We are asked to calculate the following limit as $x \to 0$:

$\lim_{x \to 0} \frac{25 - x^2}{4 - \sqrt{2x + 6}}$

Step-by-Step Solution:

1. Substitute $x = 0$ into the function:

$\frac{25 - 0^2}{4 - \sqrt{2(0) + 6}} = \frac{25}{4 - \sqrt{6}}$

This substitution doesn't give an indeterminate form (like $0/0$ or $\infty/\infty$). So, we can directly compute it:

2. Simplify the square root:

$\sqrt{2(0) + 6} = \sqrt{6}$

3. Substitute back:

$\frac{25}{4 - \sqrt{6}}$

This is the value of the limit as $x \to 0$.

So the limit is:

$\boxed{\frac{25}{4 - \sqrt{6}}}$

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Do you need further details or have any questions?

Here are five related questions to expand on this topic:

1. How would you simplify $\frac{a}{b - \sqrt{c}}$ using the conjugate?
2. What happens if you try L'Hopital's Rule in this case, even if it's not indeterminate?
3. How would the limit change if the square root term was different, such as $\sqrt{2x + 4}$?
4. Can we approximate $\sqrt{6}$ to get a numerical answer to the limit?
5. What other forms of limits can be solved by direct substitution?

Tip: When calculating limits, always check if direct substitution works first. This saves time when there is no indeterminate form.