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 $\epsilon =\frac{1}{4 \delta}$

Let $\epsilon = 1$, then for any $\delta > 0$, you can find an $x$ such that: $|x-\dfrac{1}{2}| < \delta$, and $\dfrac{1}{\left|4x-2\right|} \geq 1\iff \left|x-\dfrac{1}{2}\right| \leq \dfrac{1}{4}$. You can take $x$ such that $|x-\dfrac{1}{2}| < \min{{\dfrac{1}{4}, \delta}}$
