If `lim_(xtooo)x-x^(2) ln(1+1/x)=l` then evaluate `(25)^(l)`.

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to \infty} \left( x - x^2 \ln\left(1 + \frac{1}{x}\right) \right) = l \]

Then, we will find \( 25^l \).

### Step-by-step Solution:

1. **Expand \( \ln(1 + \frac{1}{x}) \)**:
We use the Taylor series expansion for \( \ln(1 + x) \):
\[ \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
For \( \ln(1 + \frac{1}{x}) \), we substitute \( x \) with \( \frac{1}{x} \):
\[ \ln\left(1 + \frac{1}{x}\right) = \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} - \frac{1}{4x^4} + \ldots \]

2. **Substitute the expansion into the limit**:
Now, we substitute this expansion into our limit:
\[ x - x^2 \left( \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} - \frac{1}{4x^4} + \ldots \right) \]
This simplifies to:
\[ x - \left( x^2 \cdot \frac{1}{x} - x^2 \cdot \frac{1}{2x^2} + x^2 \cdot \frac{1}{3x^3} - x^2 \cdot \frac{1}{4x^4} + \ldots \right) \]
Which becomes:
\[ x - \left( x - \frac{1}{2} + \frac{1}{3x} - \frac{1}{4x^2} + \ldots \right) \]

3. **Simplify the expression**:
Now we simplify:
\[ x - x + \frac{1}{2} - \frac{1}{3x} + \frac{1}{4x^2} - \ldots \]
The \( x \) terms cancel out, leaving:
\[ \frac{1}{2} - \frac{1}{3x} + \frac{1}{4x^2} - \ldots \]

4. **Take the limit as \( x \to \infty \)**:
As \( x \to \infty \), the terms \( \frac{1}{3x} \), \( \frac{1}{4x^2} \), etc., approach 0. Thus:
\[ \lim_{x \to \infty} \left( \frac{1}{2} - \frac{1}{3x} + \frac{1}{4x^2} - \ldots \right) = \frac{1}{2} \]
Therefore, we have:
\[ l = \frac{1}{2} \]

5. **Evaluate \( 25^l \)**:
Now we need to evaluate \( 25^l \):
\[ 25^l = 25^{\frac{1}{2}} = \sqrt{25} = 5 \]

### Final Answer:
Thus, the final answer is:
\[ \boxed{5} \]