Prove using only the epsilon , delta - definition Prove using only the epsilon , delta - definition 
That :
$$ \lim_{x\to 3} \dfrac{\left( \dfrac{1}{x} - \dfrac{1}{3} \right)}{x-3}=L \in \mathbb{R} $$

My Try :
Given $\epsilon > 0 $, there exists a delta such that 
$$ 0<|x-3|< \delta$$  implies $$\left|\dfrac{\left( \dfrac{1}{x} - \dfrac{1}{3} \right)}{x-3} – L\right| < \epsilon \ \ \ \ $$  Or $$ \ \ \ \ \left|\dfrac{-(x-3)}{3x (x-3)}-L\right|< \epsilon $$ .
Now what ?
 A: 
My Try :
Given $\epsilon > 0 $, there exists a delta such that $ 0<|x-3|< \delta$  implies $|\dfrac{( \dfrac{1}{x} - \dfrac{1}{3} )}{x-3} – L| < \epsilon$

You're starting in the wrong place.  Since you are responsible for establishing that the definition is satisfied, you're not allowed to just conjure up the $\delta$.  You need to find the $\delta$ that matches a given $\epsilon$.
You also haven't given a value to $L$.  Notice that the definition requires you to say what $L$ is.  We sometimes think about limits as a property of a function, but it's more like a statement about a function, a point, and a value.
The good news is that although you are required to prove the limit with the definition, you can guess the value of the limit with any method you want.
So here is an outline for the process of proving $\lim_{x\to a}f(x) = L$ by definition

*

*Find the value of $L$.


*Use algebra to express $|f(x) - L|$ in terms of $|x-a|$.


*Use the fact that $|x-a|$ can be assumed to be as small as we like (smaller than any $\delta$ we want), to prove that $|f(x)-L|$ can be made as small as necessary (smaller than any $\epsilon$ given us).
But that is just the scratch work.  Your proof should follow an outline like this:

We claim $\lim_{x\to 3} \dfrac{( \dfrac{1}{x} - \dfrac{1}{3} )}{x-3}  = -\dfrac{1}{9}$. [using the answer from mfl].  Given $\epsilon > 0$, let $\delta = \dots$ [here is where you have to insert the $\delta$ you found in scratch work].  Suppose that $0 < |x-3| < \delta$ for some $x$.  Then $|f(x) - (-\frac{1}{9})| = \dots < \epsilon$. [Again, what goes in place of $\dots$ is the work you've already done.]

A: If $x\ne 3$ then
$$\dfrac{\dfrac{1}{x} - \dfrac{1}{3}}{x-3}=\dfrac{\dfrac{3-x}{3x} }{x-3}=\dfrac{3-x}{3x(x-3)}=\dfrac{-1}{3x}.$$ So $L=-\dfrac19.$ Now, for $x>0$ it is
$$\left|\dfrac 19-\dfrac {1}{3x}\right|=\dfrac{|x-3|}{9x}.$$
Consider $t <1.$ Then 
$$|x-3|<t \implies \dfrac{|x-3|}{9x}<\dfrac{t}{18}$$ (note that $|x-3|<t<1\implies x>2$.)
So,
$$\forall \epsilon>0 \exists \delta=\min\{1,18\epsilon\}|0<|x-3|<\delta \implies  \dfrac{|x-3|}{9x}<\epsilon.$$
