Video:Limit: Difference between revisions

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Full timed transcript: <toggledisplay>
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Vipul: Ok, so this talk is going to be about
why under certain circumstances limits don't exist
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We are going to take this example of a function
which is defined like this: sin of one over x
0:00:47.270,0:00:47.699
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Obviously, that definition doesn't work
when x equals zero.
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So this is a function defined only for all non-zero
reals.
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The goal is to figure out what the limit as
x approaches 0 of f(x) is.
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Here is a graph of the function. This is a
y axis, and x axis.
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The function looks like this.
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It is oscillatory.
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As you approach zero it oscillates more, faster
and faster.
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What are the upper and lower limits of oscillation?
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Actually all these things should be the same
height.
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My drawing wasn't good, but, it should all
be the same height, above and below.
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What are these upper and lower limits? [ANSWER!]
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Rui: 1 and -1.
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Vipul: So the lower limit is negative one
and the upper limit is one. Ok, good.
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So what does it mean, what is the limit at
zero for this function? [ANSWER!]
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This is where...you need to really think, so
I might say ok the limit is, looks like it's zero.
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At zero, you say that looks neat, that looks
right because you see when the x value approaches,
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comes close to zero, the f(x) value also comes
close to zero.
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It keeps oscillating between -1and 1,
and it keeps coming.
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I draw a very small ball around zero, like
that.
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The function is going to keep entering this
ball.
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A ball or a square one or whatever.
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A very small neighborhood of this origin point
here in this two-dimensional picture.
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The function graph is going to enter that
repeatedly.
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Do you think the limit is zero? [ANSWER!]
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Rui: No.
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Vipul: No? Why not? Isn't it coming really
close to zero?
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Rui: Sometimes.
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Vipul: What do you mean "sometimes?"
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Rui: It means sometimes it is real close to
zero and then it flies away.
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Vipul: Ok, "flies away." [Hmm] So what's
your objection? What is not happening?
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Rui: We can not trap.
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Vipul: We cannot trap...
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Rui: ...trap it in a neighborhood of zero.
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Vipul: Function not trapped.
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What should the limit be if it is not zero?
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Should it be half, two-thirds, what should
the limit be? [ANSWER!]
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(I'll explain this later), what do you think
the limit should be?
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Rui: It doesn't have a limit.
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Vipul: It doesn't have a limit.
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Ok, so what does that mean?
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Whatever limit you claim the function has
you are wrong...If you claim the function had
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any numerical limit, if you claim if it is half you
are wrong.
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If you claim minus half you are wrong.
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If you claim the limit is 50, you are wrong.
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Whatever claim you make about the limit,
you are wrong.
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So let's try to think of this in terms of the
game between a prover and a skeptic.
0:04:00.780,0:04:02.730
(You should go and review that video
0:04:02.730,0:04:09.730
or read the corresponding material to understand
what I am going to say.)
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It's good if you have also seen the video
on the definition of limit statement being
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false, which builds on that.
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What I am now asking you, what does it mean
to say the limit does not exist?
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As x approaches c [limit] of f(x) does not exist.
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Here c is zero, but that is not relevant...
that is not necessary for the definition.
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Well it is the usual way we say that the
limit statement is false except we need to
