Engineering Mathematics Exam > Engineering Mathematics Videos > Epsilon-delta limit definition (Part - 2) - Mathematics
Epsilon-delta limit definition (Part - 2) - Video Lecture - Engineering
FAQs on Epsilon-delta limit definition (Part - 2) - Mathematics
| 1. What is the epsilon-delta limit definition? |  |
Ans. The epsilon-delta limit definition is a mathematical concept used to formally define the limit of a function. It states that a function f(x) has a limit L as x approaches a point c if and only if for every positive number ε, there exists a positive number δ such that if 0 < |x - c| < δ, then |f(x) - L| < ε.
| 2. How is the epsilon-delta limit definition used in mathematics? | |
Ans. The epsilon-delta limit definition is used to rigorously prove and analyze properties of limits in mathematics. It provides a precise way to define and calculate limits, ensuring that they exist and are well-defined. This definition is commonly used in calculus and analysis to establish theorems and solve problems related to limits.
| 3. Can you provide an example of applying the epsilon-delta limit definition? | |
Ans. Sure! Let's consider the function f(x) = 2x + 3. To prove that the limit of f(x) as x approaches 2 is 7 using the epsilon-delta limit definition, we need to show that for every ε > 0, there exists δ > 0 such that if 0 < |x - 2| < δ, then |f(x) - 7| < ε. Given ε > 0, let's choose δ = ε/2. Now, if 0 < |x - 2| < δ, we have: |f(x) - 7| = |(2x + 3) - 7| = |2x - 4| = 2|x - 2| Since 0 < |x - 2| < δ = ε/2, it follows that 0 < 2|x - 2| < 2(ε/2) = ε. Therefore, we have shown that |f(x) - 7| < ε for any ε > 0, which proves that the limit of f(x) as x approaches 2 is indeed 7.
| 4. Why is the epsilon-delta limit definition important in calculus? | |
Ans. The epsilon-delta limit definition is important in calculus because it allows us to rigorously define and analyze the concept of a limit. It provides a precise framework for understanding and calculating limits, ensuring that they exist and are well-defined. This definition is essential for proving theorems and establishing fundamental principles in calculus, such as continuity and differentiability.
| 5. Are there any alternative methods to the epsilon-delta limit definition? | |
Ans. Yes, there are alternative methods to the epsilon-delta limit definition. One such method is the concept of sequential limits, where the limit of a function is defined in terms of the limit of a sequence. Another approach is the use of limit laws and algebraic manipulations to evaluate limits. However, the epsilon-delta limit definition is considered the most rigorous and widely accepted method for defining limits in mathematics, particularly in calculus and analysis.
Text Transcript from Video
It is not unusual in calculus
textbooks to see something-- to
see problems along the lines
of, prove that the limit as x
approaches-- I don't know,
say, as x approaches 1.
Let me think of an interesting
function when x approaches 1.
Let's see.
3x times x minus 1
over x minus 1.
And you might say, hey Sal, why
did you do this x minus 1 in
the numerator and denominator?
Well, I did that because
obviously you can't
have a 0 down here.
So this function is actually
not defined as x is
equal to 1, right?
If I had just put a 3x,
this cancels out for
all the other values.
So this is essentially the
line 3x, except it has a
hole at x is equal to 1.
And that's why I did it.
Because I want to take the
limit, and it's interesting to
take a limit a to a point where
the function doesn't exist.
But the books will say, hey,
prove that this limit is equal
to-- well, it would actually
be equal to 3, right?
Prove that.
And you essentially have
to use the epsilon delta
definition of a limit.
So to prove this, you're
essentially proving
what this means.
And this means-- proving
this is the same thing as
proving-- so you could
just prove that given--.
So if you give me some epsilon
greater than 0-- and remember,
that's how close you want f
of x to get to the limit.
Given some epsilon greater than
0, in order for this to be
true, I need to prove that I
can give you-- that there
exists-- that, you know, I can
give you a delta
greater than 0.
Where as long as the distance
between x and our limit point,
x and 1, is less than delta--
remember, we don't want x to
ever get right on top
of the limit point.
Because a function might
not be defined there.
