WEBVTT
Kind: captions
Language: en
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>> Welcome to the Cypress
College math review
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on epsilon-delta
proofs of limits.
00:00:08.096 --> 00:00:12.986
The limit as x approaches c
of f of x is equal to L means
00:00:12.986 --> 00:00:15.116
that for every epsilon
greater than zero,
00:00:15.676 --> 00:00:19.946
there exists a delta greater
than zero such that if x is
00:00:19.946 --> 00:00:24.766
within delta of c, that implies
that f of x is within epsilon
00:00:24.856 --> 00:00:29.456
of L. Here's our diagram and
the function is y equals f of x.
00:00:30.786 --> 00:00:33.836
So no matter how
small an epsilon,
00:00:33.840 --> 00:00:39.400
which is a positive radius that
we choose around our limit L,
00:00:41.040 --> 00:00:46.300
we can find a delta, which is
our positive radius around c.
00:00:48.536 --> 00:00:52.736
Such that if x is
within delta of c,
00:00:53.766 --> 00:01:00.356
then the function values have
to be within epsilon of L. Now,
00:01:00.546 --> 00:01:03.386
this point right
here does not have
00:01:03.466 --> 00:01:05.676
to be an actual point
on the function.
00:01:06.056 --> 00:01:07.516
There could be a hole there.
00:01:16.796 --> 00:01:20.146
For an epsilon delta proof the
first thing you want to do is
00:01:20.146 --> 00:01:23.836
to find the limit, even if
the limit is given to you,
00:01:23.866 --> 00:01:26.356
check that you wrote the
problem down correctly.
00:01:27.626 --> 00:01:30.626
Most of the proofs you will
be doing will involve linear
00:01:30.626 --> 00:01:32.136
or quadratic functions.
00:01:32.666 --> 00:01:33.646
For both of these type
00:01:33.646 --> 00:01:36.906
of functions to find the
limit you simply substitute
00:01:36.906 --> 00:01:41.256
in the value of c. The limit
as x approaches c of f of x,
00:01:41.256 --> 00:01:45.796
is simply f of c. Next thing
you want to do is to have a copy
00:01:46.096 --> 00:01:48.006
of a proof that your
instructor did
00:01:48.006 --> 00:01:49.926
as a format for your own proof.
00:01:50.646 --> 00:01:54.096
Instructors vary on how they
want these proofs formatted.
00:01:55.206 --> 00:01:59.076
What is right is the way your
instructor wants the proof
00:01:59.076 --> 00:02:02.736
formatted, not the way someone
else tells you how to do it.
00:02:03.556 --> 00:02:05.926
You can learn from
others the math involved,
00:02:06.336 --> 00:02:08.796
but you must use your
instructor's format
00:02:08.920 --> 00:02:09.860
for your proof.
00:02:14.760 --> 00:02:17.280
Alright, I'm going to show
two different approaches
00:02:17.286 --> 00:02:20.206
for each of these proofs.
00:02:21.176 --> 00:02:23.746
Once again you're
going to want to look
00:02:23.746 --> 00:02:29.266
at your instructors formatting
to see how they want it done.
00:02:29.916 --> 00:02:33.596
So the first thing we do
is substitute in the six,
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and we get a limit of seven.
00:02:39.096 --> 00:02:40.416
So here we go.
00:02:42.006 --> 00:02:49.286
Given epsilon greater
than zero we choose delta
00:02:49.936 --> 00:02:53.976
to be some value, we don't
know what that is yet.
00:02:53.976 --> 00:02:56.836
It's going to be a function
of epsilon, we'll figure
00:02:56.836 --> 00:02:59.146
that out later, you can
just leave this blank,
00:02:59.146 --> 00:03:01.486
you can put an underline,
something,
00:03:02.396 --> 00:03:03.406
just leave it blank for now.
00:03:05.196 --> 00:03:11.136
Then, for, and then
what is x approaching?
00:03:13.406 --> 00:03:17.736
X is approaching six,
yes, x is approaching six.
00:03:18.096 --> 00:03:21.566
So x has to be within
delta of six.
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We get what?
