# Epsilon-Delta proofs

Hi, right now Im struggeling with this (calc1).

To be honest, I nearly don't understand a thing. What's going on, and when am I done with the proof?

I can plug in the limits and the approached value into the formal definiton of a limit, but that's as far as I get. (I semi-get the easy ones such as Lim(x->1)(3x+1)=4)

So I wonder: Could any of you write an explanation (with a few easy and advanced examples), and feed it all to me with a teaspoon? Can you please explain every single step you make? If you can do it in words in addition to mathematical expressions I would be very grateful. A problem I have is not seeing the connection between Epsilon and Delta, so a baby's description in plain english would help (not that we "easily" see that blablabla imples blabla :P)

And can you please explain how and why you pick certain values for epsilon and delta? I don't understand a thing of what it says in my solution guide (Robert A. Adams)...

I'm going to philosophy class now, but I really won't to understand this thourougly :)

And at the end: are we supposed do use this teqniue for evaluating lim(x->infinity)sinx ? Maybe I'll see it after I've "processed" your replys (if you make any)

Thanks alot!

Cheers!

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quasar987
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Although the $\epsilon-\delta$ concept is perhaps the most crucial concept in calculus, it is the hardest to understand.

The first thing you should do to get you in the zone for this concept is to think of CLOSENESS. Two points are CLOSE together if the distance between the two is small. This is obvious.

Lets begin.

In most of these sorts of problems we want to work out if a function's limit approaches a certain point. If it does, then we say the limit of the function is that point. Technically speaking:

$$\lim_{x\rightarrow a} f(x) = L$$

If you think of $f(x)$ as a graph on the xy-plane, then as you increase $x$ (ie. move right along the $x$-axis) the graph of the function approaches some value of $y$ (which we call $L$; the limit).

Before we can talk about functions we MUST define exactly where we are working. So the first thing you might see in such a problem is a function $f(x)$ that is DEFINED on some open interval. For example, (0,1) is an open interval - but [0,1] is not. We exclude closed intervals because endpoint can cause headaches in this sort of analysis - more on this later.

Ok, so we have our function defined on an open interval - that is, we can clearly see its graph scrawled across the plane. Now here comes the important bit:

A function $f(x)$ approaches a limit $L$ as $x$ approaches some point $a$ in the open interval (see why it was important to define the open interval!) if

$$\mbox{for every } \epsilon > 0 \mbox{ there exists a } \delta > 0 \mbox{ such that } 0 < |x-a| < \delta \, \Leftrightarrow \, |f(x) - L| < \epsilon$$

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arildno
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Gold Member
Dearly Missed
Just an important correction, oxymoron:
The limit definition is concerned with a PUNCTUATED region about x=a, i.e, x=a is excluded from the region we're looking at.
If not, only functions continuous at x=a, for example, would have a well-defined limit at a.

Secondly, it is important to exclude x=a from our region of interest in order for our function to possibly have limits at points not included in the function's domain.

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You should grind this definition into your head until you can say/write it without a second thought. It is that important as it will appear in your final exam and it will pop up in every analysis course that you take. In my opinion the first hurdle in understanding this concept is to actually be able to write the logic out.

This defintion, is a very complex logical statement. It has "for every" ... "there exists" ... "such that" ... "if" ... "then. Because of this, so many people have problems coming to grips with what you actually have to do!

But, like I said before, the main concern is that of closeness or distance between two numbers (in our case a function and a point).

Here is the statement again: A function $f(x) \rightarrow L$ as $x \rightarrow a$ if

$$\forall \, \epsilon > 0 \, \exists \, \delta \mbox{ such that } 0 < |x-a| <\delta \, \Leftrightarrow \, |f(x) - L | < \epsilon$$

Lets go through the statement very slowly.

1. $0 < |x-a| < \delta$. What is this? This part of the statement is where you should begin. It simply says that the distance from $x$ to $a$ is positive and strictly less than $\delta$. In other words, $x \neq a$. The reason for this is the same as why we chose an open interval to define our function. Remember, our function is only defined on an open interval, so we DON'T CARE what happens to the function at the endpoints, and so we don't allow $x$ to get so close to $a$ so that $x = a$.

