$\varepsilon$-$\delta$-proof: $s(z)=(z-3)\cdot t(z)$ is continuous at $z_{0}=3$

Question

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

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

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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.

Answer 1

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 $\delta = \epsilon$.