To compute the limit

\[
\lim_{x \to 3} \frac{\sqrt{2x + 3}}{9 - x^2},
\]

we first need to evaluate the expression as \(x\) approaches 3.

1. Substitute \(x = 3\):
\[
\sqrt{2(3) + 3} = \sqrt{6 + 3} = \sqrt{9} = 3,
\]
and
\[
9 - (3^2) = 9 - 9 = 0.
\]

Since the numerator approaches 3 and the denominator approaches 0, we have a form of \(\frac{3}{0}\), leading to \(+\infty\) or \(-\infty\) depending on the direction from which \(x\) approaches 3. To analyze the limit further, let's check from both sides.

2. Check the behavior as \(x\) approaches 3 from the left (\(x \to 3^-\)):
\[
9 - x^2 \to 0^+\quad (as\; x < 3),
\]
thus the expression approaches \(+\infty\).

3. Check the behavior as \(x\) approaches 3 from the right (\(x \to 3^+\)):
\[
9 - x^2 \to 0^-\quad (as\; x > 3),
\]
thus the expression approaches \(-\infty\).

Since the left-hand limit goes to \(+\infty\) and the right-hand limit goes to \(-\infty\), we conclude:

\[
\lim_{x \to 3} \frac{\sqrt{2x + 3}}{9 - x^2} \text{ does not exist}.
\]

Therefore, we write:

\[
\text{The limit does not exist.}
\]