# Definition of a limit

1. Sep 11, 2013

### Chas3down

1. The problem statement, all variables and given/known data
lim (x - 4) = 2
as x -> 6

eps > 0
delta > 0

No eps given

2. Relevant equations

3. The attempt at a solution
Not sure how to apply equation..
Usually I would do

I assumed you would do..

6 - eps < x - 4 - 2 < 6 + eps
12 - eps < x < 12 + eps

So, it would be min, (12-eps,12+eps)

But what do I do from here?

Last edited: Sep 11, 2013
2. Sep 11, 2013

### Zondrina

You seem to be confused with the core concept of this question ( The limit definition ). Here's an easy way to remember the definition in symbols:

$$\forall \epsilon > 0, \exists \delta(\epsilon) > 0 \space | \space 0 < |x-a| < \delta \Rightarrow |f(x) - L| < \epsilon$$

It basically reads : Forall positive epsilon, there exists a positive delta in terms of epsilon such that we can make $f(x)$ as close to $L$ as we like. A more in depth explanation of what's going on is below.

Considering $0 < |x-a| < \delta$, we know that $|x-a|$ must always be positive, so we never actually consider what happens at $x=a$, only what happens as we approach it. That's where $|x-a| < \delta$ comes into play.

Expanding we get $-\delta < x-a < \delta$ and then $x + \delta > a > x -\delta$.

So there exists a $\delta > 0$ such that $a$ is bounded between $x + \delta$ and $x -\delta$ and we can only arbitrarily get close to $a$.

Using this, what does it say about $|f(x) - L| < \epsilon$?

Well first, lets consider $-\epsilon < f(x) - L < \epsilon$ which yields $f(x) + \epsilon > L > f(x) - \epsilon$. So $\forall \epsilon > 0$ we can make $f(x)$ as close to $L$ as we like. How close do we need to be you might ask? Sufficiently close.

What defines sufficiently close ( You probably hear this a lot )? Well, $f(x)$ varies according to the values of $x$. How far away $x$ is from $a$ depends on $\delta$. So the values of how far $f(x)$ is away from $L$ ( which change according to $\epsilon$ ) will also vary according to $\delta$. Hence $\epsilon$ will change according to the $\delta$ we choose.

So in conclusion we know we can choose a $\delta(\epsilon)$ as to make $f(x)$ as close to $L$ as we like.

So as for your question, start with the $|f(x) - L|$ portion and massage it a bit to get it into a suitable form.

Last edited: Sep 11, 2013
3. Sep 11, 2013

### Seydlitz

I just want to add this picture to supplement Zondrina explanation.

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4. Sep 11, 2013

### Chas3down

Thanks a lot to both of you! When I get back from class I will thoroughly read through all of it and hopefully able to conceptually understand it.

EDIT: I'm an idiot..

i forgot delta = eps in linear functions with a slope of one ;D

Last edited: Sep 11, 2013