Explaining the Epsilon-Delta Concept in Limits

[spandan]

I'm studying limits now (for the first time) and though have understood the intuitive concept of limit, I didn't get at all the epsilon-delta concept.

What is epsilon and delta? What is x-2? I didn't get anything at all.

So please explain me these in detail.


Thanking you in advanced for your kind help

 

[Fredrik]

Consider the sequence 1/2, 1/3, 1/4,... If you pick a small number epsilon, I can always (no matter what small number you picked) find an integer N such that 1/n is smaller than your epsilon for all n≥N. That's what we mean when se say that the limit of the sequence is 0.

Consider the function f defined by f(x)=x. (Let's keep it as simple as possible). If you pick a small number epsilon, I can always (no matter what number you picked) find a positive number delta such that |f(x)-0| is less than epsilon for all x such that |x-0| is less than delta. That's what we mean when we say that the function goes to 0 as x goes to 0.

The point is that by choosing x "close enough" to 0 (the value that x goes to), we can make f(x) be "close enough" to 0 (the limit of f(x)).

I don't know what you mean by "x-2".

 

[snipez90]

f(x) approaches a limit L (f(x) -> L) as x approaches a (x -> a) if we can make f(x) as close to L as we want provided that x is sufficiently close, but unequal to, a. This is probably the intuitive notion you are familiar with. Note we are not concerned with whether f is even defined at a since we only care about behavior of f as x gets arbitrarily close to a.

This graph helps connect the intuitive definition with the formal definition: http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/precise.gif

Let's be very specific to begin with. It is not hard to convince yourself that the function f(x) = $$\sqrt{|x|}cos(x)$$ approaches 0 as x approaches 0. We want to make f(x) close to 0, so why not make f(x) within $$\frac{1}{10}$$ of 0?

We want
$$\frac{-1}{10} < \sqrt{|x|}cos(x) < \frac{1}{10}$$,
which is the same as
$$|\sqrt{|x|}cos(x)| < \frac{1}{10}$$.
But $$|cos(x)| \leq 1$$,
so $$|\sqrt{|x|}cos(x)| = \sqrt{|x|}\cdot|cos(x)|\leq \sqrt{|x|}$$.

Now we need to make x sufficiently close to a.
By inspection, if we make x within $$\frac{1}{100}$$ of 0, i.e.,
$$|x| < \frac{1}{100}$$,
then $$|\sqrt{|x|}cos(x)| \leq \sqrt{|x|} < \sqrt{\frac{1}{100}} = \frac{1}{10}$$.

This means that if x is within 1/100 of 0, but unequal to 0 (again, don't be concerned with f(0) for the moment), then |f(x) - 0| < 1/10. Now try to generalize this example. Take another look at the definition and the picture, both provided here: http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html

What epsilon was used in the above example? Could we have chosen any number to be epsilon? Which delta was chosen and how does it relate to the given epsilon? If you can relate each component of the specific example shown with those of the precise definition, then you prove the more general statement (bolded above).

 

[JG89]

I liked Fredrik's explanation. It might also help to remember that when speaking about functions, the entire delta-neighborhood must be mapped into the epsilon-neighborhood.

 

[Tac-Tics]

The epsilon-delta definition is very weird! Don't be discouraged if you have trouble understanding it at first.

First, let's get the definition out there so we have something to work with:

$$\lim_{x \to a}f(x) = L \text{ means } \forall \epsilon: \exists \delta: \text{ if } |x - a| < \delta \text{ then } |f(x) - L| < \epsilon$$

That's definitely not an intuitive or simple definition. In one sentence, you have some of the hardest parts of logic put together: mixed quantifiers and implication (and the implication is often written "backwards" using the phrasing "whenever"). But nevermind all the details. Just get a feel for it. You never need the definition ever again once you've proven the continuity of the important classes functions. But it's a good workout for the brain!

In real life, you can never have perfect precision. In any measurement you make, there will always be some amount of error. If you are careful, you can make the bounds of that error very small.

In the definition of a limit, epsilon and delta are error bounds. That is why the absolute values are there. When you say $$|x - a| < \delta$$, what you mean is that x and a are equal within an error tolerance of delta. Similarly with epsilon, f(x), and L.

A continuous function has to do with accuracy of input and output of the function. There needs to be a relationship between accurate output and accurate input. More specifically, a continuous function must be one that allows you get an output as accurate as you desire... simply by inputting accurate enough data.

