Can I apply formal definition to $\lim\limits_{x\to1} \left[2x+1\right]$?

Question

$\lim_\limits {x\to a} f(x) = L$

If the limit exists, then $f(x)$ must meet the definition.

$\forall \epsilon > 0,\exists \delta > 0: |x-a|<\delta \implies |f(x) - L|<\epsilon$

Answer 1

You can already guess from your function that it will converge to $3$, since it is defined at $1$ and $f(1) = 3$. So if you want to be formal you can do

Given $\epsilon > 0$ choose $\delta = \frac{\epsilon}{2}$.

Then, given any $x$ such that $|x - 1| < \delta$ we know that $|f(x) - 3| = |2x + 1 - 3| = 2|x - 1| < 2\delta < \epsilon$

And you're done.

Answer 2

$$\forall \epsilon > 0,\exists \delta=\frac{\epsilon}2 > 0: |x-1|<\delta \implies |(2x+1) - 3|<\epsilon$$