choosing the some instant value for epsilon for limit of function for this proof i choose some $\epsilon$ values by myself 
I want to know this is possible 
Here is the question 
**

$Let E \subset R$ and p is a limit point of E and f:E $\to R$ suppose
  there exist a constant $M>0$ and let $L\in R$ such that $|f(x)-L| <=
> M|x-p|$ for all x $\in E$, prove that $ lim{x \to p} =L $

**
Here is my proof
since p is a limit point of E there should be {pn} in $E $
should exist  with Pn =p , for all n $\in N$ pn $\to$ p. 
Let $\epsilon $ be given 
if Lim(x$\to \infty$) f(x) = L is exist 
there exist  $\delta$ such that 
$$|(f(x)-L| <\epsilon  ,     for all x \in E, 0 < |x-p| < \delta  $$
Let choose   $$\epsilon < \delta /2 (is\ this\ possible?)$$
then 
$$|f(x) -L |< \epsilon < \delta/2  < |x-p| < \delta $$
since  $\delta  < 2|x-p|$
$$|f(x) -L |< \epsilon < \delta/2  < |x-p| < \delta < 2|x-p| $$
since 2 from 2|x-P| , 2 is some M , satisfy the condition M> 0. 
there exist a const M >0 . 
 A: A lot of this didn't make sense to me; there seem to be a lot of typos and left-out information.
Is this the right problem statement?

Let $E \subset \mathbb R$, let $p$ be a limit point of $E$, and let $f : E \to \mathbb R$. Suppose there exists a constant $M > 0$ and $L \in \mathbb R$ such that $|f(x) - L| \leq M |x - p|$ for all $x \in E$. Prove that $\lim_{x\to p}f(x) = L$.

If so, I'll try and correct your proof, but otherwise, I can't even really begin to understand what is happening.
Since $p$ is a limit point of $E$ there should be a sequence $\{p_n\} \subset E\,$ 
 such that the limit $\lim_{n\to\infty} p_n =p$ ( you can't just say $p_n = p$!), for all n in N pn -> p. (you already said this, basically)
Let $\epsilon $ be given. $(*)$
if Lim(x$\to \infty$) f(x) = L is exist. You can't assume this; this is what you're (sort of) trying to conclude. Also, why is $x$ going to $\infty$ now?
there exist  $\delta$ such that 
$$|f(x)-L| <\epsilon  ,     \forall x \in E, 0 < |x-p| < \delta  $$
This is close to what you want to conclude; i.e. you want to prove that there exists $\delta > 0$ so that $|f(x) - L| < \epsilon$ if $|x-p| < \delta$
Let choose   $$\epsilon < \delta /2 (is\ this\ possible?)$$
No, this is not possible; you already chose $\epsilon > 0$ back on line $(*)$. Basically, the rest of the argument fails from this point, though this
$$|f(x) -L |< \epsilon < \delta/2  < |x-p| < \delta $$
is almost the right idea.
Here's the idea for how to fix this.
$\epsilon > 0$ is already set. You can't change it after the beginning of the proof. But you can choose $\delta$ as small as you want, and because $p_n \to p$, you can choose $N$ large enough that $|p_n - p| < \delta$ for $n>N$. So choose $\delta < \epsilon / M$. Then if $$
|p_n - p| < \delta
$$
and
$$
|f(p_n) - L| < M|p_n-p|
$$
what can you conclude?