00:03:27.356 --> 00:03:29.036
Ok, so now we look
at the function,
00:03:29.236 --> 00:03:30.876
and what does the
function need to do?
00:03:32.676 --> 00:03:36.456
Function minus limit, ok
and what is that function?
00:03:37.106 --> 00:03:41.656
That function is 4/3x
minus one, right.
00:03:41.656 --> 00:03:42.526
That's the function.
00:03:43.306 --> 00:03:44.146
Now, what's the limit?
00:03:44.646 --> 00:03:47.356
The limit is seven, right.
00:03:47.496 --> 00:03:51.856
So minus seven, so there's the
function, and minus the limit.
00:03:53.436 --> 00:03:59.176
So we simplify that, we
have 4/3x minus eight,
00:04:00.216 --> 00:04:01.766
so what do we want
to make it look like?
00:04:01.866 --> 00:04:04.956
We want to look, make it
look like x minus six, yes.
00:04:05.806 --> 00:04:08.076
So how we going to make
it look like x minus six?
00:04:08.106 --> 00:04:10.866
By getting rid of the 4/3,
we're going to pull out 4/3;
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it's om absolute value, so
we're going to have to pull it
00:04:13.046 --> 00:04:14.360
out in absolute value.
00:04:17.700 --> 00:04:19.360
And then you simply divide.
00:04:19.756 --> 00:04:21.106
Will it end up being a six?
00:04:21.106 --> 00:04:22.546
Yes, it will end up being six.
00:04:23.946 --> 00:04:25.126
How do you figure that out?
00:04:25.476 --> 00:04:26.006
You divide.
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So we're taking eight and
you're dividing by 4/3.
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Well how do you divide by 4/3?
00:04:32.226 --> 00:04:35.326
That's the same thing
as multiply by 3/4, yes.
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So eight times 3/4.
00:04:39.296 --> 00:04:42.616
So eight times 3/4,
what do you get?
00:04:43.156 --> 00:04:44.526
Six, surprise, surprise.
00:04:45.446 --> 00:04:54.036
Ok, so what do we get, so that
is less than 4/3 times what?
00:04:55.126 --> 00:04:57.086
Well, what is x minus six?
00:04:57.146 --> 00:05:01.556
X minus six is less
than delta, yes.
00:05:02.676 --> 00:05:07.896
Ok, and what do we
need delta to be then?
00:05:08.276 --> 00:05:19.236
We need delta to be
3/4 epsilon, yes.
00:05:19.596 --> 00:05:22.636
Why? Because we need
that to be equal
00:05:22.636 --> 00:05:24.786
to epsilon the whole
thing, right?
00:05:26.046 --> 00:05:30.666
Because eventually we need the
absolute value of f of x minus L
00:05:31.086 --> 00:05:34.966
to be, is less than
epsilon, bingo.
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So what goes in the box?
00:05:38.716 --> 00:05:40.096
3/4 epsilon.
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So 3/4 epsilon goes in the
box and then our conclusion
00:05:48.346 --> 00:05:54.216
at the end is therefore, which
is the three dots in a triangle,
00:05:55.986 --> 00:05:59.616
the limit as x approaches six
00:06:00.546 --> 00:06:09.636
of our function 4/3x minus
one is equal to seven.
00:06:21.636 --> 00:06:23.556
Here's another way
to write the proof.
00:06:29.386 --> 00:06:41.106
For every epsilon greater
than zero let delta equal
00:06:41.876 --> 00:06:44.346
and then we don't know what the
delta value will be right now,
00:06:45.986 --> 00:06:58.206
such that if zero is less than
absolute value of x minus,
00:06:58.786 --> 00:07:00.496
ok now what is that
x approaching?
00:07:00.496 --> 00:07:05.326
X is approaching six
is less than delta.
00:07:06.806 --> 00:07:11.536
So here is we're telling
that x is within delta of six
00:07:11.536 --> 00:07:14.306
so if x is within delta of six,
00:07:14.866 --> 00:07:17.926
and then we go down
to the bottom.
