:32:34,600
formally, but read it
colloquially, such that

590
00:32:34,600 --> 00:32:39,950
whenever 'x' is within delta
of three, but not equal to

591
00:32:39,950 --> 00:32:44,340
three, then 'x squared
minus 2x' will be

592
00:32:44,340 --> 00:32:47,080
within epsilon of three.

593
00:32:47,080 --> 00:32:48,350
Now here's the whole point.

594
00:32:48,350 --> 00:32:52,360
We know, algebraically, how
to handle absolute values.

595
00:32:52,360 --> 00:32:56,700
Namely, the absolute value of
'x squared minus 2x minus

596
00:32:56,700 --> 00:32:59,730
three' being less than epsilon,
is the same as saying

597
00:32:59,730 --> 00:33:04,180
that 'x squared minus 2x minus
three' itself must be between

598
00:33:04,180 --> 00:33:07,100
epsilon and minus epsilon.

599
00:33:07,100 --> 00:33:10,300
Now again, we rewrite things
in fancy ways if we wish.

600
00:33:10,300 --> 00:33:11,850
There other ways
of doing this.

601
00:33:11,850 --> 00:33:13,760
I call this completing
the square.

602
00:33:13,760 --> 00:33:16,215
Well, people for 2,000 years
have called this completing

603
00:33:16,215 --> 00:33:17,280
the square.

604
00:33:17,280 --> 00:33:21,630
Namely, I rewrite minus three as
one minus four, so that 'x

605
00:33:21,630 --> 00:33:24,580
squared minus 2x plus one'
can be factored as

606
00:33:24,580 --> 00:33:26,500
'x minus one squared'.

607
00:33:26,500 --> 00:33:28,750
I'm now down to this
form here.

608
00:33:28,750 --> 00:33:32,360
Then I add four to all sides
of my inequality, and have

609
00:33:32,360 --> 00:33:36,210
that 'x minus one squared' is
between four plus epsilon and

610
00:33:36,210 --> 00:33:37,460
four minus epsilon.

611
00:33:37,460 --> 00:33:41,330
Some fairly elementary Algebra
of inequalities here.

612
00:33:41,330 --> 00:33:42,660
Now here's the key point.

613
00:33:42,660 --> 00:33:46,060
Remembering my diagram, I said
to you, how do we know whether

614
00:33:46,060 --> 00:33:48,300
we're near three or
near minus one?

615
00:33:48,300 --> 00:33:51,300
How do we distinguish, when we
draw the lines to the curve,

616
00:33:51,300 --> 00:33:52,890
what neighborhood we want?

617
00:33:52,890 --> 00:33:55,510
And notice that all we're saying
is that if 'x' is near

618
00:33:55,510 --> 00:33:58,560
three, that means what?
'x' is close to three.

619
00:33:58,560 --> 00:34:01,910
If 'x' is reasonably close to
three, then 'x minus one' is

620
00:34:01,910 --> 00:34:06,190
going to be reasonably close to
two, and hence be positive.

621
00:34:06,190 --> 00:34:08,920
Point is, that as long as you're
dealing with positive

622
00:34:08,920 --> 00:34:11,940
numbers over here, see if
epsilon is sufficiently small,

623
00:34:11,940 --> 00:34:14,070
these will then all be
positive numbers.

624
00:34:14,070 --> 00:34:18,100
For positive numbers, if the
squares obey a certain

625
00:34:18,100 --> 00:34:22,730
inequality, the square roots
will obey the same inequality,

626
00:34:22,730 --> 00:34:26,090
as we have emphasized in
one of our exercises.

627
00:34:26,090 --> 00:34:31,139
In other words, from here, we
can now say this, and from

628
00:34:31,139 --> 00:34:34,409
this, we can now conclude, that
if you want 'x squared

629
00:34:34,409 --> 00:34:39,710
minus 2x' to be within epsilon
of three, 'x' itself is going

630
00:34:39,710 --> 00:34:42,850
to have to be between 'one plus
the square root of 'four

631
00:34:42,850 --> 00:34:45,699
plus epsilon'', and 'one
plus the square root

632
00:34:45,699 --> 00:34:47,190
of 'four minus epsilon''.

633
00:34:47,190 --> 00:34:50,030
And by the way, this is quite
rigorous, but I don't think it

634
00:34:50,030 --> 00:34:50,639
turns you on.

