It strikes me that the very statement is simply paraphrasing via the definition that $\lim\limits_{x\to 1} f(x)= 3$ and there is nothing really to prove.
Note: $x\in (1 -\frac \epsilon 2, 1 + \frac \epsilon 2)$ is the exact same thing as saying $|x - 1| < \frac \epsilon 2$[1]. So if we set $\delta = \frac \epsilon 2$, we can put this into the "magic incantation"
For any $\epsilon > 0$, then there exists a $\delta$ (namely $ \frac \epsilon2$) so that whenever $|x - 1|< \delta$ we have $|f(x) - 3| < \epsilon$.
So $\lim\limits_{x\to 1} f(x)= 3$.
....
[1] For all $\epsilon$
$x \in (1 - \frac \epsilon 2, 1+\frac \epsilon 2) \iff$
$1- \frac \epsilon 2 < x < 1 + \frac \epsilon 2 \iff$
$-\frac \epsilon 2 < x -1 < \frac \epsilon 2 \iff$
If $x-1 \ge 0$ then $0 \le x -1 < \frac \epsilon 2$; and if $x-1 < 0$ then $0 < -(x-1) < \frac \epsilon 2 \iff$
$0 \le |x-1| <\frac \epsilon 2 \iff$
$|x-1| < \frac \epsilon 2$