Proving that the limit of a function doesn't exist using negation of epsilon delta

Question

$\lim \limits_{x \to \frac{1}{2}}\frac{1}{4x-2}$

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

Answer 1

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 $x \to 1/2$.

Answer 2

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 $|x-\dfrac{1}{2}| < \min{{\dfrac{1}{4}, \delta}}$