635
00:34:50,639 --> 00:34:52,989
I don't think there's any
picture that you associate

636
00:34:52,989 --> 00:34:56,159
with this, and so what I thought
I'd like to do for our

637
00:34:56,159 --> 00:35:01,960
next little thing, is to come
back to our diagram here, and

638
00:35:01,960 --> 00:35:04,800
show you what this
really means.

639
00:35:04,800 --> 00:35:09,850
Namely, notice that if epsilon
is a small positive number,

640
00:35:09,850 --> 00:35:11,060
let's take a look
at this again.

641
00:35:11,060 --> 00:35:14,260
If epsilon is a small positive
number, this is

642
00:35:14,260 --> 00:35:16,420
slightly less than four.

643
00:35:16,420 --> 00:35:19,280
Therefore, the square root is
slightly less than two,

644
00:35:19,280 --> 00:35:22,560
therefore, this number will be
slightly less than three.

645
00:35:22,560 --> 00:35:25,300
On the other hand, 'four plus
epsilon' is slightly more than

646
00:35:25,300 --> 00:35:29,080
four, so its square root will be
slightly more than two, and

647
00:35:29,080 --> 00:35:32,320
therefore, one plus that square
root will be slightly

648
00:35:32,320 --> 00:35:33,840
more than three.

649
00:35:33,840 --> 00:35:36,890
And what these two numbers, as
abstract as they look like,

650
00:35:36,890 --> 00:35:39,660
correspond to, is nothing
more than this.

651
00:35:39,660 --> 00:35:43,870
Coming back to our diagram and
choosing an epsilon, and

652
00:35:43,870 --> 00:35:47,980
coming over to the curve like
this, and projecting down like

653
00:35:47,980 --> 00:35:51,990
this, all we're saying is,
see, what are we saying

654
00:35:51,990 --> 00:35:52,970
pictorially?

655
00:35:52,970 --> 00:35:58,130
That for 'f of x' to be within
epsilon of 'l', 'x' itself has

656
00:35:58,130 --> 00:36:00,260
to be in this range here.

657
00:36:00,260 --> 00:36:05,950
And all this says is, this very
simple point to compute,

658
00:36:05,950 --> 00:36:09,740
just by projecting down like
this, its rigorous name would

659
00:36:09,740 --> 00:36:13,560
be 'one plus the square root
of 'four minus epsilon''.

660
00:36:13,560 --> 00:36:16,420
That's this number over here.

661
00:36:16,420 --> 00:36:20,460
And the furthest point, namely
this point here, which again,

662
00:36:20,460 --> 00:36:24,490
in terms of the picture is very
easy to find, that point

663
00:36:24,490 --> 00:36:28,220
corresponds to one plus
the square root

664
00:36:28,220 --> 00:36:31,660
of 'four plus epsilon'.

665
00:36:31,660 --> 00:36:35,390
And again notice, in terms
of the Algebra, I need no

666
00:36:35,390 --> 00:36:36,870
recourse to a picture.

667
00:36:36,870 --> 00:36:40,460
But in terms of the picture, I
get a heck of a good feeling

668
00:36:40,460 --> 00:36:43,960
as to what these abstract
symbols are telling me.

669
00:36:43,960 --> 00:36:48,830
Well let's continue on with the
solution of this exercise.

670
00:36:48,830 --> 00:36:51,760
You see, we didn't want to find
where 'x' was, we wanted

671
00:36:51,760 --> 00:36:54,360
to see where 'x minus
three' had to be.

672
00:36:54,360 --> 00:36:56,400
In other words, we're trying
to find the delta.

673
00:36:56,400 --> 00:37:00,290
We know that 'x minus three'
is between these

674
00:37:00,290 --> 00:37:01,830
two extremes here.

675
00:37:01,830 --> 00:37:04,460
Consequently, and here's a place
we have to be a little

676
00:37:04,460 --> 00:37:07,840
bit careful here, this of
course, is a positive number,

677
00:37:07,840 --> 00:37:10,450
because this is more than two.

678
00:37:10,450 --> 00:37:13,200
This, on the other hand, is a
negative number, because you

679
00:37:13,200 --> 00:37:16,640
see 'four minus epsilon' is less
than four, so its square

680
00:37:16,640 --> 00:37:18,940
root is less than two.

