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

• The negation of the statement would be there exists an $\epsilon$ such that for every $\delta$... and the rest of the statement that you have written is fine.
– R_D
Nov 7 '15 at 4:47
• So I can let $\epsilon = \frac{1}{4 \delta}$ Nov 7 '15 at 4:49

No, behold! The choice of $\varepsilon$ should not depend on that of $\delta$.
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 $x \to 1/2$.
• No. :) I proved that the map $x \mapsto |\frac{1}{4x-2}|$ grows indefinitely as $x \to 1/2$, and this implies what you want; the Kf-Sansoo's answer is a particular case. Nov 7 '15 at 5:13
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 $|x-\dfrac{1}{2}| < \min{{\dfrac{1}{4}, \delta}}$