So the distance between
x and 1 has to be
greater than 0, right?
That's what this tells us.
So I have to prove, you
give me an epsilon, that
I can give you a delta.
And as long as x is within
delta of 1-- and this is all
this is saying-- as long as
the distance between
x and 1, right?
That's when you subtract
2 things and take
the absolute value.
That's just distance.
So as long as the distance
between x and 1 is less than
delta then the distance between
3x-- let me write that up
there-- the distance between 3x
times x minus 1 over x minus 1,
that's the function, right?
The distance between that and
the limit point-- and 3--
is going to be less
than the epsilon.
A lot of calculus text
books, you know, they'll
write it generally.
They're like, the limit of f of
x as x approaches a of equals
l, and so this would be a.
This would be your f of
x, and this would be
your l, right there.
But anyway, when someone says
they want you to prove this,
they essentially want
you to prove this.
I'll give you an epsilon.
You have to prove that any
epsilon that you're given, or
that you give me, that I--
let's say I'm the one who has
to prove it-- any epsilon you
give me, I have to be able to
give you a delta where, as long
as x is within delta of 1,
then the function is within
epsilon of the limit point.
And how do we do that?
And I'll just say, you know, a
lot of proofs, it's just kind
of-- there's not a-- well,
there's not a systematic
way to do proofs.
And that's kind of what
makes them so interesting
on a lot of levels.
But these, there kind of tends
to be a systematic way.
And this problem I've given
you, to some degree, is kind of
the easy one that they'll give
when they ask you
to prove these.
But they usually involve
starting from this point right
here and algebraically
manipulating it until you get--
until this expression
right here looks something
like this expression.
And then you'll have
this expression is less
than, you know, epsilon
divided by something.
Or, you know, some
function of epsilon.
And it's like, oh, well as long
as delta is that, you give me
an epsilon, I can just apply
that, you know, take epsilon
and divide it by
4, or whatever.
And use that as my delta.
Then I've proved the theorem.
And so we'll do that here.
So let's start with this.
Let's start with kind of
where we want to get to.
And then figure out a delta
that's essentially a
function of epsilon.
So that whenever I'm given an
epsilon, I can say, hey,
whatever number you give me,
I'm going to give you back
a number that's, you know,
that divided by whatever.
And then this will all work.
So let's do that.
So immediately we
can simplify this.
We can cross these out.
And when you cross that out,
you still have to say, well, we
can't do that as long as x--
you can only do this as long as
you assume that x
does not equal 1.
Because if x equals 1, this
function is undefined, right?
And that's OK because we only
care about the intervals
as x approaches 1.
We only care about the
intervals as x approaches 1.
We don't care exactly
when x is equal to 1.
So that's fine.
So then this simplifies to--
this right there simplifies
to-- the absolute value of 3x
minus 3 is less than epsilon.
And then we could factor out a
3, so that's, you know, we
could say the absolute value of
3 times x minus 1 is
less than epsilon.
And you might already see, we
already have an x minus 1
there and an x minus 1 there.
If we can get this 3 over
on this side, then we'll
be done, essentially.
So the absolute value of 3
times some other number, it's
going to be the same thing as
the absolute value of 3 times
the absolute value
of x minus 1.
And if you don't believe me,
I mean, try it out with
a bunch of numbers.
This is going to be
positive no matter what.
This is going to be
positive no matter what.
It's going to have the
same magnitude, right?
And of course, that is
less than epsilon.
And this is what we say-- you
know, a lot of times a calculus
teacher will say, if this is
true, this is true, if and
only if this is true.
And that's just a
fancy way of saying.
So sometimes they'll
do a two way arrow.
Sometimes they'll write IF
and only IF, kind of an IFF.
An IF with 2 f's.
But that just means if
that's true, that's true.
And if this is true,
then that's true.
And that's generally the case,
whenever you just do some
algebraic manipulation.
You can manipulate this back to
get that, or that to get that.
And then they'll say, well, if
this is true, this is true
if and only if this is true.
But this is just
algebraic manipulation.
Right?
Because you can go back and
forth between these steps.
And that's true, if and
only if that is true.