2. $\epsilon$ and $\delta$ are simply positive, non-zero real numbers. The general idea of these sorts of problems are, given an $\epsilon$ can you find a $\delta$ which satisfies $0 < |x-a| <\delta \, \Leftrightarrow \, |f(x) - L | < \epsilon$. More on this in a minute, but all you should realise is that $\epsilon$ and $\delta$ are just positive, non-zero real numbers.

3. Draw a graph of $f(x)$ on the xy-plane such that $f(x)$ is defined on some open interval, say $(0,x)$ for simplicity. Now choose a point $a \in (0,x)$. Draw a line from $a$ straight up to the graph of $f(x)$ and then a straight line to the $y$-axis, so that it meets the point $f(a)$.

This is the general set up for understanding what is involves in formulating $\epsilon-\delta$ proofs. If you would rather you can try to just picture this, but I suggest you draw it.

The point of these proofs is to prove that there always exists a $\delta > 0$ for any given $\epsilon$. This is crucial. There must always exist a $\delta > 0$. But what is $\epsilon$ and $\delta$?

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I see Arildno has posted. So I want to just say that in doing these sorts of problems, we are not concerned with what happens at the limit point. ONLY WHAT HAPPENS AS THE FUNCTION APPROACHES THE LIMIT POINT. This is a very important idea.

Continuing...

$\epsilon$ is a special kind of positive, nonzero real number. It actually defines a width! The same goes for $\delta$. They are merely widths or intervals. More on this in a minute.

Ok, so you should have your function $f(x)$. As you move along the function $f(x)$ to the right, ie toward your point $a$ inside the open interval, the distance between $x$ and $a$ is decreasing right? Of course it is.

What I want you to do is take an $\epsilon > 0$, say 1, it can be anything at all! The point $f(a) = L$ along the $y$-axis (which corresponds to the point $a$) now has associated with it a width of length $\epsilon$.

|
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- f(a) + $\epsilon$
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- f(a)
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- f(a) - $\epsilon$
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+---------------

Here is the picture

arildno
Homework Helper
Gold Member
Dearly Missed
Do not mix up f(a) and L, oxymoron!

So this is all $\epsilon$ is. It just defined how wide our interval is.

Notice that not all of the graph of $f(x)$ will lie between $L + \epsilon$ and $L - \epsilon$, depending on how you draw your graph. It doesn't have to.

Ok, now as we move along the graph to the right $f(x)$ moves around the plane, perhaps moving into and out of our interval.

|
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- L + $\epsilon$ ---------------------------------------
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- L FUNCTION MOVES ALONG HERE ->
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- L - $\epsilon$ ----------------------------------------
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+-----------------------|$a - \delta$----|a----|$a+\delta$---- x-axis

The definition of $f(x) \rightarrow L$ as $x \rightarrow a$ means that once we get within the interval $[a-\delta, a+\delta]$ (as shown on the diagram above), the graph of the function MUST BE CONTAINED WITHIN $[L+\epsilon,L-\epsilon]$. This is the crucial idea!!

Remember, that we have TOTAL CONTROL over how big $\delta$ is. So even for function which is very wild, (moves up and down all over the place) I can take a very small $\delta$ (which makes the interval very small) such that the function lies within $[L+\epsilon,L-\epsilon]$.

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A few things I want to stress:

1. It doesnt matter what happens at $a$. In fact $f(a)$ might not even be defined there!. This is exactly why $\delta$ is STRICTLY BIGGER THAN $|x-a|$. Limits are about what happens near the limit point, not AT the limit point.

2. There is nothing sacred about $\epsilon$ and $\delta$. So long as they are positive, nonzero.

3. However, typically think of $\epsilon , \delta$ as SMALL numbers. It is only what happens for small $\epsilon$ that matters. Remember the definition states that for "any given $\epsilon > 0$" you want to find a $\delta > 0$ so that the rest of the definition holds.

Lets have a look at an example.

