<s>[INST] 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 ?
 [/INST] 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,
$$\forall \epsilon>0 \exists \delta=\min\{1,18\epsilon\}|0<|x-3|<\delta \implies  \dfrac{|x-3|}{9x}<\epsilon.$$
</s>