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add one step in the beginning, which is for
every L in R [the reals].
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It says that for every L in R [the reals] the statement
limit x approaches c, f(x) equals L, is false.
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So how does it read?
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It says, for every L in R [the reals] there exists epsilon
greater than zero such that for every delta
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greater than zero there exists x, within the
delta neighborhood of c such that f(x) is
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not in the epsilon neighborhood of L.
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How would you interpret this in terms of a
game between a prover and a skeptic?[ANSWER, THINKING ALONG!]
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Rui: For every limit the prover proposes...
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Vipul: This is not quite the same as the limit
game which you may have seen in a previous
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video which was assuming that the limit was
already given as a part of the game.
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This is sort of a somewhat more general game or
a more meta game where part of the game
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is also the prover trying to specify what
the limit should be.
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The first step the prover plays, the prover
is in black, skeptic is in red.
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The first step the prover plays, proposes
a value of the limit. Then?
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Rui: The skeptic chooses an epsilon.
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Vipul: What's the goal of the skeptic in choosing
the epsilon?
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The goal of the skeptic is.. so let's say
the prover chose a limit value L here, that's
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numerical value L here.
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The skeptic picks epsilon.
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The skeptic will pick epsilon, which means
the skeptic is picking this band from L minus
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epsilon to L plus epsilon.
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Now what does the prover try to do?
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The prover tries to pick a delta. What is
the prover trying to do?
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Find a neighborhood of c, such that the
function in that neighborhood of c the function
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is trapped within epsilon of L.
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So in our case, c is zero in this example,
so the prover will be trying to pick a neighborhood
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of zero, is something like... zero plus delta
on the right and zero minus delta on the left.
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What's the goal of the prover?
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To say that whenever x is in this interval,
for all x,
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The prover is trying to say that all for x
in here, the function [difference from L] is less than epsilon.
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The skeptic who is trying to disprove that.
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What does the skeptic need to do?
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Rui: Every time the prover finds an x.
0:07:03.900,0:07:07.540
Vipul: Well the prover finds, picks the delta,
what does the skeptic try to do?
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Rui: Just pick an x.
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Vipul: Picks an x such that the function...
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Rui: Is out of the...
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Vipul: Is outside that thing.
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Let me make this part a little bit more...so
here you have... the same colors.
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This is
the axis...The skeptic...The prover has picked
this point and the skeptic has picked epsilon.
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So this is L plus epsilon, L minus epsilon.
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The prover is now, it so happens that c is
zero here.
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So that everything is happening near the y
axis.
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Now, the prover wants to pick a delta, the
prover wants to pick, like this, should be
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the same.
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So this is c plus delta which c is zero, so
zero plus delta and zero minus delta.
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Now, under what conditions...What happens
next?
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The prover is implicitly trying to claim that
the function, when the x value is close here,
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the function value is trapped here.
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What the skeptic wants to show is that, that's
not true.
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If it isn't true, in order to do that, the
skeptic should pick a value of x.
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So the skeptic needs to pick a value of x
somewhere in this interval such that at that
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value of f(x)...let me just make the x axis...so
the skeptic wants to pick a value of x, maybe
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its somewhere here, such that when you evaluate
the function at x it lies outside.
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If when you evaluate the function at x, and it lies
