Is my choice of $\delta$ under the given bounds for M correct in this $\epsilon-\delta$ proof? Are the choices of < or = correct? 
Prove that
$$\lim_{x \to 9} \sqrt{x} = 3$$


Let $\epsilon > 0$.
$\color{red}{\text{Choose} \ \delta = \min\{\epsilon(\sqrt{9-M} + 3), M\} \ \text{where} \ M \in (0,9)}$.
Case 1: $\epsilon \le \frac{M}{\sqrt{9-M} + 3}$
Then $\delta = \epsilon(\sqrt{9-M} + 3)$
$$|\sqrt{x} - 3| = \frac{|x - 9|}{|\sqrt{x} + 3|}$$
$$= \frac{|x - 9|}{\sqrt{x} + 3}$$
$$< \frac{\delta}{\sqrt{x} + 3}$$
$$= \frac{\epsilon(\sqrt{9-M} + 3)}{\sqrt{x} + 3}$$
$$< \frac{\epsilon(\sqrt{9-M} + 3)}{\sqrt{9-M} + 3} = \epsilon$$
because $|x-9| < \epsilon(\sqrt{9-M} + 3) \to |x-9| < M \to \frac{1}{\sqrt{x} + 3} < \frac{1}{\sqrt{9-M} + 3}$
Case 2: $\epsilon \ge \frac{M}{\sqrt{9-M} + 3}$
Then $\delta = M$
$$|\sqrt{x} - 3| = \frac{|x - 9|}{|\sqrt{x} + 3|}$$
$$ = \frac{|x - 9|}{\sqrt{x} + 3}$$
$$< \frac{\delta}{\sqrt{x} + 3}$$
$$ = \frac{M}{\sqrt{x} + 3}$$
$$< \frac{M}{\sqrt{9-M} + 3} \le \epsilon$$
because $|x-9| < M \to \frac{1}{\sqrt{x} + 3} < \frac{1}{\sqrt{9-M} + 3}$
QED
Is that right?

To be honest, I do not quite understand what you wrote. What I can tell you is that $\lim\limits_{x\to a^2}\sqrt x = a$, because we can choose $\delta = \epsilon$ and write $$\vert \sqrt x-a\vert =\frac{\vert x-a^2\vert}{\vert \sqrt x+a\vert}< \vert x-a^2\vert <\delta=\epsilon.$$ $\require{enclose} \enclose{horizontalstrike}{\textrm{This will  always work.}}$

Edit: As Jyrki Lahtonen pointed out, this will of course only work if we know that $\vert \sqrt x+a\vert>1$. If this is not the case, then we have to resort to other (more cumbersome, for as far as I can see) methods. 
