Let `f : R to R` be a real function. The function `f` is double differentiable. If there exists `ninN` and `p in R` such that `lim_(x to oo)x^(n)f(x)=p` and there exists `lim_(x to oo)x^(n+1)f(x)` , then `lim_(x to oo)x^(n+1)f'(x)` is equal to

Explanation

To solve the problem step by step, we need to analyze the given limits and apply L'Hôpital's Rule appropriately. Here's the detailed solution:

### Step-by-Step Solution:

1. **Understanding the Given Limits**:
We are given that:
\[ \lim_{x \to \infty} x^n f(x) = p \]
This means that as \( x \) approaches infinity, the function \( x^n f(x) \) approaches the constant \( p \).

2. **Finding the Limit of \( x^{n+1} f(x) \)**:
We also know that:
\[ \lim_{x \to \infty} x^{n+1} f(x) \]
exists. We can express this limit in terms of the previous limit:
\[ \lim_{x \to \infty} x^{n+1} f(x) = \lim_{x \to \infty} x \cdot (x^n f(x)) = \lim_{x \to \infty} x \cdot p = \infty \text{ (if } p \neq 0\text{)} \]
However, since the limit exists, we conclude \( p = 0 \).

3. **Applying L'Hôpital's Rule**:
We can apply L'Hôpital's Rule to find \( \lim_{x \to \infty} x^{n+1} f(x) \):
\[ \lim_{x \to \infty} \frac{f(x)}{1/x^{n+1}} \text{ (as } x \to \infty, 1/x^{n+1} \to 0\text{)} \]
Differentiating the numerator and denominator gives:
\[ \lim_{x \to \infty} \frac{f'(x)}{-\frac{(n+1)}{x^{n+2}}} \]
Simplifying this, we have:
\[ \lim_{x \to \infty} -x^{n+2} f'(x) = 0 \]

4. **Finding the Limit of \( x^{n+1} f'(x) \)**:
Now, we need to find:
\[ \lim_{x \to \infty} x^{n+1} f'(x) \]
From the previous step, we know:
\[ \lim_{x \to \infty} x^{n+1} f'(x) = -n \cdot p \]
Since we established that \( p = 0 \), we have:
\[ \lim_{x \to \infty} x^{n+1} f'(x) = -n \cdot 0 = 0 \]

5. **Final Result**:
Therefore, the limit we are looking for is:
\[ \lim_{x \to \infty} x^{n+1} f'(x) = 0 \]

### Final Answer:
\[ \lim_{x \to \infty} x^{n+1} f'(x) = 0 \]