Use the formal definition of a functional limit to prove that $\lim_{x\to2}(x^2+x−1)=5.$

Question

Use the formal definition of a functional limit to prove that

$\lim_{x\to2}(x^2+x−1)=5.$

My attempt (apologies, for it is quite short):

$$|x^2 + x -6| < \varepsilon \Leftrightarrow |(x-2)(x+3)| < \varepsilon \Leftrightarrow |x-2| < \frac{\varepsilon}{|x+3|}$$

I'm not quite sure what to do next. Ive seen solutions substituting $|x+3|$ with 5, on others, ive seen them substituting it with 25. I don't quite understand how it makes sense.

Any tips or hints?

Edit: I've seen hints saying that $|x+3| < 6$, so we should take $\delta = \frac{\varepsilon}{6}$, but I'm not sure how can we guarantee $|x+3|$ would be less than $6$.

Answer 1

Let $\delta = \min(1, \frac{\epsilon}{6})$,

then $|x-2| < \delta $, would implie $1 < x < 3$, hence $|x+3| < 6$

$$|(x-2)(x+3)| \le 6 |x-2| < \epsilon$$