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encuentra el valor de a para que la función dada sea continua en todo su dominio

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.

Explore how to determine the value of 'a' for which a function remains continuous across its entire domain. Step-by-step solution provided with detailed analysis of limits and continuity. Suitable for advanced high school students and college-level calculus learners.

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