$lim_{x\rightarrow y}\frac{x^{y}-y^{x}}{x^{x}-y^{y}}=$

Question

$lim_{x\rightarrow y}\frac{x^{y}-y^{x}}{x^{x}-y^{y}}=$

Accepted Answer

To solve the limit problem $ \lim_{x \rightarrow y} \frac{x^y - y^x}{x^x - y^y} $, we need to analyze the behavior of the expression as $ x $ approaches $ y $.

### Explanation for Option 2 being correct:

1. **Expression Analysis:**
   - As $ x \rightarrow y $, both the numerator $ x^y - y^x $ and the denominator $ x^x - y^y $ approach 0, resulting in an indeterminate form $ \frac{0}{0} $.
   - To resolve this indeterminate form, we can apply L'Hôpital's Rule, which is applicable for limits of the form $ \frac{0}{0} $.

2. **Applying L'Hôpital's Rule:**
   - Differentiate the numerator: The derivative of $ x^y $ with respect to $ x $ is $ yx^{y-1} $, and the derivative of $ y^x $ with respect to $ x $ is $ y^x \ln y $.
   - Differentiate the denominator: The derivative of $ x^x $ with respect to $ x $ is $ x^x (\ln x + 1) $, and the derivative of $ y^y $ with respect to $ x $ is 0 since it is a constant with respect to $ x $.

3. **Limit Evaluation:**
   - The limit becomes:
     $$
     \lim_{x \rightarrow y} \frac{yx^{y-1} - y^x \ln y}{x^x (\ln x + 1)}
     $$
   - As $ x \rightarrow y $, substitute $ x = y $ in the derivatives:
     $$
     \frac{yy^{y-1} - y^y \ln y}{y^y (\ln y + 1)}
     $$
   - Simplifying, we get:
     $$
     \frac{y^y (1 - \ln y)}{y^y (1 + \ln y)} = \frac{1 - \ln y}{1 + \ln y}
     $$

4. **Conclusion:**
   - This matches Option 2: $ \frac{1 - \log_e y}{1 + \log_e y} $.

### Explanation for why other options are incorrect:

- **Option 1: $ \frac{1 + \log_e y}{1 - \log_e y} $**
  - This option suggests a reversal of the signs in the numerator and denominator compared to the correct expression. The correct application of L'Hôpital's Rule and simplification does not lead to this form.

- **Option 3: $ \frac{-1 + \log_e y}{1 + \log_e y} $**
  - This option incorrectly introduces a negative sign in the numerator. The derivation and simplification process do not support this alteration.

- **Option 4: $ \frac{-1 - \log_e y}{1 - \log_e y} $**
  - This option incorrectly introduces negative signs in both the numerator and denominator. The correct limit evaluation does not yield this result.

In summary, the correct application of L'Hôpital's Rule and subsequent simplification confirms that Option 2 is the correct answer.