To solve the limit, we need to find the value of $\lim_{x \to 2^+} \frac{x - 2}{\sqrt{4x - x^2 - 2}}$.

Step-by-step solution:

1. Substituting $x = 2$ into the function:
   • The numerator becomes $2 - 2 = 0$.
   • The denominator becomes $\sqrt{4(2) - (2)^2 - 2} = \sqrt{8 - 4 - 2} = \sqrt{2}$.

Thus, substituting directly gives $\frac{0}{\sqrt{2}} = 0$.

2. Analyzing the limit from the right: Since we are approaching 2 from the right, we should check the behavior of both the numerator and the denominator for values slightly larger than 2.

   For $x \to 2^+$, the numerator $x - 2$ is positive but very small, and the denominator $\sqrt{4x - x^2 - 2}$ approaches $\sqrt{2}$ from a positive direction.

3. Conclusion: The limit exists and is equal to 0: $\lim_{x \to 2^+} \frac{x - 2}{\sqrt{4x - x^2 - 2}} = 0.$

Therefore, the value of $a$ is such that the function is continuous at this point, implying that $a = 0$.

If you have any questions or want further details, feel free to ask!

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Here are five related questions you might consider:

1. What are the conditions for a function to be continuous at a point?
2. How can we use L'Hôpital's rule to evaluate limits involving indeterminate forms like $\frac{0}{0}$?
3. How does the behavior of limits differ when approaching from the left versus the right?
4. How do square roots and other radical expressions affect the continuity of a function?
5. Can the function have discontinuities elsewhere in its domain?

Tip: When handling limits involving square roots, always check the domain of the function to ensure that the expression inside the root remains non-negative.