Proving that the limit of a function doesn't exist using negation of epsilon delta $\lim \limits_{x \to \frac{1}{2}}\frac{1}{4x-2}$
I want to use the negation, $\exists \epsilon>0$ such that  $ \forall \delta>0$ , $\lvert\frac{1}{4x-2}-L \rvert \ge \epsilon$, $\forall x$ with $0<\lvert x-\frac{1}{2} \rvert <\delta$
So can I say that because $\lvert \frac{1}{4x-2}\rvert =\lvert \frac{1}{4(x-\frac{1}{2})} \rvert = \frac{1}{4}\lvert \frac{1}{x-\frac{1}{2}} \rvert \ge \epsilon $
Then $\frac{1}{4} \ge \epsilon \lvert x-\frac{1}{2} \rvert$
Can I then let $\epsilon =\frac{1}{4 \delta}$

No, behold! The choice of $\varepsilon$ should not depend on that of $\delta$.
In fact we can prove something stronger than necessary:
If $x \neq 1/2$, then
$$
\bigg| \frac{1}{4x-2} \bigg| = \frac{1}{4}\frac{1}{|x- \frac{1}{2}|};
$$
If $\varepsilon > 0$, then
$\frac{1}{4}\frac{1}{|x - \frac{1}{2}|} > \varepsilon$ if $|x-\frac{1}{2}| < \varepsilon/4$; hence $0 < |x-\frac{1}{2}| < \varepsilon/4$ only if 
$$
\bigg| \frac{1}{4x-2} \bigg| > \varepsilon,
$$
which says that
$$
\bigg| \frac{1}{4x-2} \bigg| \to \infty
$$
as $x \to 1/2$.
