Evaluate the following limits :
`Lim_( x to a) (x^(a) - a^(a))/(a^(x) - a^(a))`

Answer

To evaluate the limit

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

we start by substituting \( x = a \):

1. **Substitution**:
\[ \frac{a^a - a^a}{a^a - a^a} = \frac{0}{0}. \]
This is an indeterminate form \( \frac{0}{0} \).

**Hint**: When you encounter a \( \frac{0}{0} \) form, you can apply L'Hôpital's Rule.

2. **Applying L'Hôpital's Rule**:
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}. \]

Here, \( f(x) = x^a - a^a \) and \( g(x) = a^x - a^a \).

3. **Differentiate the numerator**:
\[ f'(x) = \frac{d}{dx}(x^a) - \frac{d}{dx}(a^a) = a x^{a-1} - 0 = a x^{a-1}. \]

4. **Differentiate the denominator**:
\[ g'(x) = \frac{d}{dx}(a^x) - \frac{d}{dx}(a^a) = a^x \ln a - 0 = a^x \ln a. \]

5. **Re-evaluate the limit**:
Now substituting the derivatives back into the limit gives:
\[ \lim_{x \to a} \frac{a x^{a-1}}{a^x \ln a}. \]

6. **Substituting \( x = a \)**:
\[ = \frac{a \cdot a^{a-1}}{a^a \ln a} = \frac{a^a}{a^a \ln a} = \frac{1}{\ln a}. \]

Thus, the final answer is:

\[ \lim_{x \to a} \frac{x^a - a^a}{a^x - a^a} = \frac{1}{\ln a}. \]

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