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.
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.