Consider the function

$$f(x) = \left\{ \begin{array}{ll} 1 & \mbox{if } x \neq 2 \\ 3 & \mbox{if } x = 2 \end{array} \right.$$

What is $\lim_{x\rightarrow 2}f(x)$?

Does it even exist? Think about this for a second. Of course the limit exists. Remember, WE DO NOT CARE WHAT HAPPENS AT THE LIMIT POINT!! Only what happens near it. So even if $f(x) = 10000$ when $x = 2$, the limit still exists.

Now what is the limit? Is it 3? It makes sense because at $x=2$, $f(x) = 3$. WRONG! Why? Because I said "at x equals 2". Remember, we dont care what happens AT the limit point!!! Only what happens near it.

The limit is actually 1. Because as $f(x)$ comes closer and closer to $x=2$ BUT NEVER AT THE POINT ITSELF, the function $f(x)$ comes closer and closer to 1 (actually it never leaves 1!).

So, where does the $\epsilon - \delta$ bit come into play. Well, you still have to prove the limit is 1.

To start these proofs always begin with:

Fix $\epsilon > 0$. If you dont start with this you will get 0 marks.

Now, no matter how small $\epsilon$ is I can ALWAYS find a $\delta > 0$ such that the function is contained within that width. Think about it - it is a very smooth, flat function. It is obvious that I can pick a $\delta > 0$ to satisfy the definition. In fact I could choose $\delta = 1 million$ and the function will still remain within the width $[1+\epsilon, 1-\epsilon]$.

Then

$0 < |x-2| < 1,000,000 \, \Leftrightarrow \, |f(x) - 1| < \epsilon$

So we have

$$\lim_{x\rightarrow 2} f(x) = 1$$

Continuing on with that example.

Lets say I dont believe the proof. Lets say I try to be tricky and give you

$$\epsilon = 10^{-100}$$

Can you still find a $\delta > 0$ such that $0 < |x-2| < \delta \, \Leftrightarrow \, |f(x) - 1| < \epsilon$?

Well, this epsilon is mighty small. But we will have no problem

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-$1+10^{-100}$-------------------------
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-1 ___________________ ________
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-$1-10^{-100}$--------------------------
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+------------|$2 -\delta$------|2--------|$2+\delta$

See, even if I choose a stupidly small $\epsilon$ (remember, I am entitled to do so because the statement is FOR ANY $\epsilon > 0$), I can still find a $\delta > 0$ such that the function remains within the width determined by our choice of $\epsilon$. In this case it is easy to see this because the function is flat. Note that I can simply remove the point of the graph at $x=2$ because we dont even need to consider it.

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Consider the function

$$f(x) = \left\{ \begin{array}{cc} 3 & \mbox{if } x \geq 2 \\ 1 & \mbox{if } x < 2 \end{array}\right.$$

This function is VERY similar to the previous one. HOWEVER, in this case the limit does not exist!

In other words, for any $\epsilon > 0$ you give me I cant guarantee a $\delta > 0$.

For example. Say we fix $\epsilon = 1/2$.
-..............................~~~~~~~~~~~ <-function jumps at $x=2$
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- 1.5 -----------------------------------
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- 1 ~~~~~~~~~~~~
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- 0.5 -----------------------------------
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+----------|$2-\delta$-------|2-------|$2+\delta$----

Obviously for any $\delta > 0$ that I choose, the function will NOT be contained with the width (because after the jump at $x=2$ the function is not in the width!). Even if I try my hardest and choose $\delta = 10^{-100}$ the function still jumps out of the width. So this function has no limit at $x=2$.

Notice that if $\epsilon = 10$ then I CAN find a $\delta > 0$. BUT!!! The fact that for at least one $\epsilon > 0$ I cant find a $\delta > 0$ means that the limit doesn't exist. So you MUST be certain that you only consider very small $\epsilon$ because, as in this case, you MAY be able to find a $\delta$ which works - but dont let that fool you. Be certain that the $\epsilon$ you consider is small enough that it may cause concern. In this case $\epsilon = 1/2$ was small enough.

