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To solve this limit problem, let’s analyze each part step-by-step:

1. **Expression Overview**: We are given:
   \[
   \lim_{x \to \infty} \left( \frac{\frac{1}{1 + \frac{5}{x}} - \frac{5}{x^2}}{\frac{5}{x} + 1 - \frac{1}{5x^2}} \right) \cdot \frac{x^2}{x^2}
   \]
   The goal is to simplify the limit as \( x \to \infty \) by multiplying both the numerator and the denominator by \( x^2 \).

2. **Multiplying by \( x^2 \)**:
   - When multiplying the numerator by \( x^2 \):
     \[
     x^2 \left(\frac{1}{1 + \frac{5}{x}} - \frac{5}{x^2}\right) = \frac{x^2}{1 + \frac{5}{x}} - 5
     \]
   - When multiplying the denominator by \( x^2 \):
     \[
     x^2 \left(\frac{5}{x} + 1 - \frac{1}{5x^2}\right) = 5x + x^2 - \frac{1}{5}
     \]

3. **Evaluating Each Term as \( x \to \infty \)**:
   - In the **numerator**: 
     \[
     \frac{x^2}{1 + \frac{5}{x}} - 5 \approx x^2 - 5
     \]
     because \(\frac{5}{x} \to 0\) as \( x \to \infty \).
   - In the **denominator**:
     \[
     5x + x^2 - \frac{1}{5} \approx x^2
     \]
     since \(5x/x^2 \to 0\) and \(\frac{1}{5x^2} \to 0\).

4. **Taking the Limit**:
   As \( x \to \infty \), the expression simplifies to:
   \[
   \lim_{x \to \infty} \frac{x^2 - 5}{x^2} = \lim_{x \to \infty} \left(1 - \frac{5}{x^2}\right)
   \]
   Since \(\frac{5}{x^2} \to 0\) as \( x \to \infty \), the limit is:
   \[
   1
   \]

So, the answer for the limit is **1**.

Would you like more details on any specific part of this solution?

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### Related Questions
1. What is the significance of multiplying by \( x^2 \) in limit problems?
2. How would this approach change if we were dealing with a limit as \( x \to 0 \) instead?
3. What other techniques are commonly used to simplify limits at infinity?
4. Can you apply L'Hôpital's Rule to this type of problem? Why or why not?
5. How does the limit change if additional higher-degree terms are in the numerator?

### Tip
When simplifying limits at infinity, always consider dividing by the highest power of \( x \) in the expression to reveal the dominant terms as \( x \to \infty \).

We can now find the limit of ln(y) by multiplying the numerator and denominator by x^2.

This solution explains how to find the limit of ln(y) as x approaches infinity by simplifying the rational expression through multiplication by x^2. Key concepts include understanding limits at infinity and simplifying complex expressions. Suitable for students in Grades 11-12.

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