When coming up with proofs of continuity, you are usually looking for a value for delta in terms of epsilon that satisfy the logical implication. These proofs also make heavy use of the triangle rule for absolute value.

The

 

[Count Iblis]

If the sequence a_n tends to a limit a, then we usually formulate this by saying that for every epsilon>0 there exists a N such that for all
n > N we have that |a_n -a| < epsilon.

A math prof. of a course I was following once said in class that he didn't understand why we always let the small n be larger than the big N

 

[spandan]

I still don't get it very nicely. I would appreciate if you could put it in simpler words.

The $$\epsilon - \delta$$ definition of limit is given as follows:

Let $$f(x)$$ be defined in the neighbourhood of $$a$$. Then $$f(x)$$ is said to tend to limit $$L$$ as $$x$$ approaches to $$a$$, or symbolically,
$$lim_{x \rightarrow a} f(x) =L$$
If for every positive number $$\epsilon$$ - however small it be - there exists a corresponding positive number $$\delta$$ such that
$$0 < |x - a| < \delta \mbox{then} |f(x) - L| < \epsilon$$

I have some questions:
1. Can $$x$$ be equal to $$a$$?
2. Is is that we can select $$\epsilon$$ and $$\delta$$ ourselves? How? Can any small number serve as $$\epsilon$$?

I am also confused about the "Existence of a funstion at a point" part:

A function $$f(x)$$ will exist at a point $$x = a$$ if $$\mbox{left hand limit = right hand limit}$$
$$lim_{x \rightarrow a-h} f(x) = lim_{x \rightarrow a+h} f(x) \mbox{ for } |h| < \delta$$
$$lim_{h \rightarrow 0} f(a-h) = lim_{h \rightarrow 0} f(a+h) \mbox{ for } |h| < \delta$$

Explain this one to me please.


What is h?

Consider the sequence 1/2, 1/3, 1/4,... If you pick a small number epsilon, I can always (no matter what small number you picked) find an integer N such that 1/n is smaller than your epsilon for all n≥N. That's what we mean when se say that the limit of the sequence is 0.

Say if I pick epsilon to be 0.0009, find me the integer N.

 

[lubuntu]

The existence of a function at point has nothing to do with its limit at that point. You mean to say the limit at that point only exists if the left and right limits exist and agree.

 

[Fredrik]

Say if I pick epsilon to be 0.0009, find me the integer N.

Since 1/0.0009=1111.111..., the smallest integer we can use for this purpose is 1112. So I pick N=1112, and you say "ha-ha, that N is useless when I choose epsilon to be 0.000007 instead". But then I calculate 1/0.000007=142857.142... and choose N=142858.

The meaning of the statement "the limit of this sequence is 0" is that I will always win this game if I get to choose my number (N) after you choose yours (epsilon).

 

[Fredrik]

I have some questions:
1. Can $$x$$ be equal to $$a$$?
2. Is is that we can select $$\epsilon$$ and $$\delta$$ ourselves? How? Can any small number serve as $$\epsilon$$?

1. The function f doesn't have to be defined at a, so if we allow x=a, the inequality $|f(x)-L|<\epsilon$ doesn't always make sense. So $0<|x-a|<\delta$ is right.

2. All positive numbers serve as $\epsilon$.

I am also confused about the "Existence of a funstion at a point" part:

That should be "the limit of the function f as x goes to a exists if..."

What is h?

Any number satisfying $0<h<\delta$.

f(x) is said to go to y as x goes to a from the right (x→a+) if

$$\forall\epsilon>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } \exists \delta>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } 0<x-a<\delta\implies |f(x)-y|<\epsilon$$

f(x) is said to go to y as x goes to a from the left (x→a-) if

$$\forall\epsilon>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } \exists \delta>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } 0<a-x<\delta\implies |f(x)-y|<\epsilon$$

 

[lion0001]

I'm studying limits now (for the first time) and though have understood the intuitive concept of limit, I didn't get at all the epsilon-delta concept.

What is epsilon and delta? What is x-2? I didn't get anything at all.