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And you kind of get an idea
of how many steps it's going
00:07:22.006 --> 00:07:24.666
to take so you go down to
the bottom of the proof
00:07:26.176 --> 00:07:30.266
and we have our final statement.
00:07:33.146 --> 00:07:39.406
Ok, so we have our function,
which is 4/3 x minus one.
00:07:40.056 --> 00:07:42.786
And then we have our
limit, minus seven,
00:07:43.886 --> 00:07:47.656
all of that's in absolute
value, is less than epsilon.
00:07:49.116 --> 00:07:53.396
So that's 4/3x minus eight.
00:07:57.656 --> 00:08:00.286
The advantage to this
is I can go through
00:08:00.286 --> 00:08:02.556
and I can multiply
both sides by 3/4,
00:08:11.316 --> 00:08:26.216
so I get x minus six is less
than 3/4 epsilon, which tells me
00:08:26.636 --> 00:08:31.206
that delta needs
to be 3/4 epsilon.
00:08:33.516 --> 00:08:36.226
It's just another way
to do it, and it's not
00:08:36.226 --> 00:08:37.326
like you get to choose.
00:08:38.326 --> 00:08:40.536
You choose the way your
instructor wants it done.
00:08:43.300 --> 00:08:45.000
This is just another formatting,
first we find the limit,
00:08:51.880 --> 00:08:53.680
First we find the limit
00:08:59.260 --> 00:09:02.080
so the limit is -5, so L is -5.
00:09:05.006 --> 00:09:14.086
So given epsilon greater than
zero choose delta equal to,
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we'll figure this out
a little bit later,
00:09:20.680 --> 00:09:26.980
then for absolute x minus
what is x approaching four.
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In other words, if x is
within delta of four we get,
00:09:38.466 --> 00:09:41.516
now we go down and we look at
what our function is doing.
00:09:44.726 --> 00:09:50.826
f of x minus L, ok
what do we get?
00:09:50.826 --> 00:09:54.176
Our function is three
minus two x
00:09:54.546 --> 00:09:56.146
and our limit is negative five,
00:09:56.146 --> 00:09:57.486
so we have a double
negative there.
00:10:00.476 --> 00:10:06.406
So we have three minus two x
plus five, which simplifies
00:10:06.406 --> 00:10:14.926
to eight minus two x. This time,
to make it into x minus four,
00:10:15.546 --> 00:10:16.496
we're going to have to take
00:10:16.496 --> 00:10:18.916
out the absolute
value of negative two.
00:10:23.086 --> 00:10:25.686
So that simplifies to just
two because the absolute value
00:10:25.686 --> 00:10:26.876
of negative two is two.
00:10:30.806 --> 00:10:34.966
That is less than two times
now what is the absolute value
00:10:34.966 --> 00:10:36.736
of x minus four smaller than?
00:10:37.216 --> 00:10:41.366
From our hypothesis it's
smaller than delta, yes.
00:10:41.666 --> 00:10:43.836
That was our hypothesis.
00:10:48.346 --> 00:10:50.936
So what is delta?
00:10:51.416 --> 00:10:57.976
Delta is epsilon over two,
why is it epsilon over two?
00:10:58.626 --> 00:11:02.816
Because this product needs
to be equal to epsilon.
00:11:03.246 --> 00:11:07.366
Our conclusion needs to be
that f of x is within epsilon
00:11:07.556 --> 00:11:12.446
of L. Here we go, which
means that delta needs
00:11:12.446 --> 00:11:16.566
to be epsilon over two.
00:11:19.376 --> 00:11:25.186
Therefore, the limit
as x approaches four
00:11:26.906 --> 00:11:33.806
of our function 3-2x
is equal to -5.
00:11:41.386 --> 00:11:43.226
Let's do this with
the other formatting
00:11:43.226 --> 00:11:44.676
that I suggested earlier.
00:11:49.856 --> 00:11:59.066
For every epsilon greater
than zero let delta equal,
00:11:59.446 --> 00:12:05.566
once again we'll figure that
out later, if zero is less
00:12:05.566 --> 00:12:08.606
than absolute x minus,
what is x approaching?