And then the absolute value
of 3 is just 3, right?
We know that.
Let's divide both sides by 3
and you get-- well, I can
just get rid of the absolute
value signs right there.
Because we know that's 3.
If you divide both sides of the
equation by 3, you get the
absolute value of x minus 1
is less than epsilon over 3.
So what have we
just done so far?
We've just proved that the
absolute value-- I need to
write this in another color.
We've just proved that the
absolute value of essentially
3x times x minus 1 over x minus
1 minus 3 is less than epsilon,
if and only if x minus 1 is
less than epsilon over 3.
Right?
So it's like, this is true, as
long as the distance between x
and 1 is less than
epsilon over 3.
So we can just use epsilon
over 3 as our delta.
Remember the whole point of
this was like, you give me
an epsilon, you give me a
distance, or you say, I want
to be within this distance
of my limit point.
I want the function to be that
close to the limit point.
And this is just the distance
between the function and
the limit point, right?
This is f of x.
And this is the limit
that it approaches.
So you want to be that close.
And I say, well, you're going
to be that close if and only if
x-- the distance between x and
1, or the distance between x
and the value it's
approaching-- is
less than this.
So we just prove
this algebraically.
So whatever number you give
me-- if you give me a-- if
you say, Sal, I want to be
within 1 of the limit point.
I want f of x to be no further
than 1 from the limit point,
then I'll say, OK, and
I'll give you 1/3, right?
Because this is true.
Let me write that out.
So let's say you pick
1 as an epsilon.
So you say, hey, you need to
prove to me that there's some x
value where, as long as the x
value is no more-- there's
some value of delta.
So as long as we're no more
than delta away from 1, then
the function itself will
be no more than 1 away
from the limit value.
Right?
And then I say, well,
you, gave me an epsilon
that's equal to 1.
So I could just make-- I can
just say this is, you know, we
just proved, if and only if
absolute value of x minus 1 is
less than whatever number
you gave me divided by 3.
And that actually makes
a lot of sense if you
graph this equation.
Let me see.
Let me draw the x-axis, y-axis.
We already established.
This thing looks just like the
equation of 3x except it has
a hole at x is equal to 1.
Let me draw the graph.
It'll have slope.
It'll have the slope of 3.
So it'll be pretty steep.
It'll look something like that.
And when we're at 1--
this point right
here-- there's a hole.
Right?
This is 3.
So all we're saying, we just
proved that-- so if you said
that you wanted to be within 1
of our limit point, that's like
you giving me an epsilon of 1.
So this distance
right here is 1.
Which would be a
pretty big number.
So you want the function
to be within 1 of our
limit point, right?
Because this is
the limit point.
Then all we're saying is, you
just have to take 1/3 of that.
So we just have to be within
1/3-- so this distance right
here is 1/3 away from 1.
Or we could have said it more
generally, this is epsilon.
This is epsilon.
This would be epsilon over 3.
This is epsilon over 3.
And the reason why this is a
proof, is because it shows no
matter what epsilon you give
me, I can always find a delta.
Because I can just take
whatever number you gave
me, and divide it by 3.
And I can divide any real
number greater than 0.
I can divide any
real number by 3.
So I can always-- no matter
what you give me-- I can always
find you something else.
And therefore, I have proven by
the epsilon delta definition of
limits that the limit
as x approaches 1, of
this, is equal to 3.
textbooks to see something-- to
see problems along the lines
of, prove that the limit as x
approaches-- I don't know,
say, as x approaches 1.
Let me think of an interesting
function when x approaches 1.
Let's see.
3x times x minus 1
over x minus 1.
And you might say, hey Sal, why
did you do this x minus 1 in
the numerator and denominator?
Well, I did that because
obviously you can't
have a 0 down here.
So this function is actually
not defined as x is
equal to 1, right?
If I had just put a 3x,
this cancels out for
all the other values.
So this is essentially the
line 3x, except it has a
hole at x is equal to 1.
And that's why I did it.
Because I want to take the
limit, and it's interesting to
take a limit a to a point where
the function doesn't exist.