So please explain me these in detail.Thanking you in advanced for your kind help

Consider this example , f(x)_as x approaches 1_= 2x + 3 = 5
We must prove that:
for each positive number ", there is a positive number delta such that

| (2x + 3 ) - 5| < epsilon for all x satisfying 0 < | x -1 | < delta

that is | 2(x - 1 ) | < epsilon , for all x satisfying 0 < | x -1 | < delta

now 2|x - 1 | < epsilon , now divide by 2 to get

| x - 1 | < (epsilon / 2) , here is the thing, look at this and look at

0 < | x - 1 | < delta

both look identical except for the " 0 < " in the delta part

since | x - 1 | < (epsilon/2)

and | x -1 | < delta ( note here i didnt include the " 0 < " part
which is fine because that just tell us that | x -1 | has to be greater than 0

so we can make | x - 1 | < delta = (epsilon/2)

we choose delta = (epsilon/2) , so the statement

$$\forall\epsilon>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } \exists \delta>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } 0<x-a<\delta\implies |f(x)-y|<\epsilon$$

holds since for 0 < | x -1 | < (epsilon/2) we have

delta = (epsilon/2) ==> epsilon = 2*delta

|2(x - 1 ) | < 2(delta) here epsilon = 2 delta ,see above

|2(x -1 ) | < 2 ( epsilon /2 ) here we just simplify

| 2(x-1) | < epsilon as required NOTE: look at this website, several examples in detail about proving with epsilon delta are given

http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx

 

[spandan]

Can I take epsilon = 600 billion as well?

And I remember my Calculus teacher mentioning that delta should be greater than or equal to epsilon. Why is this?

 

[Fredrik]

Can I take epsilon = 600 billion as well?

Look at the definition again:

We say that f(x)→y when x→a if

$$\forall\epsilon>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } \exists \delta>0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } 0<|x-a|<\delta\implies |f(x)-y|<\epsilon$$

($\forall$ means "for all". $\exists$ means "there exists"). The statement "there exists a number $\delta>0$ such that $0<|x-a|<\delta\implies |f(x)-y|<\epsilon$" is going to be true when [tex]\epsilon=6\cdot 10^{11}[/itex] for most functions f that you will encounter. But this tells you nothing, since we only say that f(x)→y when x→a if that statement holds for all $\epsilon>0$, including every member of (for example) the sequence 1/2ⁿ.

And I remember my Calculus teacher mentioning that delta should be greater than or equal to epsilon. Why is this?

You probably just heard him wrong or misunderstood him. Consider the example $f(x)=\sqrt x$ and choose $\epsilon=\frac 1 2$. We obviously want to define the limit in a way that means that this f(x)→0 as x→0, but do you think it's possible to chose a $\delta\geq\epsilon=\frac 1 2$ such that $0<|x-0|<\delta \implies |f(x)-0|<\epsilon$? When x is slightly less than 1/2, f(x) is approximately 1.4 which is $>\epsilon=0.5$, so $|f(x)-0|<\epsilon$ doesn't hold.

 

[Tac-Tics]

Can I take epsilon = 600 billion as well?

The equation has to hold for all epsilon, including 600 billion. However, the challenge is in small numbers.

And I remember my Calculus teacher mentioning that delta should be greater than or equal to epsilon. Why is this?

There is no need for delta to be greater than epsilon. They will be related to each other, but not necessarily in a simple way like this.

 

[spandan]

So if I pick epsilon=0.00001, what would delta be? Say $$lim_{x \rightarrow 2} (2x^2 + 3x - 14) = 0$$. What would be delta? How do we find out? How are epsilon and delta related to each other? So delta is not smaller than or equal to epsilon?
This is how my book solves this question:
Find the limit $$lim_{x \rightarrow 3} (3x - 4)$$ and verify the result.
Solution:
$$lim_{x \rightarrow 3} (3x - 4)$$ = 3*3-4 = 5. To verify that result, we have to show that the corresponding to $$\epsilon > 0$$, there exists $$\delta$$ such that $$|(3x - 4) - 5| < \epsilon$$ whenever $$|x-3| < \delta$$.

Here we have to find $$\delta$$ in terms of $$\epsilon$$, considering $$\epsilon$$ is given. Now

$$|(3x-4)-5| < \epsilon \Rightarrow |3x-9| < \epsilon \Rightarrow |x-3| < (\epsilon / 3)$$
Hence $$\delta \leq (\epsilon / 3)$$.How is delta found out in terms of epsilon in this solution? Is |x - 3| equal to delta? How? I didn't get this solution. Please help me out.

 

[Hurkyl]

Maybe an alternate description will help.

Fred and George are going to play a game. The game goes as follows:
(1) Fred writes a positive number. We will call that number $\epsilon$.
(2) George writes a positive number. We will call that number $\delta$.
(3) Fred writes down another number. We will call that number $x$.

