How to write this non-linear Epsilon-Delta proof w/ a radical?

Question

The question: Prove that $\lim \limits_{x \to 9} 2+\sqrt{x} = 5$.

After reading this link from Milefoot on how to construct a non-linear delta-epsilon proof (second example), this is what I came up with (sorry if this isn't the correct way of writing it):

$|x - 9| < \delta \Rightarrow 0 < |f(x) - 5|< \epsilon$

$|(2 + \sqrt{x}) -5| = |\sqrt{x} -3| < \epsilon$

$-\epsilon < \sqrt{x}-3 < \epsilon$

$(-\epsilon+3)^2 < x < (\epsilon + 3)^2$

$(-\epsilon+3)^2 -9 < x -9< (\epsilon + 3)^2-9$

$\epsilon^2 - 6\epsilon < x -9< \epsilon^2 + 6\epsilon $

Therefore, this works if you define $\delta$ as $\min{(\epsilon^2 - 6\epsilon,\epsilon^2 + 6\epsilon )}$.

However, the solution on the UC Davis website was substantially different. Their final answer was $\delta$ as $\min{(1,(8+\sqrt{3})\epsilon)}$. I realize that there may be several valid answers for a delta-epsilon proof, but the way I did is was so different, and honestly it's a concept I'm struggling with. Could anyone please look over my solution to check if I'm wrong, and if so, where did I go wrong?

Thanks.

Answer 1

What you've done is correct up to $$\epsilon^2 - 6\epsilon < x -9< \epsilon^2 + 6\epsilon.$$ In the given guide, $|x-c|<\delta \iff -\delta < x-c < \delta$.

Observe that $$\epsilon^2 - 6\epsilon = \epsilon (\epsilon -6) \begin{cases} \ge 0 &\text{if } x \le 0 \text{ or } x \ge 6,\\ <0 &\text{if } 0 < x < 6. \end{cases}$$

Therefore, if $0 < \epsilon < 1$, the $\delta$ should be defined as $\min\{6\epsilon - \epsilon^2,\epsilon^2 + 6\epsilon \}$. Then $|x-9|<\delta$ implies $$|x-9| < 6\epsilon - \epsilon^2 \text{ and } |x-9| < \epsilon^2 + 6\epsilon$$ $$\implies 9-x < 6\epsilon - \epsilon^2 \text{ and } x-9 < \epsilon^2 + 6\epsilon$$ $$\iff \epsilon^2 - 6\epsilon < x-9 < \epsilon^2 + 6\epsilon$$

Answer 2

Squaring inequalities involving potentially negative numbers is a recipe for disaster; consider for instance squaring both sides of the inequality $-1 < 0$...

Instead note that $|\sqrt x - 3| < \epsilon \iff |\sqrt x-3|^2 < \epsilon^2$, and that $|\sqrt x-3|^2 \leq |\sqrt x-3||\sqrt x+3| = |x - 9|$. Thus it suffices to take $\delta = \epsilon^2$.

Note that our choice of $\delta$ doesn't even depend on the particular number $9$ here; this says that the square root function is uniformly continuous.