Prove using only the epsilon , delta - definition 

That :

$$ \lim\_{x\to 3} \dfrac{\left( \dfrac{1}{x} - \dfrac{1}{3} \right)}{x-3}=L \in \mathbb{R} $$

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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 ?