Epsilon-delta for a linear function

The definition. $\lim_{x\to a}f(x)=\ell$ means: for every $\varepsilon>0$ there exists $\delta>0$ such that for all $x$ with $|x-a|<\delta$ (and $x\ne a$), $|f(x)-\ell|<\varepsilon$.

Scratch work. We want to make $|(3x-1)-5|$ small. We have
$$|(3x-1)-5|=|3x-6|=3|x-2|.$$
This is smaller than $\varepsilon$ as soon as $|x-2|<\frac\varepsilon3$. So $\delta=\frac\varepsilon3$ should work.

Proof. Let $\varepsilon>0$ and set $\delta=\dfrac\varepsilon3$. If $|x-2|<\delta$, then
$$|(3x-1)-5|=3|x-2|<3\delta=\varepsilon.$$
This proves that $\lim_{x\to2}(3x-1)=5$.

Remark. The scratch work (finding $\delta$) and the proof (checking that it works) are different things. In a written proof, you only present the second one, starting with "Let $\varepsilon>0$".