Main content

## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 1

Lesson 17: Optional videos# Formal definition of limits Part 4: using the definition

Explore the epsilon-delta definition of limits in calculus, as we rigorously prove a limit exists for a piecewise function. Dive into the process of defining delta as a function of epsilon, and learn how to apply this concept to validate limits with precision. Created by Sal Khan.

## Want to join the conversation?

- the video was really awesome, thank you so much

just a little question:

is this "ε = 2∂" true for any kind of function or just this example??(78 votes)- Just this function! Because he defined f(x) to be 2x when x is not 5. Another function f(x) will yield a different function for ε.(154 votes)

- I'm so lost from this video(44 votes)
- At3:20-3:35, Sal says that x is not equal to 5 as if the expression |x-c| < ∂ exactly implies that x is within range of c but not equal to c. For the latter proposition, shouldn't there also be a 0 < at left-hand side of the inequality ? ( 0 < |x-c| < ∂ ). Because we are trying to approach c, not to get there, given the definition of limit.(31 votes)
- the inequality shown @3:25will include c. The inequality should read:

0<|x-c|<delta. When you remove the absolute value braces to evaluate

this expression, you will create 2 inequality expressions: one describing

the range x<c, the other the range x<c. The value x=c is excluded.(3 votes)

- where did the 5 come from in |x-5| ?(10 votes)
- In the last video, Sal wrote down the equation |x - c| < ∂, which means x is within ∂ of c. |x - 5| < ∂ is the same thing, but we are defining c to be 5.(24 votes)

- What would it look like if the limit you were trying to prove was actually wrong? For example, if you tried to prove that for f(x) = x^2, that as x => 3 that L = 10.(11 votes)
- To disprove a limit, we can show that there is some ∈>0 such that there is no δ>0 such that for all c such that |x-c|<δ, |f(x)-L|<∈.

Let's say ∈<1 (because 3^2=9 and |9-10|=1).

We can always pick c=3 so that |x-c|<δ (because |x-c|=0 and 0<δ), but |f(c)-L|>∈ (because |f(c)-L|=|9-10|=1 and 1>∈).

This means that if ∈<1, there is no δ>0 such that for all c such that |x-c|<δ, |f(x)-L|<∈.

Thus, we have disproved that the limit of x^2 as x approaches 3 is 10.

I hope this helps you disprove limits with the epsilon-delta definition!(17 votes)

- What if it's a function that grows in a non linear way, like in an exponential function?

For exemple, in the function f(x) = 2^x. Given E > 0, would the delta in the left side of a value c be different from the delta in the right side of c?(16 votes)- but the proof still stands ,that's where the one sided limit proof comes from i think?(0 votes)

- What does he mean when he writes, '|x - c| < delta'?(3 votes)
- This is a formal way of writing that the difference between x and c is less than some extremely small number, delta.(15 votes)

- Why should we prove that for all epsilon if we have a delta then the limit at that point (at which we have to prove the limit) is going to be equal to L(Here L =limf(x) x->a). We can just take the casewhen delta->0 and see whether the epsilon->0.

If epsilon->0 then we can say that within the range of a+delta and a-delta for every x(such that x not equal to a) the value of f(x) approaches L or is approximately equal to L. Isn't that what a limit is?(2 votes) - Well, if x<y and x<z how y becomes equal to z. Such as if 2<3 and 2<4 than 3<4. Shouldn't the equation be y=kz.(4 votes)
- I'll try to explain it another way.

We want to choose such an y so that x < y. We know that there is a z for which x < z. Therefore we can choose y = z, because z fits the description.(8 votes)

- I don't understand how |2x-10|=2delta can be replaced by |fx-L|=2delta. Can someone please explain? Why are we able to get rid of the 2 on the left side of the equation?(5 votes)
- First off, consider the limit we're taking. It's lim (x-->5) of 2x = 10. Now, from this we see that x = x, c = 5, f(x) = 2x and L = 10. From the statement of epsilon-delta, we have |x-5|<δ (which comes from |x-c|<δ) and |2x-10|<ε (which comes from |f(x)-L|<ε. Now, take |x-5|<δ and multiply both sides by 2. We get |2x-10|<2δ. Now, 2x and 10 are basically f(x) and L. So, substituting them, we have |f(x)-L|<2δ.

Hope that made sense.(3 votes)