681
00:37:18,940 --> 00:37:21,490
So if I subtract two from--
it's a negative number--

682
00:37:21,490 --> 00:37:24,580
and since delta has to be
positive, if this thing is

683
00:37:24,580 --> 00:37:28,880
negative, two minus the square
root of 'four minus epsilon'

684
00:37:28,880 --> 00:37:30,730
will be positive.

685
00:37:30,730 --> 00:37:35,040
So what I do is, to solve this
problem, is I simply choose

686
00:37:35,040 --> 00:37:39,530
delta to be the minimum
of these two numbers.

687
00:37:39,530 --> 00:37:42,330
And then by definition,
what do I have?

688
00:37:42,330 --> 00:37:45,280
That when the absolute value
of 'x minus three' is less

689
00:37:45,280 --> 00:37:48,920
than delta, and at the same
time, this is crucial,

690
00:37:48,920 --> 00:37:49,770
greater than 0.

691
00:37:49,770 --> 00:37:52,600
In other words, we never
let 'x' equal three.

692
00:37:52,600 --> 00:37:56,570
For this choice of delta, by how
we worked backwards, this

693
00:37:56,570 --> 00:37:59,920
will come out to be
less than epsilon.

694
00:37:59,920 --> 00:38:03,200
And that is exactly what you
mean by our formal definition

695
00:38:03,200 --> 00:38:06,140
of saying that the limit of 'x
squared minus 2x' as 'x'

696
00:38:06,140 --> 00:38:08,980
approaches three, in this
case, equals three.

697
00:38:08,980 --> 00:38:12,000
For someone who wants more
concrete evidence, I think all

698
00:38:12,000 --> 00:38:17,860
one has to do is, for example,
take a number like 'one plus

699
00:38:17,860 --> 00:38:20,480
the square root of 'four plus
epsilon'', that is the widest

700
00:38:20,480 --> 00:38:24,240
point on our interval, and
actually plug that in for 'x',

701
00:38:24,240 --> 00:38:27,830
and go through this computation
of squaring 'one

702
00:38:27,830 --> 00:38:30,190
plus the square root of
'four plus epsilon''.

703
00:38:30,190 --> 00:38:32,890
Subtract twice, 'one plus
the square root

704
00:38:32,890 --> 00:38:34,390
of 'four plus epsilon''.

705
00:38:34,390 --> 00:38:37,720
Carry out that computation
and you get what?

706
00:38:37,720 --> 00:38:39,750
'Three plus epsilon'.

707
00:38:39,750 --> 00:38:43,150
That using the upper extreme,
you wind up with what?

708
00:38:43,150 --> 00:38:48,570
In excess of three by epsilon,
which by the way is exactly

709
00:38:48,570 --> 00:38:49,740
what's supposed to happen.

710
00:38:49,740 --> 00:38:52,560
In fact, I hope this doesn't
give you an eye sore, I will

711
00:38:52,560 --> 00:38:56,180
pull down the top board again
to show you also what these

712
00:38:56,180 --> 00:38:59,180
two distances mean, and
what this means.

713
00:38:59,180 --> 00:39:02,390
You see, all these two numbers
mean is the following, and

714
00:39:02,390 --> 00:39:04,910
I'll pull this down just
far enough to see it.

715
00:39:04,910 --> 00:39:10,140
That these two numbers were the
widths from this end point

716
00:39:10,140 --> 00:39:14,550
to three on the one extreme,
and from this end point to

717
00:39:14,550 --> 00:39:17,270
three on the other extreme.

718
00:39:17,270 --> 00:39:21,170
In other words, those two
numbers that we took the

719
00:39:21,170 --> 00:39:25,250
minimum of were the half-width
intervals

720
00:39:25,250 --> 00:39:26,880
that surrounded three.

721
00:39:26,880 --> 00:39:29,550
You see the problem, as we
mentioned before is, is that

722
00:39:29,550 --> 00:39:33,330
even though epsilon and minus
epsilon are symmetric with

723
00:39:33,330 --> 00:39:37,340
respect to 'l', when you
translate over to the curve,

724
00:39:37,340 --> 00:39:40,260
because this curve is not a
straight line when you project

725
00:39:40,260 --> 00:39:43,610
down, these two widths
will not be equal.

726
00:39:43,610 --> 00:39:46,750

727
00:39:46,750 --> 00:39:49,030
By picking the minimum of
these two widths, you

728
00:39:49,030 --> 00:39:52,740
guarantee that you have
locked yourself inside

729
00:39:52,740 --> 00:39:54,000
the required area.

