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

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

  • $\begingroup$ 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. $\endgroup$
    – R_D
    Nov 7 '15 at 4:47
  • $\begingroup$ So I can let $\epsilon = \frac{1}{4 \delta}$ $\endgroup$ Nov 7 '15 at 4:49

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 $x \to 1/2$.

  • $\begingroup$ You used a contradiction there, correct? $\endgroup$ Nov 7 '15 at 5:09
  • $\begingroup$ 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. $\endgroup$
    – Megadeth
    Nov 7 '15 at 5:13
  • $\begingroup$ Why to downvote this answer? $\endgroup$
    – Megadeth
    Nov 7 '15 at 5:22

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.