But the books will say, hey,
prove that this limit is equal
to-- well, it would actually
be equal to 3, right?
Prove that.
And you essentially have
to use the epsilon delta
definition of a limit.
So to prove this, you're
essentially proving
what this means.
And this means-- proving
this is the same thing as
proving-- so you could
just prove that given--.
So if you give me some epsilon
greater than 0-- and remember,
that's how close you want f
of x to get to the limit.
Given some epsilon greater than
0, in order for this to be
true, I need to prove that I
can give you-- that there
exists-- that, you know, I can
give you a delta
greater than 0.
Where as long as the distance
between x and our limit point,
x and 1, is less than delta--
remember, we don't want x to
ever get right on top
of the limit point.
Because a function might
not be defined there.
So the distance between
x and 1 has to be
greater than 0, right?
That's what this tells us.
So I have to prove, you
give me an epsilon, that
I can give you a delta.
And as long as x is within
delta of 1-- and this is all
this is saying-- as long as
the distance between
x and 1, right?
That's when you subtract
2 things and take
the absolute value.
That's just distance.
So as long as the distance
between x and 1 is less than
delta then the distance between
3x-- let me write that up
there-- the distance between 3x
times x minus 1 over x minus 1,
that's the function, right?
The distance between that and
the limit point-- and 3--
is going to be less
than the epsilon.
A lot of calculus text
books, you know, they'll
write it generally.
They're like, the limit of f of
x as x approaches a of equals
l, and so this would be a.
This would be your f of
x, and this would be
your l, right there.
But anyway, when someone says
they want you to prove this,
they essentially want
you to prove this.
I'll give you an epsilon.
You have to prove that any
epsilon that you're given, or
that you give me, that I--
let's say I'm the one who has
to prove it-- any epsilon you
give me, I have to be able to
give you a delta where, as long
as x is within delta of 1,
then the function is within
epsilon of the limit point.
And how do we do that?
And I'll just say, you know, a
lot of proofs, it's just kind
of-- there's not a-- well,
there's not a systematic
way to do proofs.
And that's kind of what
makes them so interesting
on a lot of levels.
But these, there kind of tends
to be a systematic way.
And this problem I've given
you, to some degree, is kind of
the easy one that they'll give
when they ask you
to prove these.
But they usually involve
starting from this point right
here and algebraically
manipulating it until you get--
until this expression
right here looks something
like this expression.
And then you'll have
this expression is less
than, you know, epsilon
divided by something.
Or, you know, some
function of epsilon.
And it's like, oh, well as long
as delta is that, you give me
an epsilon, I can just apply
that, you know, take epsilon
and divide it by
4, or whatever.
And use that as my delta.
Then I've proved the theorem.
And so we'll do that here.
So let's start with this.
Let's start with kind of
where we want to get to.
And then figure out a delta
that's essentially a
function of epsilon.
So that whenever I'm given an
epsilon, I can say, hey,
whatever number you give me,
I'm going to give you back
a number that's, you know,
that divided by whatever.
And then this will all work.
So let's do that.
So immediately we
can simplify this.
We can cross these out.
And when you cross that out,
you still have to say, well, we
can't do that as long as x--
you can only do this as long as
you assume that x
does not equal 1.
Because if x equals 1, this
function is undefined, right?
And that's OK because we only
care about the intervals
as x approaches 1.
We only care about the
intervals as x approaches 1.
We don't care exactly
when x is equal to 1.
So that's fine.
So then this simplifies to--
this right there simplifies
to-- the absolute value of 3x
minus 3 is less than epsilon.
And then we could factor out a
3, so that's, you know, we
could say the absolute value of
3 times x minus 1 is
less than epsilon.
And you might already see, we
already have an x minus 1
there and an x minus 1 there.
If we can get this 3 over
on this side, then we'll
be done, essentially.
So the absolute value of 3
times some other number, it's
going to be the same thing as
the absolute value of 3 times
the absolute value
of x minus 1.
And if you don't believe me,
I mean, try it out with
a bunch of numbers.
This is going to be
positive no matter what.
This is going to be
positive no matter what.
It's going to have the
same magnitude, right?