730
00:39:54,000 --> 00:39:57,710
You see, that's all this
particular thing means.

731
00:39:57,710 --> 00:40:01,550
Now the trouble is that one
might now get the idea that

732
00:40:01,550 --> 00:40:05,120
there are no shorter ways of
doing the same problem.

733
00:40:05,120 --> 00:40:07,580
You see, I showed you
a rigorous way.

734
00:40:07,580 --> 00:40:10,770
There was no law that said we
had to find the biggest delta.

735
00:40:10,770 --> 00:40:14,485
In other words, if delta equals,
a half will work, in

736
00:40:14,485 --> 00:40:16,670
other words, if you're within
a half of three, something's

737
00:40:16,670 --> 00:40:19,060
going to happen, then certainly
any smaller number

738
00:40:19,060 --> 00:40:21,410
will work just as well.

739
00:40:21,410 --> 00:40:23,020
In other words, I can
get estimates.

740
00:40:23,020 --> 00:40:24,790
Let me show you what
I mean by that.

741
00:40:24,790 --> 00:40:28,660
Another way of tackling how to
make 'x squared minus 2x minus

742
00:40:28,660 --> 00:40:32,580
three' smaller than epsilon is
to use our properties of

743
00:40:32,580 --> 00:40:35,180
absolute values, and
to factor this.

744
00:40:35,180 --> 00:40:37,880
In other words, we know that
'x squared minus 2x minus

745
00:40:37,880 --> 00:40:41,040
three' is 'x minus three'
times 'x plus one'.

746
00:40:41,040 --> 00:40:44,560
We know that the absolute
value of a product is a

747
00:40:44,560 --> 00:40:47,360
product of the absolute values,
so these two things

748
00:40:47,360 --> 00:40:54,050
here are synonyms. Then we
know that near 'x' equals

749
00:40:54,050 --> 00:40:56,410
three, which we're interested
in here, 'x plus

750
00:40:56,410 --> 00:40:59,550
one' is near four.

751
00:40:59,550 --> 00:41:02,700
Well what we mean more
rigorously is this, choose an

752
00:41:02,700 --> 00:41:06,430
epsilon, and if the absolute
value of 'x minus three' is

753
00:41:06,430 --> 00:41:09,560
less than epsilon, in other
words, if 'x minus three' is

754
00:41:09,560 --> 00:41:13,040
less than epsilon but greater
than minus epsilon, then by

755
00:41:13,040 --> 00:41:16,400
adding on four to all three
sides of the inequality here,

756
00:41:16,400 --> 00:41:20,390
we see that 'x plus one' is less
than 'four plus epsilon',

757
00:41:20,390 --> 00:41:22,140
but greater than 'four
minus epsilon'.

758
00:41:22,140 --> 00:41:31,650

759
00:41:31,650 --> 00:41:34,550
If the size of epsilon is less
than one, then certainly this

760
00:41:34,550 --> 00:41:38,490
is less than five, and this
is greater than three.

761
00:41:38,490 --> 00:41:41,200
Now of course, somebody may say
to us, but what if epsilon

762
00:41:41,200 --> 00:41:42,070
is more than one?

763
00:41:42,070 --> 00:41:47,770
Well obviously, if a guy says
make this thing within 10, and

764
00:41:47,770 --> 00:41:50,190
I can make it within one,
certainly within

765
00:41:50,190 --> 00:41:51,600
10 is within one.

766
00:41:51,600 --> 00:41:53,870
I can always pick the
smaller number.

767
00:41:53,870 --> 00:41:57,140
It's like that nonsense of the
fellow asking for an ice cream

768
00:41:57,140 --> 00:41:59,240
cone, and the waitress said
what flavor, and he said

769
00:41:59,240 --> 00:42:00,460
anything except chocolate.

770
00:42:00,460 --> 00:42:01,880
And she said, I'm all out of
chocolate, will you take

771
00:42:01,880 --> 00:42:03,540
anything except vanilla?

772
00:42:03,540 --> 00:42:05,600
The idea here is that if
somebody says make this less

773
00:42:05,600 --> 00:42:09,240
than 15, if you've made it less
than 10, in particular,

774
00:42:09,240 --> 00:42:10,840
you've made it less than 15.

775
00:42:10,840 --> 00:42:13,140
And so all we do over
here, you see, is

776
00:42:13,140 --> 00:42:13,890
something like this.

