It looks like you uploaded an image with mathematical limits problems. I will go ahead and extract and solve them step by step. Let's start with Problem 12:

12. Evaluate each limit, if it exists.

a) $\lim_{x \to 0} \frac{\sqrt{9 + x} - 3}{x}$

This is an indeterminate form, $\frac{0}{0}$, so we can rationalize the numerator.

To do this, multiply both the numerator and denominator by $\sqrt{9 + x} + 3$:

$\frac{\sqrt{9 + x} - 3}{x} \cdot \frac{\sqrt{9 + x} + 3}{\sqrt{9 + x} + 3} = \frac{(\sqrt{9 + x})^2 - 3^2}{x(\sqrt{9 + x} + 3)} = \frac{9 + x - 9}{x(\sqrt{9 + x} + 3)} = \frac{x}{x(\sqrt{9 + x} + 3)}$

Simplifying:

$= \frac{1}{\sqrt{9 + x} + 3}$

Now, as $x \to 0$, this expression becomes:

$\frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

So the limit is:

$\lim_{x \to 0} \frac{\sqrt{9 + x} - 3}{x} = \frac{1}{6}$

Would you like me to continue solving the rest of the problems or focus on any specific part in more detail?