What is the formal epsilon-delta definition and how do you use to prove a limit Could someone explain this:


*Use the formal epsilon-delta of limit to show


$$\lim_{x\to 3}(x+5)=8$$
Be sure to include a concluding statement.
 A: To paraphrase Hagen von Eitzen's definition in the comments above, the FORMAL definition of a limit is:
We say that the limit of a function $f$ exists at a point $a$ if, there is a real number $L$ such that:
For any positive real number $\epsilon$, we can find ANOTHER positive real number $\delta$ (which may, in fact, depend on which $\epsilon$ we are considering), such that whenever the distance from $a$ to $x$ is between $0$ and $\delta$, the distance between $f(x)$ and $L$ is less than $\epsilon$. In this case we say the limit of $f$ at $a$ is $L$.
In symbols:
$\forall \epsilon > 0, \exists \delta > 0: 0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$.
Colloquially, if $x$ is "near" $a$, then $f(x)$ is "near" $L$. We avoid the question of what happens AT $a$, where $f$ may be undefined (but $f$ might be defined at $a$, there are "special cases" where this is important).
For your particular question, you are already told what $L$ is, namely: $L = 8$. So we start by letting $\epsilon$ be "any old positive real number". 
Now that we have $\epsilon$ (which we imagine as being "very small", so we want to be "arbitrarily close" to $L$), we need to find some suitable $\delta$ to make the magic happen (how close to $a$ does $x$ actually have to be?).
How can we ensure that $|f(x) - L| < \epsilon$, for our $f$, and our $L$? Our function $f$ here is given by $f(x) = x + 5$, so:
$|f(x) - L| = |x+5 - 8| = |x - 3|$.
Well, this is convenient, since we are looking to limit the size of $|x - 3|$ in the first place (since our value of "$a$" is $3$, in this problem).
So it appears that $0 < |x - 3| < \epsilon$ will do the trick, that is, we pick $\delta = \epsilon$.
Then, no matter what $\epsilon > 0$ is, there exists our $\delta( = \epsilon)$ (which is, of course, likewise greater than $0$) so that if:
$0 < |x - 3| < \delta$ then that means:
$|x - 3| < \epsilon$, that is:
$|(x + 5) - 8| < \epsilon$, as desired, thus proving $\lim\limits_{x \to 3} (x + 5) = 8$.
A: Given $\epsilon>0$, we have
$$|(x+5)-8|<\epsilon$$
whenever $0<|x-3|<\delta = \epsilon$.
