For this limit $\lim_{x \to 5}{\sqrt{x-1}}=2$ find $ δ$

$\newcommand{\R}{\mathbb{R}}$ I will show that for all $ a > 0$, $$ \lim_{x \to a} \sqrt{x} = \sqrt{a} $$

We have to show that for any $\epsilon > 0$, there is $\delta > 0$ such that for all $x \ge 0$, $$ |x - a| < \delta \Rightarrow |\sqrt{x} - \sqrt{a}| < \epsilon $$

Let $|x - a| < \delta$. Then,

$$ |\sqrt{x} - \sqrt{a}| = \left| \dfrac{(\sqrt{x} - \sqrt{a})(\sqrt{x} + \sqrt{a})}{\sqrt{x} + \sqrt{a}} \right| = \dfrac{|x-a|}{\sqrt{x} +\sqrt{a}} < \dfrac{\delta}{\sqrt{a}} $$

Choose $\epsilon \ge \delta/\sqrt{a}$ or equivalently $\delta \le \epsilon \sqrt{a}$.

In your question, you can apply the change of variable $x - 1 = y$ to have exactly the same form I introduced, where $a = 4$. The change of variable (shifting) does not alter the dependence of $\delta$ over $\epsilon$.

Therefore, you can choose whatever $0 < \delta \le 1 \cdot\sqrt{4} = 2$.