So $\epsilon-\delta$ proofs are all about determining if a function stays within a predetermined width (by a choice of $\epsilon$) for a variable width at your control.

A function approaches a limit point if for any $\epsilon > 0$ I can find a $\delta > 0$ such that the distance between $a$ and $x$ (which is non-zero because we do not allow $x=a$) is strictly less than $\delta$ which implies that $|f(x) - L| < \epsilon$.

See how easy this proof seems to be now.

We are endeavouring to find a $\delta$ for any possibility $\epsilon$ such that the distance between $x$ and $a$ is less than $\delta$ IMPLIES the distance between $f(x)$ and $L$ is less than $\epsilon$.

Remember that $|x-a| < \delta$ is the same thing as $[a-\delta,a+\delta]$.

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First: Thanks a lot to all 3 of you for your replys!

Can you check that I'm doing this right:?

$$\lim_{x\rightarrow \-1} x+1/x^2-1= -1/2$$

$$|f(x)-L| = |x+1/2(x-1)| < |x+1| < \delta\$$

If DELTA = EPSILON then: |f(x)-L| < EPSILON

I'm sorry I didn't write it all completely out. I'm not good with latex.

Well I would use the algebra of limits to work this out - not the $\epsilon-\delta$ method.

$$\lim_{x\rightarrow 1} \left(x+\frac{1}{x^2}-1\right) = \lim_{x\rightarrow 1} x + \lim_{x\rightarrow 1} \frac{1}{x^2} - \lim_{x\rightarrow 1} 1 = 1 + 1 - 1 = 1$$

So I would say that

$$\lim_{x\rightarrow 1} \left(x+\frac{1}{x^2}-1\right) = 1$$

Which doesn't explain why you've written

"$$\lim_{x\rightarrow 1} \left(x+\frac{1}{x^2}-1\right) = -\frac{1}{2}$$"

And also

$$|f(x) - L| = \left|x+\frac{1}{x^2}+\frac{1}{2}\right| = \left|x+\frac{1}{2}\left(\frac{2}{x^2} -1 \right)\right|$$

Not what you have written.

Generally you don't use the $\epsilon-\delta$ method when the function is nice and continuous.

If you wanted to use the $\epsilon-\delta$ method I would do something like this: (assuming what I've written is correct and the limit is actually 1, not -1/2).

Fix $\epsilon > 0$. Then there must be a number $\delta > 0$ such that

$$0 < |x-1| < \delta \, \Leftrightarrow \, |f(x) - 1| < \epsilon$$

Notice that

$$|f(x) - L | = \left|x+\frac{1}{x^2}-\frac{1}{2}\right| = -\frac{1}{2}+\epsilon < x+\frac{1}{x^2} < \frac{1}{2}+\epsilon$$

and if I choose $\delta = \epsilon/2$ for reasonably large $\epsilon$ then the definition of the limit is satisfied by inspection (if you have Maple this is possible). But notice how this got messy very quickly, and I didn't prove this for EVERY possible $\epsilon$. So my first method is highly recommended.

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I'm so sorry, it should be lim(x->-1)

It says in the text that I should verify the limit (-1/2) with the Delta-Epsilon proof.

Oh, and its (x+1) / (x^2-1)

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But

$$\lim_{x\rightarrow -1} \left(x+\frac{1}{x^2}-1\right) = -1$$

not -1/2.

Perhaps you should type the exact question as it is given in the book.

I'm sorry of it wasn't clear but the expression is (x+1) / (x^2-1)

That's better.

Now

$$\lim_{x\rightarrow -1}\left(\frac{x+1}{x^2-1}\right) = \lim_{x\rightarrow -1}\left(\frac{1}{x-1}\right)$$

by factoring the denominator as a difference of squares. Then

$$\lim_{x\rightarrow -1}\left(\frac{1}{x-1}\right) = \frac{\lim_{x\rightarrow -1}(1)}{\lim_{x\rightarrow -1}(x-1)} = \frac{1}{-2} = -\frac{1}{2}$$

So the $\epsilon-\delta$ method will work now. I'll post later as I've got to go.

Ok, thanks.

quasar987