## Video transcript

In the last video, we
took our first look at the epsilon-delta definition
of limits, which essentially says if you claim that the limit
of f of x as x approaches C is equal to L, then that
must mean by the definition that if you were given any
positive epsilon that it essentially tells us how close
we want f of x to be to L. We can always find
a delta that's greater than 0, which is
essentially telling us our distance from C such that
if x is within delta of C, then f of x is going to
be within epsilon of L. If we can find a
delta for any epsilon, then we can say
that this is indeed the limit of f of x
as x approaches C. Now, I know what
you're thinking. This seems all very abstract. I want to somehow
use this thing. And what we will do in
this video is use it and to rigorously prove that
a limit actually exists. So, right over here, I've
defined a function, f of x. It's equal to 2x everywhere
except for x equals 5. So it's 2x everywhere for
all the other values of x, but when x is equal to
5, it's just equal to x. So I could have
really just written 5. It's equal to 5 when
x is equal to 5. It's equal to itself. And so we've drawn
the graph here. Everywhere else, it
looks just like 2x. At x is equal to 5, it's
not along the line 2x. Instead, the function
is defined to be that point right over there. And if I were to ask you
what is the limit of f of x as x approaches
5, you might think of it pretty intuitively. Well, let's see. The closer I get
to 5, the closer f of x seems to
be getting to 10. And so, you might fairly
intuitively make the claim that the limit of f of x
as x approaches 5 really is equal to 10. It looks that way. But what we're
going to do is use the epsilon-delta definition
to actually prove it. And the way that most of
these proofs typically go is we define delta
in the abstract. And then we essentially
try to come up with a way that given any epsilon, we can
always come up with a delta. Or another way is
we're going to try to describe our delta as
a function of epsilon, not to confuse you too much. But maybe I shouldn't
use f again. But delta equals
function of epsilon that is defined for
any positive epsilon. So you give me an
epsilon, I just put into our little formula
or little function box. And I will always
get you a delta. And if I can do that for any
epsilon that'll always give you a delta, where this is true,
that if x is within delta of C, then the corresponding f of x
is going to be within epsilon of L. Then the limit
definitely exists. So let's try to do that. So let's think about being
within delta of our C. So, let's think about this
right over here is 5 plus delta, this is 5 minus delta. So that's our range we're
going to think about. We're going to think about
it in the abstract at first. And then we're going
to try to come up with a formula for delta
in terms of epsilon. So how could we describe all of
the x's that are in this range but not equal to 5 itself? Because we really
care about the things that are within delta of 5,
but not necessarily equal to 5. This is just a
strictly less than. They're within a range
of C, but not equal to C. Well, that's going to be all of
the x's that satisfy x minus 5 is less than delta. That describes all of
these x's right over here. And now, what we're going to
do, and the way these proofs typically go, is
we're going to try to manipulate the left-hand
side of the inequality, so it starts to look
something like this, or it starts to look
exactly like that. And as we do that,
the right-hand side of the inequality is going to
be expressed in terms of delta. And then we can
essentially say well, look. If the right-hand
side is in terms of delta and the
left-hand side looks just like that, that really defines
how we can express delta in terms of epsilon. If that doesn't make
sense, bear with me. I'm about to do it. So, if we want x minus 5 to
look a lot more like this, when x is not equal to 5-- in all
of this, this whole interval, x is not equal to 5--
f of x is equal to 2x, our proposed limit
is equal to 10. So if we could somehow get
this to be 2x minus 10, then we're in good shape. And the easiest
way to do that is to multiply both sides
of this inequality by 2. And 2 times the absolute
value of something, that's the same thing as the
absolute value of 2 times that thing. If I were to say 2 times
the absolute value of a, that's the same thing as
the absolute value of 2a. So on the left-hand
side right over here, this is just going to be the
absolute value of 2x minus 10. And it's going to be less
than on the right-hand side, you just end up with a 2 delta. Now, what do we have here
on the left-hand side? Well, this is f of x as
long as x does not equal 5. And this is our limit. So we can rewrite this as f of
x minus L is less than 2 delta. And this is for x
does not equal 5. This is f of x, this
literally is our limit. Now this is interesting. This statement right
over here is almost exactly what we want
right over here, except the right sides
are just different. In terms of epsilon, this
has it in terms of delta. So, how can we define delta
so that 2 delta is essentially going to be epsilon? Well, this is our chance. And this is where we're defining
delta as a function of epsilon. We're going to make 2
delta equal epsilon. Or if you divide
both sides by 2, we're going to make delta
equal to epsilon over 2. Let me switch colors just
to ease the monotony. If you make delta
equal epsilon over 2, then this statement
right over here becomes the absolute
value of f of x minus L is less than,
instead of 2 delta, it'll be less than 2
times epsilon over 2. It's just going to
be less than epsilon. So this is the key. If someone gives you any
positive number epsilon for this function, as long as
you make delta equal epsilon over 2, then any x
within that range, that corresponding
f of x is going to be within epsilon
of our limit. And remember, it has to be
true for any positive epsilon. But you could see how
the game could go. If someone gives
you the epsilon, let's say they want to be
within 0.5 of our limit. So our limit is up here,
so our epsilon is 0.5. So it would literally be the
range I want to be between 10 plus epsilon would be 10.5. And then 10 minus
epsilon would be 9.5. Well, we just came
up with a formula. We just have to make
delta equal to epsilon over 2, which is equal to 0.25. So that'll give us a range
between 4.75 and 5.25. And as long as we pick an
x between 4.75 and 5.25, but not x equals 5, the
corresponding f of x will be between 9.5 and 10.5. And so, you give
me any epsilon, I can just apply this
formula right over here to come up with the delta. This would apply
for any real number. I mean, especially
any positive number. For any epsilon
you give me, I just get a delta defined
this way, and then I can go through this arc. If the absolute value of x
minus 5 is less than delta, if delta is defined
in this way, which I could define for
any epsilon, then it will be the case
that f of x will be within epsilon of our limit.