00:12:09.146 --> 00:12:14.776
Four. Is less than delta, some
instructors want this zero,
00:12:15.226 --> 00:12:17.856
some don't care about it,
it's up to your instructor.
00:12:18.866 --> 00:12:25.006
Go down to the bottom,
absolute f of x,
00:12:25.206 --> 00:12:28.916
minus L is less than epsilon.
00:12:30.236 --> 00:12:36.756
The function was three
minus 2x, the limit is -5.
00:12:41.946 --> 00:12:46.736
So we simplify that and
we get eight minus 2x.
00:12:52.766 --> 00:12:57.296
We need x minus four, so we
factor out the absolute value
00:12:57.716 --> 00:13:00.576
of -2, you can't
take our just two.
00:13:02.056 --> 00:13:05.636
We have to take out the absolute
value of -2, which leaves us
00:13:05.636 --> 00:13:08.386
with the absolute
value of x minus four.
00:13:08.786 --> 00:13:10.796
Let's be careful with
our absolute values.
00:13:11.686 --> 00:13:13.986
But the absolute value
of -2 is simply 2.
00:13:19.536 --> 00:13:28.006
Divide both sides by
two, which tells us
00:13:28.376 --> 00:13:31.816
since the absolute value of x
minus four is less than delta,
00:13:31.816 --> 00:13:34.936
and the absolute value of x
minus is less than epsilon
00:13:34.936 --> 00:13:41.416
over two, that tells us that
delta must be epsilon over two.
00:13:42.236 --> 00:13:48.936
We've just proved that the
limit as x approaches four
00:13:50.556 --> 00:13:56.606
of the function three
minus 2x is -5.
00:14:04.576 --> 00:14:07.446
Alright let's find this limit
on this quadratic function.
00:14:08.296 --> 00:14:13.836
So we have two squared plus
three times two, minus five,
00:14:14.876 --> 00:14:16.006
so I have a limit of five.
00:14:18.036 --> 00:14:27.966
So given epsilon greater than
zero, we choose delta to equal,
00:14:28.736 --> 00:14:31.426
now here with a quadratic
function we have
00:14:31.426 --> 00:14:32.826
to have two things going on.
00:14:33.886 --> 00:14:37.266
So for quadratic functions
you're going to have delta equal
00:14:37.676 --> 00:14:43.886
to a minimum of the set that
contains one and an unknown.
00:14:44.296 --> 00:14:46.896
So like we did before,
you'll have a blank line
00:14:46.896 --> 00:14:48.166
or a box or something.
00:14:48.956 --> 00:14:52.796
So delta to be the minimum
of the set that contains one
00:14:52.836 --> 00:14:55.576
and your unknown, we'll
fill that in later.
00:14:58.966 --> 00:15:06.446
Then for x within delta of
two, since x is approaching two.
00:15:10.436 --> 00:15:18.656
We get absolute value
of f of x minus L equal
00:15:19.336 --> 00:15:24.086
and our function is x
squared plus 3x minus five.
00:15:25.386 --> 00:15:32.616
Our limit is five, simplify
that x squared plus 3x minus 10.
00:15:33.586 --> 00:15:38.096
Factor then, be sure to put
each factor in absolute value.
00:15:41.446 --> 00:15:44.546
Now we need a bound on the
absolute value of x plus five.
00:15:46.276 --> 00:15:48.066
We know that x is
approaching two.
00:15:49.516 --> 00:15:51.476
So we need to take
this to the side
00:15:51.886 --> 00:15:54.946
and find a bound on
the other factor.
00:15:58.106 --> 00:16:01.876
So we draw a number line and we
know that x is approaching two.
00:16:03.366 --> 00:16:05.896
So our delta value
of one is going
00:16:05.896 --> 00:16:09.206
to be what helps us with this.
00:16:11.696 --> 00:16:17.226
So if delta is one, that says
we're between one and three.
00:16:17.546 --> 00:16:20.166
So we go one each way from
what x is approaching,
00:16:20.166 --> 00:16:21.396
x is approaching two.
00:16:21.816 --> 00:16:26.326
And we need a bound on the
absolute value of x plus five.