777
00:42:13,890 --> 00:42:17,780
We say look it, if we want to
make this number here very

778
00:42:17,780 --> 00:42:22,170
small, we know that this number
here is less than five,

779
00:42:22,170 --> 00:42:25,710
we know that 'epsilon over five'
is a positive number if

780
00:42:25,710 --> 00:42:29,640
epsilon is small, so why don't
we just pick the absolute

781
00:42:29,640 --> 00:42:34,100
value of 'x minus three' to be
less than 'epsilon over five?'

782
00:42:34,100 --> 00:42:36,260
In fact, that's what this delta
will be in that case.

783
00:42:36,260 --> 00:42:38,590
In other words, if the absolute
value of x minus

784
00:42:38,590 --> 00:42:48,820
three is within 'epsilon over
five', if 'x' is within

785
00:42:48,820 --> 00:42:51,620
'epsilon over five' of
three, notice what

786
00:42:51,620 --> 00:42:53,040
this product becomes.

787
00:42:53,040 --> 00:42:56,160
This is less than 'epsilon over
five', this we know from

788
00:42:56,160 --> 00:43:00,640
before is less than five, and
therefore, this product is

789
00:43:00,640 --> 00:43:02,080
less than epsilon.

790
00:43:02,080 --> 00:43:05,640
In other words, we have
exhibited the delta such that

791
00:43:05,640 --> 00:43:09,040
when this happens, the absolute
value of what we want

792
00:43:09,040 --> 00:43:10,860
to make less than
epsilon indeed

793
00:43:10,860 --> 00:43:13,090
becomes less than epsilon.

794
00:43:13,090 --> 00:43:18,210
Now because I recognize that
this is a hard topic, you'll

795
00:43:18,210 --> 00:43:21,290
notice in the reading
assignments that these

796
00:43:21,290 --> 00:43:24,120
problems are covered
in great detail.

797
00:43:24,120 --> 00:43:28,220
Everything that I've said in the
lesson so far is repeated

798
00:43:28,220 --> 00:43:31,920
in great computational depth
in our learning exercises.

799
00:43:31,920 --> 00:43:35,350
The point is that even if I
can make the epsilons and

800
00:43:35,350 --> 00:43:39,230
deltas seem a little bit more
meaningful for you than by the

801
00:43:39,230 --> 00:43:43,400
formal definition, notice that
it's simple only in comparison

802
00:43:43,400 --> 00:43:44,870
to what we had before.

803
00:43:44,870 --> 00:43:46,940
But it's still very,
very difficult.

804
00:43:46,940 --> 00:43:51,370
The beauty of Calculus is that
in many cases, we do not have

805
00:43:51,370 --> 00:43:54,500
to know what delta looks like
for a given epsilon.

806
00:43:54,500 --> 00:43:57,400
What we shall do there
therefore, in our next

807
00:43:57,400 --> 00:44:01,430
lecture, is to develop recipes
that will allow us to get the

808
00:44:01,430 --> 00:44:04,970
answers to these limit problems
without having to go

809
00:44:04,970 --> 00:44:09,580
through this genuinely difficult
problem of finding a

810
00:44:09,580 --> 00:44:12,550
given delta to match
a given epsilon.

811
00:44:12,550 --> 00:44:15,940
Though we must admit, in real
life, in many cases where

812
00:44:15,940 --> 00:44:19,460
you're making approximations, we
will want to know what the

813
00:44:19,460 --> 00:44:20,810
tolerance limits are.

814
00:44:20,810 --> 00:44:24,140
I am not belittling the epsilon
delta approach.

815
00:44:24,140 --> 00:44:27,790
All I'm saying is that in
certain problems, you do not

816
00:44:27,790 --> 00:44:31,360
need the epsilon delta approach
to get nice results,

817
00:44:31,360 --> 00:44:34,560
and that will be the topic
of our next lecture.

818
00:44:34,560 --> 00:44:36,090
So until next time, goodbye.

819
00:44:36,090 --> 00:44:39,210

820
00:44:39,210 --> 00:44:41,740
GUEST SPEAKER: Funding for the
publication of this video was

821
00:44:41,740 --> 00:44:46,460
provided by the Gabriella and
Paul Rosenbaum Foundation.

822
00:44:46,460 --> 00:44:50,630
Help OCW continue to provide
free and open access to MIT

823
00:44:50,630 --> 00:44:54,840
courses by making a donation
at ocw.mit.edu/donate.

824
00:44:54,840 --> 00:44:59,583