1
$\begingroup$

Function $t: \mathbb{R}\rightarrow\mathbb{R}$ is given with $|t(z)| < 1$ for all $z \in\mathbb{R}$. It's unknown wheter $t(z)$ is continuous.

Let $s(z) = (z-3) \cdot t(z)$ and now prove that $s$ is continuous at $z_{0}=3$

It was recommended from task to use epsilon-delta, so I will use it.

Let $\varepsilon > 0$, let $\delta > 0$ and let $|z-z_{0}|<\delta$

$$|(z-3)\cdot t(z)- (z_{0}-3)\cdot t(z_{0})|$$

Since task says $|t(z)| < 1$, I'm allowed to write:

$$|(z-3)\cdot t(z)- (z_{0}-3)\cdot t(z_{0})| < |(z-3)-(z_{0}-3)|$$

And this is same as

$$|z-3-z_{0}+3| = |z-z_{0}|< \delta = \varepsilon$$


I have troubles with epsilon-delta still and I think this task is different level for me because there is another function included and we only know it's smaller 1, maybe not continuous etc.

Anyway I tried and I'd like to know if I did correct and if not please tell me how to do it correct and my mistake.

$\endgroup$
2
$\begingroup$

You cannot get rid of the $t(z)$ inside the absolute values like that since you don't know the signs. However, instead note that the point $z_0=3$ and you get that you need to show that

$$|(z-3)t(z)-0|<\epsilon$$

however this is not difficult, as you know that

$$|(z-3)t(z)| = |z-3||t(z)|\le |z-3|<\delta$$

so just choose $\delta = \epsilon$.

$\endgroup$
  • $\begingroup$ My result same as your, i mean end $\delta = \varepsilon$. I have question, my solve is complete wrong or only a bit wrong? Say this task give me 3 point. How many point you give me? Ty answer I try to understand it now. $\endgroup$ – tenepolis Sep 19 '16 at 12:23
  • $\begingroup$ @tenepolis I'd probably give you 2/3 for what you wrote. You sort of get the main idea, and you structure things well, but you make a significant logical jump which is false. $\endgroup$ – Adam Hughes Sep 19 '16 at 12:24
  • $\begingroup$ We need show at 3 that it's continuous. Why you put $0$ for $z_{0}$? Because $s(3) = 0$ ? $\endgroup$ – tenepolis Sep 19 '16 at 12:26
  • $\begingroup$ Oh haha in exam 2/3 very good I'm very happy with it :D I hope professor same as your :D $\endgroup$ – tenepolis Sep 19 '16 at 12:26

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.