00:16:26.866 --> 00:16:29.816
Substituting in one we
get the absolute value
00:16:29.816 --> 00:16:32.096
of six, which is six.
00:16:32.636 --> 00:16:34.306
You just do this
on the end points.
00:16:34.536 --> 00:16:37.906
The absolute value of x
plus five when I substitute
00:16:37.906 --> 00:16:42.186
in three would be the absolute
value of eight which is eight.
00:16:43.496 --> 00:16:47.066
Choose the larger value
so the absolute value
00:16:47.066 --> 00:16:49.526
of x plus five is
smaller than eight.
00:16:50.776 --> 00:16:52.316
We know that the absolute value
00:16:52.316 --> 00:16:54.756
of x minus two is
less than delta.
00:16:55.506 --> 00:16:57.866
So we have a bound on that.
00:16:59.056 --> 00:17:08.036
We know that we need that to be
smaller than epsilon which means
00:17:08.466 --> 00:17:16.496
that delta needs to
be epsilon over eight.
00:17:19.596 --> 00:17:22.456
That tells us what to put
in our unknown quantity.
00:17:23.926 --> 00:17:28.406
Now, if delta was the minimum
of the set one and epsilon
00:17:28.406 --> 00:17:33.586
over eight, the smaller of
those two we're guaranteed
00:17:33.586 --> 00:17:39.046
that if x is within delta
of two, if x is within delta
00:17:39.046 --> 00:17:46.486
of two then for sure
both the y values will be
00:17:46.486 --> 00:17:49.486
within epsilon of the limit.
00:17:52.876 --> 00:17:59.356
Therefore, the limit
as x approaches two
00:18:00.640 --> 00:18:09.620
of our function is five.
00:18:18.760 --> 00:18:22.060
Here's another format
for that same proof.
00:18:22.786 --> 00:18:28.936
We know the limit's five
for every epsilon greater
00:18:28.936 --> 00:18:35.166
than zero let delta equal
the minimum of the set.
00:18:40.436 --> 00:18:45.866
One and once again and
unknown quantity such that
00:18:50.646 --> 00:18:56.246
if zero is less than
absolute x minus two is less
00:18:56.246 --> 00:19:02.156
than delta then, go down
to the bottom of the page.
00:19:04.276 --> 00:19:09.196
Absolute value of f of x
minus L is less than epsilon.
00:19:09.196 --> 00:19:11.336
In other words f of
x is within epsilon
00:19:11.336 --> 00:19:13.876
of L. What's the function?
00:19:15.296 --> 00:19:17.916
X squared plus 3x minus five.
00:19:19.606 --> 00:19:20.596
And what's the limit?
00:19:21.296 --> 00:19:27.096
Five. We need that to be less
than epsilon, simplify that.
00:19:28.466 --> 00:19:31.776
That's x squared
plus 3x minus 10.
00:19:33.796 --> 00:19:35.306
Need that to be less
than epsilon.
00:19:36.766 --> 00:19:40.886
Factor it and keep each
factor inside absolute value.
00:19:46.926 --> 00:19:48.866
And now we know that
we need a bound
00:19:49.006 --> 00:19:51.026
on the absolute value
of x plus five.
00:19:51.596 --> 00:19:59.936
At this point we know
that x is approaching two.
00:20:00.046 --> 00:20:03.766
And we do a number line working
00:20:03.766 --> 00:20:08.076
with our other delta
value of one.
00:20:11.026 --> 00:20:15.896
So we go one to the left of
two and one to the right of two
00:20:15.896 --> 00:20:18.966
since we know that x
is approaching two.
00:20:21.586 --> 00:20:22.776
Now we come up with a bound
00:20:22.776 --> 00:20:24.886
on the absolute value
of x plus five.
00:20:27.426 --> 00:20:34.336
Absolute value of x plus five
if x was one would be six.
00:20:34.726 --> 00:20:38.066
And the absolute
value of x plus five
00:20:38.066 --> 00:20:42.956
if x was three would be eight.
00:20:45.556 --> 00:20:51.296
Therefore the absolute value of
x plus five is less than eight.
