In this video, I will explain with an example
how to prove that a function has a limit, directly from the definition of limit.
For the first example, I have chosen a particularly simple function
so that I can focus on the structure of the proof.
This is the problem:
I want to prove that lim_(x → 3) (2x + 1) = 7,
and I want to do it directly from the definition of limit.
So let's begin by recalling that definition of limit.
I need to show that for EVERY positive ε, there exists a positive δ
such that IF the distance between x and 3 is between 0 and δ,
THEN the distance between 2x + 1 and 7 is smaller than ε.
Since this is what I WANT to show, before I do anything,
let me look at it piece by piece
to figure out what the proof should look like;
what the structure of the proof is.
First, I need to prove that for every positive ε, ____.
So I will begin the proof
by fixing an arbitrary value of ε positive.
Arbitrary, because I need this proof to work for ALL values of ε,
so I don't get to choose ε; I cannot put any constraints on it.
Second, I need to show
that there exists a δ with a certain property.
The easiest way to do this is to say what I am TAKING as δ.
And since I FIRST FIXED ε, I can now take δ DEPENDING on ε.
The order matters a lot:
first fix ε, second take δ as a function of ε.
If I am NOT careful with this, it won't be clear what is arbitrary
and what I am choosing as a function of what,
and I may end up writing some pseudo proof
that is not actually a proof.
And then, I need to prove an implication.
So I will ASSUME the IF part; I FIX a generic real number
and ASSUME the distance between x and 3 is between 0 and δ.
And after doing some math, hopefully I can conclude
that the distance between 2x + 1 and 7 is smaller than ε.
And that should be the structure of the proof.
There are extra things to write, particularly between step 3 and 4.
The most difficult point is to figure out what δ is.
So let's do some rough work to try to come up with it.
For now, I'm going to look ONLY at the implication.
I want to end off the proof
by concluding something about |(2x + 1) - 7|.
I begin with a bit of algebra.
THAT'S the same as |2x - 6|,
and if I take 2 as a common factor, that's 2 ⋅ |x - 3|.
In the middle of my proof,
I'm going to be ASSUMING that |x - 3| is less than δ,
so 2 times it (|x - 3|) will be less than 2δ.
And now, look at what happens:
I want to prove that the left-hand side is less than ε,
and I have HERE that it is less than 2δ.
So if I take 2δ to be ε, or δ to be ε/2, it will work.
And this is okay,
because I am ALLOWED to CHOOSE δ as a function of ε.
So choosing δ as ε/2 is legal.
Notice that this is NOT the ONLY value of δ that works.
I could also pick ANYTHING SMALLER as long as it's positive,
and it would ALSO work.
And in many other proofs,
it's going to be important to keep track of things like this, and to use them.
But in this proof, it's not necessary.
So I'm just going to keep δ as ε/2 for simplicity.
Everything I have done so far wasn't a proof.
I haven't EVEN STARTED the proof yet.
What I was doing
was the rough work I needed to FIGURE OUT how the proof goes.
But now that I've done that, I'm ready; let's write the ACTUAL proof in order.
As we said, first I fix an arbitrary value of ε;
let ε be greater than 0.
And then, I'm going to take δ to be ε/2.
How do I know this?
Because of the ROUGH WORK that I did before.
Then, I fix a generic value of x -- a real number --
and I ASSUME
that the distance between x and 3 is between 0 and δ.
And then, |(2x + 1) - 7| -- I copy the algebra from before --
is going to be equal to 2 ⋅ |x - 3|.
By my assumption, this is smaller than 2δ.
And by my choice of δ, this is ε.
And look at that! I have proven
that the distance between 2x + 1 and 7 is smaller than ε, as needed.
That's it; that completes the proof.
In the next video,
I will present a second example of a proof of the same type,
but for a much more complicated function.
The structure of the proof will be exactly the same,
but the math we need to do between the steps
will be more complex.
And specifically,
finding the right value of δ as a function of ε
will take much more work.