Fred wins if both of the following are true:
(A) $0 < |x - 2| < \delta$
(B) $|(2x^2 + 3x - 14) - 0| \geq \epsilon$
otherwise, George wins.

The definition of limit says that if $\lim_{x \rightarrow 2} (2x^2 + 3x - 14) = 0$, then there is a strategy George can use that will let him win every time. If that limit doesn't exist or is not zero, then there is a strategy Fred can use that will let him win every time.

 

[Fredrik]

This is how my book solves this question:
Find the limit $$lim_{x \rightarrow 3} (3x - 4)$$ and verify the result.
Solution:
$$lim_{x \rightarrow 3} (3x - 4)$$ = 3*3-4 = 5. To verify that result, we have to show that the corresponding to $$\epsilon > 0$$, there exists $$\delta$$ such that $$|(3x - 4) - 5| < \epsilon$$ whenever $$|x-3| < \delta$$.

Here we have to find $$\delta$$ in terms of $$\epsilon$$, considering $$\epsilon$$ is given. Now

$$|(3x-4)-5| < \epsilon \Rightarrow |3x-9| < \epsilon \Rightarrow |x-3| < (\epsilon / 3)$$
Hence $$\delta \leq (\epsilon / 3)$$.





How is delta found out in terms of epsilon in this solution? Is |x - 3| equal to delta? How? I didn't get this solution. Please help me out.

What part of it are you having problems with? The solution you posted clearly shows that if we choose δ=ε/3 (or smaller), we have |(3x-4)-5|<ε for all x that satisfies 0<|x-3|<δ, and by definition of the limit, that means that 3x-4→5 as x→3.

I don't see why you're asking if |x-3| is equal to delta. It seems that you have ignored the words "for all" in all of the replies you got. You should think about what they mean.

 

[Edgardo]

If you have problems understanding the definition of
$$\lim_{x \rightarrow a} f(x) = L$$
start with a really easy example.

Easy example:
Consider $$f(x) = x$$ and $$\lim_{x \rightarrow 5} x = ?$$

Obviously the limit is 5, i.e. $$\lim_{x \rightarrow 5} x = 5$$ but how do you prove
that? (It is obvious if you draw it. Do you know the geometric meaning of the limit?)

To show that the limit is 5, you have to show the following:
$$|x-a|< \delta \Rightarrow |f(x)-L| < \epsilon$$
(this is just taken from the definition. L is the limit)

In other words:
Proof:
Step 0. Choose an appropriate value for delta
Step 1. You start with $$|x-a|< \delta$$,
Step 2. then make some manipulations and
Step 3. arrive at $$|f(x)-L| < \epsilon$$
End of proof
--

In our example $$\lim_{x \rightarrow 5} x$$,

we have a=5 and f(x)=x and we want
to show that L=5. For this, we have to show that:

$$|x-5|< \delta \Rightarrow |x-5|< \epsilon$$
Remember the definition:
$$|x-a|< \delta \Rightarrow |f(x)-L|< \epsilon$$


Step 1. You start with $$|x-5|< \delta$$,
Step 2. then make some manipulations
Step 3. and arrive at $$|x-5|< \epsilon$$.

How do you get from Step 1 to Step 3, i.e. from $$|x-5|< \delta$$ to $$|x-5|< \epsilon$$?
See Step 0: Choose an appropriate value for delta.
An obvious choice is delta = epsilon.
Let's check that:

Proof:
Step 0: Choose delta = epsilon
Step 1: |x-5| < delta = epsilon
Step 2: No complicated manipulations necessary here.
Step 3: |x-5| < epsilon from Step 1, thus |f(x)-L| < epsilon.
End of proof

Note: If you WRITE DOWN THE PROOF always start with Step 0: Choose appropriate delta

Although working "backwards" is a good method to find delta in terms of epsilon
always start with Step 0 for the proof.


---

Another example:

f(x) = 7*x

What is $$\lim_{x \rightarrow 3} 7x$$ ?

$$\lim_{x \rightarrow 3} 7x = 21$$

Proof:
Step 0: Let delta = 1/7 epsilon, then
Step 1: |x-3| < delta = 1/7 epsilon
Step 2: Manipulations
|x-3| < 1/7 epsilon
=> 7 |x-3| < epsilon
=> |7x-21|< epsilon

Step 3: We arrived at
|7x-21| < epsilon, thus
|f(x)-L| < epsilon
End of proof

----

Note that usually you have a delta that depends on epsilon.
In the first example we had
delta = epsilon
in the second we had
delta = 1/7*epsilon

In other examples you may have
delta = epsilon^1/2,
delta = epsilon ^3/2

or just delta = 8 (or some other constant).