00:20:54.886 --> 00:20:57.026
So we have that the
absolute value
00:20:57.026 --> 00:21:00.476
of x plus five times
the absolute value
00:21:00.476 --> 00:21:06.036
of x minus two is
less than eight times.
00:21:06.636 --> 00:21:09.546
And what are we going
to put in for delta?
00:21:09.656 --> 00:21:14.376
Because we know that
the absolute value
00:21:14.376 --> 00:21:18.726
of x minus two is
less than delta.
00:21:20.126 --> 00:21:25.186
We're going to put in
epsilon over eight, yes.
00:21:25.666 --> 00:21:30.136
We're going to put in epsilon
over eight because epsilon
00:21:30.140 --> 00:21:33.240
over eight times eight
gives us epsilon.
00:21:36.300 --> 00:21:39.480
And then write your conclusion
at the very bottom of the proof.
00:21:41.220 --> 00:21:56.440
Therefore the limit as x
approaches two equals five.
00:22:04.300 --> 00:22:05.820
Let's find our limit.
00:22:07.180 --> 00:22:11.620
So we have the square of
-1 minus three times -1.
00:22:11.626 --> 00:22:15.836
So we have one plus three is
four so the limit is four.
00:22:18.656 --> 00:22:26.386
So given epsilon greater
than zero we choose delta
00:22:26.846 --> 00:22:30.486
to be the minimum of the
set that contains one
00:22:31.026 --> 00:22:33.496
and an unknown quantity
that we'll figure out later.
00:22:38.106 --> 00:22:52.696
Then for x within delta of
-1 we get absolute value
00:22:52.696 --> 00:22:59.486
of the quantity f of x minus
L equaling absolute value
00:22:59.486 --> 00:23:05.736
of x squared minus 3x minus
the limit of four, factor that,
00:23:06.756 --> 00:23:09.156
be sure to put each
factor in absolute value.
00:23:12.816 --> 00:23:14.956
And now we know that
we need a bound
00:23:15.326 --> 00:23:17.566
on the absolute value
of x minus four.
00:23:18.016 --> 00:23:20.606
If we come over here
this simplifies
00:23:20.706 --> 00:23:23.276
as the absolute of x plus one.
00:23:24.466 --> 00:23:26.246
So we know we have a bound
on the absolute value
00:23:26.246 --> 00:23:28.426
of x plus one, that
bound is delta.
00:23:29.056 --> 00:23:35.766
We know that we have a bound
on x plus one; we need a bound
00:23:36.116 --> 00:23:37.956
on the absolute value
of x minus four.
00:23:41.396 --> 00:23:44.536
So now we need to
do our number line.
00:23:53.406 --> 00:23:57.286
So delta equals one
puts us where?
00:24:00.556 --> 00:24:02.146
Between -2 and zero.
00:24:05.626 --> 00:24:08.956
So the absolute value of x
minus four would be what?
00:24:10.896 --> 00:24:16.796
-2 minus four is -6,
absolute value of -6 is six.
00:24:18.086 --> 00:24:22.896
Absolute value of x
minus four would be four.
00:24:26.476 --> 00:24:31.696
So the absolute value of x
minus four is smaller than six,
00:24:33.156 --> 00:24:34.836
and that's the bound
that we needed.
00:24:36.846 --> 00:24:40.716
So this is smaller than six
times and what's the bound
00:24:40.716 --> 00:24:42.516
on the absolute value
of x plus one?
00:24:43.136 --> 00:24:51.726
Delta. So that's six times what
was delta was smaller than?
00:24:52.116 --> 00:24:54.206
Well delta is the
smaller of the two,
00:24:54.206 --> 00:24:58.276
one or our unknown quantity
there and that needs
00:24:58.276 --> 00:25:01.446
to be epsilon over
six because we need
00:25:01.446 --> 00:25:04.146
that product to be epsilon.
00:25:05.386 --> 00:25:08.126
So that's the unknown
quantity that goes in our box.
