The value of `lim_(x->y) (x^y-y^x)/(x^x-y^y)` is:

Options

A. (a) `(1-logx)/(1+logx)`
B. (b) `(1-logy)/(1+logy)`
C. (c) `(logx-logy)/(logx+logy)`
D. (d) None of these

Answer

To solve the limit \( \lim_{x \to y} \frac{x^y - y^x}{x^x - y^y} \), we will follow these steps:

### Step 1: Identify the limit
We start with the limit:
\[ L = \lim_{x \to y} \frac{x^y - y^x}{x^x - y^y} \]

### Step 2: Substitute \( x = y \)
Substituting \( x = y \) directly into the expression gives us an indeterminate form \( \frac{0}{0} \), since both the numerator and denominator become zero. Thus, we need to apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
L'Hôpital's Rule states that if we have an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator:
\[ L = \lim_{x \to y} \frac{(x^y)' - (y^x)'}{(x^x)' - (y^y)'} \]

### Step 4: Differentiate the numerator and denominator
1. **Numerator**:
- The derivative of \( x^y \) with respect to \( x \) is \( y x^{y-1} \).
- The derivative of \( y^x \) with respect to \( x \) is \( y^x \ln y \).
- Therefore, the derivative of the numerator is:
\[ (x^y)' - (y^x)' = y x^{y-1} - y^x \ln y \]

2. **Denominator**:
- The derivative of \( x^x \) with respect to \( x \) is \( x^x (\ln x + 1) \).
- The derivative of \( y^y \) with respect to \( x \) is \( 0 \) since \( y \) is treated as a constant.
- Therefore, the derivative of the denominator is:
\[ (x^x)' - (y^y)' = x^x (\ln x + 1) \]

### Step 5: Rewrite the limit
Now we can rewrite the limit:
\[ L = \lim_{x \to y} \frac{y x^{y-1} - y^x \ln y}{x^x (\ln x + 1)} \]

### Step 6: Substitute \( x = y \) again
Substituting \( x = y \) into the limit:
- The numerator becomes:
\[ y y^{y-1} - y^y \ln y = y^y - y^y \ln y = y^y (1 - \ln y) \]
- The denominator becomes:
\[ y^y (\ln y + 1) \]

### Step 7: Simplify the limit
Thus, the limit simplifies to:
\[ L = \frac{y^y (1 - \ln y)}{y^y (\ln y + 1)} = \frac{1 - \ln y}{\ln y + 1} \]

### Final Result
Therefore, the value of the limit is:
\[ \lim_{x \to y} \frac{x^y - y^x}{x^x - y^y} = \frac{1 - \ln y}{1 + \ln y} \]

### Conclusion
The correct answer is option **b**: \( \frac{1 - \ln y}{1 + \ln y} \).