I'm having trouble with this exercise, apparently I'm wrong, however there's no process explained in the answers. The answer should be $\delta = 2\epsilon$
Prove the limit $\lim_{x \to 9} \sqrt{x - 5} = 2$
$L = 2,\ c = 9,\ f(x) = \sqrt{x - 5}$
We will work with the following inequalities
$\epsilon \lt 0$
$0 \lt |x - c | \lt \delta$
$|f(x) - L| \lt \epsilon$
$0 \lt |x - 9| \lt \delta$
$|\sqrt{x - 5} - 2| \lt \epsilon$
First we find the open inverval in which $x$ has to be.
$|\sqrt{x - 5} - 2| \lt \epsilon$
$- \epsilon \lt \sqrt{x - 5} - 2 \lt \epsilon$
$2 - \epsilon \lt \sqrt{x - 5} \lt \epsilon + 2$
$\epsilon^2 - 4\epsilon + 4\lt x - 5 \lt \epsilon^2 + 4 \epsilon + 4,\ x \neq 5$
$\epsilon^2 - 4\epsilon + 9 \lt x \lt \epsilon^2 + 4 \epsilon + 9,\ x \neq 5$
Since we will probably never reach 5, we don't care about it. We'll end with the interval $(\epsilon^2 - 4\epsilon + 9,\ \epsilon^2 + 4 \epsilon + 9)$
$\delta = \min{\{ 9 - (\epsilon^2 - 4\epsilon + 9),\ (\epsilon^2 + 4 \epsilon + 9) - 9\}} = \min{\{ \epsilon^2 - 4\epsilon,\ \epsilon^2 + 4 \epsilon\}}$