00:25:09.216 --> 00:25:13.916
So as long as delta was the
smaller of those two then
00:25:14.246 --> 00:25:20.976
if x is within delta of -1, that
guarantees that f of x will be
00:25:20.976 --> 00:25:24.036
within epsilon of
the limit of four.
00:25:25.836 --> 00:25:32.726
Which guarantees that the
limit as x approaches -1
00:25:33.340 --> 00:25:36.920
of our function is four.
00:25:51.680 --> 00:25:53.876
Alright we know the
limit's four, so we're going
00:25:53.880 --> 00:25:58.380
to do this proof again from a
different formatting approach.
00:26:00.420 --> 00:26:07.340
For every epsilon greater than
zero let delta equal the minimum
00:26:07.346 --> 00:26:16.416
of the set that contains one and
an unknown quantity such that
00:26:21.146 --> 00:26:27.616
if zero is less than absolute
value of x minus -1 is less
00:26:27.616 --> 00:26:32.736
than delta, then and we go down
to the bottom of the proof.
00:26:34.446 --> 00:26:40.506
Absolute value of f of x
minus L is less than epsilon.
00:26:41.476 --> 00:26:44.976
Our function is x
squared minus 3x.
00:26:45.546 --> 00:26:46.726
And our limit's four.
00:26:48.356 --> 00:26:51.016
We factor that, being
careful to put each factor
00:26:51.016 --> 00:26:52.436
in its own absolute value.
00:26:56.506 --> 00:27:00.816
We have a bound on the absolute
value x plus one, that's delta.
00:27:02.566 --> 00:27:07.206
We need a bound on the
absolute value of x minus four.
00:27:08.396 --> 00:27:10.296
So now we go to our number line.
00:27:19.286 --> 00:27:23.756
So delta equal one to the
left, and delta equal one
00:27:23.896 --> 00:27:25.366
to the right, that's our radius.
00:27:29.566 --> 00:27:33.636
So we need the absolute
value of x minus four.
00:27:36.206 --> 00:27:42.276
So that would be -2 minus
four, that's the absolute value
00:27:42.276 --> 00:27:44.186
of -6, which gives us six.
00:27:44.556 --> 00:27:46.076
We're substituting in zero
00:27:48.736 --> 00:27:52.366
so we get the absolute
value of -4, which is four.
00:27:53.076 --> 00:27:58.616
So the absolute value of x
minus four is smaller than six.
00:28:02.336 --> 00:28:08.956
So the absolute value of x
minus four is smaller than six.
00:28:09.666 --> 00:28:16.116
The absolute value of x plus
one is smaller than delta.
00:28:18.676 --> 00:28:25.346
We need this product; we need
this product to be smaller
00:28:25.346 --> 00:28:31.696
than epsilon which tells
us that what needs to go
00:28:31.696 --> 00:28:37.236
in the box is epsilon over six.
00:28:37.236 --> 00:28:41.546
In other words the other value
for delta is epsilon over six.
00:28:45.006 --> 00:28:47.726
That completes our proof
after we write our conclusion,
00:28:49.286 --> 00:28:52.436
the limit as x approaches -1
00:28:53.146 --> 00:28:57.076
of our function x
squared minus 3x is four.
00:28:57.076 --> 00:29:01.296
So as long as delta is
the minimum of the set one
00:29:01.296 --> 00:29:07.446
and epsilon over six, if
x is within delta of -1,
00:29:08.586 --> 00:29:13.846
that forces f of x to be within
epsilon of the limit four.
00:29:20.206 --> 00:29:21.916
Here's some additional problems
00:29:22.196 --> 00:29:24.216
that I encourage you
to try on your own.
00:29:26.776 --> 00:29:30.806
Pause the video, write out
the proofs following your
00:29:30.806 --> 00:29:37.636
instructor's format and then
restart the video and check
00:29:37.676 --> 00:29:42.206
with the delta values that I
have at the end of the video.
00:29:47.666 --> 00:29:51.406
Here are the delta values for
the extra practice problems.
00:29:52.016 --> 00:29:55.236
Although the correct answer
is a well written proof
00:29:55.236 --> 00:29:58.076
that matches the format
laid out by your instructor.