------

Here's a nice video on the delta-epsilon definition of a limit:
http://www.youtube.com/watch?v=u06Yrvt2XLc&fmt=18

 

[symbolipoint]

Anyone notice that when functions are not simple linear expressions, that the epsilon-delta limit proofs are usually not easy?

 

[Hurkyl]

Anyone notice that when functions are not simple linear expressions, that the epsilon-delta limit proofs are usually not easy?

That's because linear functions are among the simplest of functions! That's why things like derivatives and the mean value theorem are so important -- they allow you to approximate complicated functions with linear functions. Another particular example is the limit of a quadratic expression; one controls a quadratic term like

$$|x^2 - a|$$​

by factoring it into a product of linear terms

$$|(x - \sqrt{a})| |(x + \sqrt{a}) |$$​

and then finding a way to simultaneously control both (linear) factors.

 

[spandan]

In the definition of a limit, epsilon and delta are error bounds. That is why the absolute values are there. When you say $$|x - a| < \delta$$ , what you mean is that x and a are equal within an error tolerance of delta. Similarly with epsilon, f(x), and L.

Then shouldn't x-a BE delta?

Maybe an alternate description will help.

Fred and George are going to play a game. The game goes as follows:
(1) Fred writes a positive number. We will call that number $$\epsilon$$ .
(2) George writes a positive number. We will call that number $$\delta$$ .
(3) Fred writes down another number. We will call that number $$x$$ .

So then delta and epsilon are not related to each other. Is it? But the graphs I've seen suggest that they ARE related.And delta and epsilon are not related to each other means any can be bigger or smaller than other, right?

I don't see why you're asking if |x-3| is equal to delta. It seems that you have ignored the words "for all" in all of the replies you got. You should think about what they mean.

Thanks, I uinderstood it now.

The confusion is all because a friend of mine told me that the result is done when we prove that delta is smaller than or equal to epsilon. So I thought |x-3| is delta...

 

[Hurkyl]

So then delta and epsilon are not related to each other. Is it? But the graphs I've seen suggest that they ARE related.


And delta and epsilon are not related to each other means any can be bigger or smaller than other, right?

Yes and no. There is no a priori relationship between delta and epsilon in this game. But if George wants to win, he has to pick $\delta$ intelligently, taking advantage of his knowledge of what Fred picked for $\epsilon$. If George ignores $\epsilon$, he will usually lose the game.


It turns out in this particular game, if I've done the analysis correctly, George needs to pick his delta in the interval

$$0 < \delta \leq \frac{\sqrt{121 + 8 \epsilon} - 11}{4}$$

If George does not do so, then Fred can win. Note that if George picks anything in this interval, he will win!

Now, I did my analysis unnecessarily strictly; when George searches for his winning strategy, he will probably settle for an approximate solution (using a method you have probably already seen); he doesn't need to know the entire range of $\delta$ values he can use... he just needs to find a single value that works. He will probably settle on the strategy of picking $\delta$ to be the smaller of the two numbers $\epsilon / 11$ and $1$.



Each different limit yields a different game, and each one will require George to devise a different winning strategy. However, one general principle is that if George has a winning strategy for one particular game, he can always make another winning strategy by choosing a smaller (positive) value. If, for some reason, George really doesn't like having $\delta \geq \epsilon$ he does have the freedom to always make sure that $\delta < \epsilon$: if he uncovers a winning strategy that would say otherwise, he just modifies the winning strategy to give something smaller. But such a restriction is definitely unnecessary.

 

[snipez90]

Anyone notice that when functions are not simple linear expressions, that the epsilon-delta limit proofs are usually not easy?

Hmm I find that for a lot these functions, finding a certain limit just tests how well acquainted you are with manipulating inequalities. Pathological functions are much more fun to play with.

 

[spandan]

Does both $$\epsilon$$ and ε represent epsilon? My textbook uses $$\epsilon$$ (the "belongs to" / "is a member of") sign, but many use ε.

Fred wins if both of the following are true:
(A) $$0 < |x - 2| < \delta$$
(B) $$|(2x^2 + 3x - 14) - 0| \geq \epsilon$$
otherwise, George wins.

