`lim_(xtoa)(x^a-a^x)/(x^x-a^a)` is equal to

Options

A. `(1+log_ea)/(1-log_ea)`
B. `(log_e(e//a))/(log_e(ae))`
C. `(log_e(a//e))/(log_e(ae))`
D. none of these

Explanation

To solve the limit

\[ \lim_{x \to a} \frac{x^a - a^x}{x^x - a^a}, \]

we can apply L'Hôpital's Rule since both the numerator and denominator approach 0 as \( x \) approaches \( a \).

### Step 1: Verify the form of the limit
First, we check the limit:

- As \( x \to a \), \( x^a \to a^a \) and \( a^x \to a^a \), so the numerator approaches \( 0 \).
- Similarly, \( x^x \to a^a \) and \( a^a \to a^a \), so the denominator also approaches \( 0 \).

Since both the numerator and denominator approach \( 0 \), we can apply L'Hôpital's Rule.

### Step 2: Differentiate the numerator and denominator
We differentiate the numerator and denominator separately.

**Numerator:**
\[ \frac{d}{dx}(x^a - a^x) = a x^{a-1} - a^x \ln(a). \]

**Denominator:**
To differentiate \( x^x \), we can use the logarithmic differentiation:
Let \( y = x^x \).
Taking the natural logarithm:
\[ \ln(y) = x \ln(x). \]
Differentiating both sides:
\[ \frac{1}{y} \frac{dy}{dx} = \ln(x) + 1 \implies \frac{dy}{dx} = y(\ln(x) + 1) = x^x(\ln(x) + 1). \]
Thus,
\[ \frac{d}{dx}(x^x) = x^x(\ln(x) + 1). \]
So, the derivative of the denominator is:
\[ \frac{d}{dx}(x^x - a^a) = x^x(\ln(x) + 1). \]

### Step 3: Apply L'Hôpital's Rule
Now we can apply L'Hôpital's Rule:
\[ \lim_{x \to a} \frac{a x^{a-1} - a^x \ln(a)}{x^x(\ln(x) + 1)}. \]

### Step 4: Substitute \( x = a \)
Now we substitute \( x = a \):
\[ = \frac{a a^{a-1} - a^a \ln(a)}{a^a(\ln(a) + 1)}. \]
This simplifies to:
\[ = \frac{a^a - a^a \ln(a)}{a^a(\ln(a) + 1)} = \frac{a^a(1 - \ln(a))}{a^a(\ln(a) + 1)}. \]
Cancelling \( a^a \) (assuming \( a \neq 0 \)):
\[ = \frac{1 - \ln(a)}{\ln(a) + 1}. \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to a} \frac{x^a - a^x}{x^x - a^a} = \frac{1 - \ln(a)}{\ln(a) + 1}. \]