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.


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


however this is not difficult, as you know that

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

so just choose $\delta = \epsilon$.

  • $\begingroup$ 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

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