# Show that $\delta=\min\left\{1,\frac\epsilon{10}\right\}$ implies $\lim_{x\to 1}3x=3$ in the context of the $\epsilon$-$\delta$ definition

Suppose that $\epsilon>0$ and let $\delta=\min\left\{1, \dfrac{\epsilon}{10}\right\}$. Show that $\lim_{x\to 1}3x=3$.

Can someone please help me, I have no idea what to do. I know to start like this:

Let $\epsilon>0$ be given and $\delta=\min\left\{1, \dfrac{\epsilon}{10}\right\}$. If $0<|x-1|<\delta$ then:

$$|3x-3|=3|x-1|<3\delta<3\dfrac{\epsilon}{10}.$$

But how do I get to $=\epsilon$?

You have showed correctly that $$\vert 3x-3\vert<\frac 3{10}\epsilon.$$ But you also know that $\frac3{10}\epsilon<\epsilon$, always. So you may simply say $$\vert 3x-3\vert<\frac3{10}\epsilon<\epsilon$$ and you're done.
N.B. I noticed you wrote you want to get $=\epsilon$ somewhere. This is not what you want. Just take another (very scrutinous) look at the definition of limit once more.