The definition of limit says that if $$lim_{x \\rightarrow 2} (2x^2 + 3x - 14) = 0$$ , then there is a strategy George can use that will let him win every time. If that limit doesn't exist or is not zero, then there is a strategy Fred can use that will let him win every time.

Shouldn't it be $$|f(x) - L| < \epsilon$$? You've written $$|f(x) - L| \geq \epsilon$$

There is no a priori relationship between delta and epsilon in this game.

What is "priori"?

It turns out in this particular game, if I've done the analysis correctly, George needs to pick his delta in the interval

$$0 < \delta \leq \frac{\sqrt{121 + 8 \epsilon} - 11}{4}$$

If George does not do so, then Fred can win. Note that if George picks anything in this interval, he will win!

Now, I did my analysis unnecessarily strictly; when George searches for his winning strategy, he will probably settle for an approximate solution (using a method you have probably already seen); he doesn't need to know the entire range of delta values he can use... he just needs to find a single value that works. He will probably settle on the strategy of picking delta to be the smaller of the two numbers epsilon / 11 and 1 .

How did you do the analysis?

Can you do the proof of this problem for me please?

 

[Fredrik]

Then shouldn't x-a BE delta?

No. You really have to think about what the words "for all" (or "for each") mean in the definition.

So then delta and epsilon are not related to each other. Is it? But the graphs I've seen suggest that they ARE related.

There's one delta for each epsilon, so the value of the biggest possible delta you can choose clearly depends on epsilon.

And delta and epsilon are not related to each other means any can be bigger or smaller than other, right?

You will sometimes be able to choose a delta that's > epsilon and sometimes not. It depends on what function you're dealing with.

The confusion is all because a friend of mine told me that the result is done when we prove that delta is smaller than or equal to epsilon. So I thought |x-3| is delta...

You don't prove what delta is. You find the biggest value of delta you can choose.

Does both $$\epsilon$$ and ε represent epsilon?

Yes. It doesn't matter if you write "epsilon" ($\epsilon$) or "varepsilon" ($\varepsilon$).

Shouldn't it be $$|f(x) - L| < \epsilon$$? You've written $$|f(x) - L| \geq \epsilon$$

No. George is the the one who gets to pick delta, so he's the one who wins when the limit exists. He only wins when Fred loses.

What is "priori"?

He meant that there's no relationship between them that holds in general. Any relationship between epsilon and the largest possible choice of delta must be caused by the specific problem we're considering.

http://dictionary.reference.com/browse/a priori

How did you do the analysis?

Can you do the proof of this problem for me please?

You already posted a complete solution to a similar but much simpler problem, and you say that you don't understand it. Would another complete solution really help? I think it would be better if you explained what it is that bothers you about the solution you posted, so you can get some help understanding that.

I think the cause of all your difficulties is that you're consistently ignoring the words "for all".

 

[spandan]

No, I understood it now. I was thinking the proof will be complete once we somehow prove delta is smaller than or equal to epsilon. I thought that in the last step you assumed |x-3| to be delta and somehow change the smaller than sign to smaller than and equals to. This what I had not got. But how do we know that epsilon/3 (and smaller) values of delta work?

But I couldn't solve that (difficult) question.

 

[snipez90]

But how do we know that epsilon/3 (and smaller) values of delta work?

Um, it follows from the proof itself?

$$\lim_{x \to 3}3x-4 = 5$$

Proof:

$$\text{ Given } \epsilon > 0, \text{ choose }\delta = \frac{\epsilon}{3}. \text{ For all x, if } 0 < |x - 3| < \delta, \text{ then } |f(x) - L| = |(3x-4)-5| = 3\cdot|x-3| < 3\cdot \delta = 3\cdot \frac{\epsilon}{3} = \epsilon.$$

$$\text{ We are requiring that } 0 < |x - 3| < \delta = \frac{\epsilon}{3}. \text{ It should be clear why smaller values work }.$$
Your question is kind of ambiguous though. For any epsilon > 0, it should be clear that picking $$\frac{\epsilon}{3}$$ to be delta will work. Try some actual values, but really take another look at the precise definition of the limit how each part relates to this specific proof.

But I couldn't solve that (difficult) question.

Hmm, this is somewhat tricky, take a look at 3) under examples here: http://en.wikibooks.org/wiki/Calculus/Formal_Definition_of_the_Limit . As Hurkyl mentioned, there are different ways to handle these types of limits. But yeah you should try it on your own. Look up the triangle inequality if you're not familiar with it.