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We are tasked with finding the value of \( p > 0 \) that makes the limit

\[
\lim_{n \to \infty} n^2 \left( \sqrt{n^p + 7} - \sqrt{n^p + 2} \right)
\]

exist and be positive. 

### Step-by-step approach:

1. **Factorization and Simplification**: The expression involves a difference of square roots, so a standard method to simplify this is to multiply by the conjugate. That is, we multiply and divide by \( \sqrt{n^p + 7} + \sqrt{n^p + 2} \), yielding:

   \[
   \sqrt{n^p + 7} - \sqrt{n^p + 2} = \frac{(n^p + 7) - (n^p + 2)}{\sqrt{n^p + 7} + \sqrt{n^p + 2}}
   \]

2. **Simplify the Numerator**: The numerator simplifies to:
   \[
   (n^p + 7) - (n^p + 2) = 5
   \]
   Thus, the original expression becomes:
   \[
   n^2 \cdot \frac{5}{\sqrt{n^p + 7} + \sqrt{n^p + 2}}
   \]

3. **Asymptotic Behavior**: For large \( n \), we can analyze the asymptotic behavior of the denominator. When \( n \to \infty \), both \( \sqrt{n^p + 7} \) and \( \sqrt{n^p + 2} \) behave like \( \sqrt{n^p} = n^{p/2} \), so:

   \[
   \sqrt{n^p + 7} + \sqrt{n^p + 2} \approx 2n^{p/2}
   \]

4. **Limit Expression**: Substituting this approximation into the expression, we get:

   \[
   n^2 \cdot \frac{5}{2n^{p/2}} = \frac{5n^2}{2n^{p/2}} = \frac{5}{2} n^{2 - p/2}
   \]

5. **Determining \( p \)**: For the limit to exist and be positive, the exponent of \( n \) must be 0 (so that \( n^{2 - p/2} \) does not blow up or tend to zero). Thus, we set:

   \[
   2 - \frac{p}{2} = 0
   \]

   Solving for \( p \), we get:
   \[
   p = 4
   \]

Thus, the value of \( p \) that makes the limit exist and be positive is \( \boxed{4} \).

---

Would you like more details or further clarification on any step? 

Here are five related questions to expand on this topic:
1. How does the method of multiplying by the conjugate simplify other expressions with square roots?
2. Why is it necessary to analyze the asymptotic behavior when computing limits as \( n \to \infty \)?
3. What other techniques can be used to compute limits involving square roots and powers?
4. How would the problem change if the power \( p \) was negative or non-integer?
5. How do we evaluate the behavior of functions involving roots and powers using L'Hopital's Rule?

**Tip**: Always consider the dominant terms in the numerator and denominator when computing limits for large values of \( n \).

Encuentre p > 0 sabiendo que existe y es positivo el límite lim_{n→∞} n^2 (√(n^p + 7) - √(n^p + 2))

This problem explores the limit of n^2 times the difference of two square roots involving n raised to power p. The solution involves multiplying by the conjugate, simplifying the expression, and using asymptotic expansions to find that the limit exists for p = 4.

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