Suppose I have $\epsilon > 0$ and $\delta$ is min of $\left\{1,\frac{\epsilon}{10}\right\}$
If $0< |x-1| < \delta$, is it then true that $|3x-3|< \epsilon$? I am saying yes and here is why.
I can deduce that that $0<|3x-3|<3\delta$ , and since $\delta$ is min of $\left\{1,\frac{\epsilon}{10}\right\}$, then $3\delta$ is min of $\left\{3,\frac{3\epsilon}{10}\right\}$
Now $0<|x-1|< 1$ which leads to $0<|3x-3|< 3$,
or $0< |x-1|<\frac{\epsilon}{10}$, which leads to $0< |3x-3|<\frac{3\epsilon}{10}$
now I conclude that $|3x-3|<\epsilon$, because obviously $\frac{3\epsilon}{10}<\epsilon$
and I realize I have a problem because 3 does not have to be less than $\epsilon$. Suppose $\epsilon$ was 0.1.
So this means my original statement is false.
If $\epsilon =0.1$, then $\delta =\min \{ 1,\frac{0.1}{10}\} =\frac{0.1}{10}=\frac{1}{100}$.
Now we suppose that $0<|x-1|<\frac{1}{100}$ and ask:
Is $|3x-3|<0.1$?
The answer is yes because:
$|x-1|<\frac{1}{100}\Rightarrow 3|x-1|<\frac{3}{100}\Rightarrow |3x-3|<\frac{3}{100}$
and $\frac{3}{100}<0.1$, thus $|3x-3|<\epsilon$ as required.