outside this strip then the skeptic wins and
0:09:11.720,0:09:16.290
if the value of the function of x is inside
the strip then the prover wins.
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Now looking back at this function, the question
is, can the prover pick an L such that regardless,
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so can the prover pick a value of L such that...Is
this whole thing coming?
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Such that regardless of the epsilon that the
skeptic picks, there exists a delta such that
0:09:37.860,0:09:44.439
for all x the function is trapped? Or is it
instead true that the skeptic will win? (i.e.) Is
0:09:44.439,0:09:50.579
it true that whatever L the prover picks there
exists an epsilon, since the skeptic picks
0:09:50.579,0:09:57.360
an epsilon, such that whatever delta the prover
picks the function in not in fact, trapped
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here. What do you think looking at the picture
here?
0:10:00.399,0:10:05.329
Can you trap the function in a rectangle
like this? [ANSWER!]
0:10:05.329,0:10:06.100
Rui: No.
0:10:06.100,0:10:09.930
Vipul: Well, not if it is a very small rectangle.
0:10:09.930,0:10:16.930
What should the skeptic's strategy be?
0:10:17.060,0:10:23.930
The claim is that the limit does not exist,
that is the claim.
0:10:23.930,0:10:25.990
The claim is that this limit doesn't exist.
0:10:25.990,0:10:29.750
What is the skeptic's strategy?
0:10:29.750,0:10:31.990
What do you mean by skeptic strategy?
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Well, the skeptic should have some strategy
that works, so the skeptic should pick an
0:10:37.370,0:10:43.290
epsilon that is smart and then the skeptic
should pick an x that works.
0:10:43.290,0:10:50.209
What epsilon should the skeptic pick? Suppose
the skeptic picks epsilon as 50 million,
0:10:50.209,0:10:52.050
is that a winning strategy?
0:10:52.050,0:10:52.790
Rui: No.
0:10:52.790,0:10:53.899
Vipul: Why not?
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Rui: He should pick something between -1 and
1, right?
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Vipul: Well epsilon is a positive number so
what do you mean?
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Rui: Oh, anything between one, smaller.
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Vipul: Smaller than...
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Rui: Less than one. Epsilon.
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Vipul: Less than one. Why will that work?
0:11:12.470,0:11:19.470
Rui: Because even if it is less than one then
anything, no matter what kind of delta...
0:11:20.930,0:11:27.930
Vipul: Whatever L the prover picked...What
is the width of this interval? The distance
0:11:28.209,0:11:29.589
from the top and the bottom is?
0:11:29.589,0:11:30.279
Rui: 2
0:11:30.279,0:11:30.980
Vipul: [2 times] epsilon.
0:11:30.980,0:11:31.680
Rui: [2 times] epsilon.
0:11:31.680,0:11:38.680
Vipul: 2 epsilon. If epsilon
is less than one, the skeptic's strategy is
pick epsilon less than one any epsilon.
0:11:43.089,0:11:50.089
The skeptic can fix epsilon in the beginning, maybe pick
epsilon as 0.1 or something, but any epsilon
0:11:50.610,0:11:52.019
less than one will do.
0:11:52.019,0:11:59.019
In fact epsilon equal to one will do. Let
us play safe and pick epsilon as 0.1.
0:11:59.810,0:12:00.999
Why does it work?
0:12:00.999,0:12:06.600
Because this 2 epsilon cannot include both
one and minus one.
0:12:06.600,0:12:12.649
It cannot cover this entire thing because
this has width two, from one to minus one.
0:12:12.649,0:12:17.589
If the skeptic picks an epsilon less than
one, regardless of the L the prover has tried,
0:12:17.589,0:12:23.079
the strip is not wide enough to include everything
from minus one to one.
0:12:23.079,0:12:27.990
Regardless of what Delta the prover picks,
we know that however small an interval we
0:12:27.990,0:12:32.180
pick around zero, the function is going to
take all values from negative one to one in
0:12:32.180,0:12:35.759
that small interval.
0:12:35.759,0:12:40.819
Now the skeptic will be able to find an x
such that the function value lies outside
0:12:40.819,0:12:42.290
the interval.
0:12:42.290,0:12:45.579
The skeptic should...the key idea is that
the skeptic pick epsilon small enough, in
0:12:45.579,0:12:50.360
this case the skeptic's choice of epsilon
doesn't depend on what L the prover chose.
0:12:50.360,0:12:51.269
It need not.
0:12:51.269,0:12:52.889
The strategy doesn't.
0:12:52.889,0:12:59.889
Then after the prover has picked a delta,
picked an x such that the function lies outside.
0:13:01.249,0:13:07.410
Regardless of the L the prover picks,
that L doesn't work as a limit because
0:13:07.410,0:13:10.550
the skeptic wins and so the limit doesn't
exist.</toggledisplay>


==Misconceptions==
==Misconceptions==

Revision as of 20:09, 22 December 2012

ORIGINAL FULL PAGE: Limit
STUDY THE TOPIC AT MULTIPLE LEVELS:
ALSO CHECK OUT: Quiz (multiple choice questions to test your understanding) |Page with videos on the topic, both embedded and linked to

Motivation and general idea

{{#widget:YouTube|id=iZ_fCNvYa9U}}

Full timed transcript: [SHOW MORE]

Definition for finite limit for function of one variable

Two-sided limit

{{#widget:YouTube|id=0vy0Fslxi-k}}

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Left hand limit

Right hand limit

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Relation between the limit notions

Definition of finite limit for function of one variable in terms of a game

Two-sided limit

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Non-existence of limit

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Misconceptions

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Conceptual definition and various cases

Formulation of conceptual definition

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Functions of one variable case

This covers limits at and to infinity.

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Real-valued functions of multiple variables case

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