And of course, that is
less than epsilon.
And this is what we say-- you
know, a lot of times a calculus
teacher will say, if this is
true, this is true, if and
only if this is true.
And that's just a
fancy way of saying.
So sometimes they'll
do a two way arrow.
Sometimes they'll write IF
and only IF, kind of an IFF.
An IF with 2 f's.
But that just means if
that's true, that's true.
And if this is true,
then that's true.
And that's generally the case,
whenever you just do some
algebraic manipulation.
You can manipulate this back to
get that, or that to get that.
And then they'll say, well, if
this is true, this is true
if and only if this is true.
But this is just
algebraic manipulation.
Right?
Because you can go back and
forth between these steps.
And that's true, if and
only if that is true.
And then the absolute value
of 3 is just 3, right?
We know that.
Let's divide both sides by 3
and you get-- well, I can
just get rid of the absolute
value signs right there.
Because we know that's 3.
If you divide both sides of the
equation by 3, you get the
absolute value of x minus 1
is less than epsilon over 3.
So what have we
just done so far?
We've just proved that the
absolute value-- I need to
write this in another color.
We've just proved that the
absolute value of essentially
3x times x minus 1 over x minus
1 minus 3 is less than epsilon,
if and only if x minus 1 is
less than epsilon over 3.
Right?
So it's like, this is true, as
long as the distance between x
and 1 is less than
epsilon over 3.
So we can just use epsilon
over 3 as our delta.
Remember the whole point of
this was like, you give me
an epsilon, you give me a
distance, or you say, I want
to be within this distance
of my limit point.
I want the function to be that
close to the limit point.
And this is just the distance
between the function and
the limit point, right?
This is f of x.
And this is the limit
that it approaches.
So you want to be that close.
And I say, well, you're going
to be that close if and only if
x-- the distance between x and
1, or the distance between x
and the value it's
approaching-- is
less than this.
So we just prove
this algebraically.
So whatever number you give
me-- if you give me a-- if
you say, Sal, I want to be
within 1 of the limit point.
I want f of x to be no further
than 1 from the limit point,
then I'll say, OK, and
I'll give you 1/3, right?
Because this is true.
Let me write that out.
So let's say you pick
1 as an epsilon.
So you say, hey, you need to
prove to me that there's some x
value where, as long as the x
value is no more-- there's
some value of delta.
So as long as we're no more
than delta away from 1, then
the function itself will
be no more than 1 away
from the limit value.
Right?
And then I say, well,
you, gave me an epsilon
that's equal to 1.
So I could just make-- I can
just say this is, you know, we
just proved, if and only if
absolute value of x minus 1 is
less than whatever number
you gave me divided by 3.
And that actually makes
a lot of sense if you
graph this equation.
Let me see.
Let me draw the x-axis, y-axis.
We already established.
This thing looks just like the
equation of 3x except it has
a hole at x is equal to 1.
Let me draw the graph.
It'll have slope.
It'll have the slope of 3.
So it'll be pretty steep.
It'll look something like that.
And when we're at 1--
this point right
here-- there's a hole.
Right?
This is 3.
So all we're saying, we just
proved that-- so if you said
that you wanted to be within 1
of our limit point, that's like
you giving me an epsilon of 1.
So this distance
right here is 1.
Which would be a
pretty big number.
So you want the function
to be within 1 of our
limit point, right?
Because this is
the limit point.
Then all we're saying is, you
just have to take 1/3 of that.
So we just have to be within
1/3-- so this distance right
here is 1/3 away from 1.
Or we could have said it more
generally, this is epsilon.
This is epsilon.
This would be epsilon over 3.
This is epsilon over 3.
And the reason why this is a
proof, is because it shows no
matter what epsilon you give
me, I can always find a delta.
Because I can just take
whatever number you gave
me, and divide it by 3.
And I can divide any real
number greater than 0.
I can divide any
real number by 3.
So I can always-- no matter
what you give me-- I can always
find you something else.
And therefore, I have proven by
the epsilon delta definition of
limits that the limit
as x approaches 1, of
this, is equal to 3.
About this Video
Apr 20, 2026 Last updated
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