The given problem involves calculating the limit:

$\lim_{{x \to 2}} \frac{4 - x^2}{3 - \sqrt{x^2} + 5}$

Let's solve this step by step.

Step 1: Simplify the expression

First, observe the function: $f(x) = \frac{4 - x^2}{3 - \sqrt{x^2} + 5}$

The square root term simplifies as $\sqrt{x^2} = |x|$, which for $x = 2$, becomes $|2| = 2$. So the denominator becomes:

$3 - \sqrt{x^2} + 5 = 3 - 2 + 5 = 6$

Thus, the expression simplifies to:

$\frac{4 - x^2}{6}$

Step 2: Evaluate the limit as $x \to 2$

Now, substitute $x = 2$ into the simplified expression:

$\frac{4 - 2^2}{6} = \frac{4 - 4}{6} = \frac{0}{6} = 0$

Final Answer:

$\lim_{{x \to 2}} \frac{4 - x^2}{3 - \sqrt{x^2} + 5} = 0$

Would you like further clarification on any of the steps or more details? Here are five related questions that might help deepen your understanding:

1. How does the absolute value affect the simplification of $\sqrt{x^2}$?
2. What happens if you evaluate the limit as $x \to -2$?
3. Can L'Hopital's Rule be applied to limits of this type?
4. How do discontinuities affect limit calculations?
5. What are some techniques to simplify complex limits involving square roots?

Tip: Always check for simplifications in both the numerator and the denominator before applying limit laws.