The problem asks to evaluate the limit:

$\lim_{x \to b} \frac{x^b - b^x}{x^x - b^b}.$

Step-by-step solution:

1. Check the form of the limit:

   When $x = b$, both the numerator and denominator become $0$:

   • Numerator: $x^b - b^x = b^b - b^b = 0$.
   • Denominator: $x^x - b^b = b^b - b^b = 0$.

   Thus, this is a $\frac{0}{0}$ indeterminate form, so we can apply L'Hôpital's Rule.

2. Apply L'Hôpital's Rule:

   L'Hôpital's Rule states that for limits of the form $\frac{0}{0}$, we take the derivatives of the numerator and the denominator.

   $\lim_{x \to b} \frac{x^b - b^x}{x^x - b^b} = \lim_{x \to b} \frac{\frac{d}{dx}(x^b - b^x)}{\frac{d}{dx}(x^x - b^b)}.$

   Now, let’s find the derivatives.

3. Derivative of the numerator: $x^b - b^x$

   • $\frac{d}{dx}(x^b) = b \cdot x^{b-1}$ (using the power rule).
   • $\frac{d}{dx}(b^x) = b^x \ln(b)$ (since $b^x$ is an exponential function).

   So, the derivative of the numerator is: $b \cdot x^{b-1} - b^x \ln(b).$

4. Derivative of the denominator: $x^x - b^b$

   • $\frac{d}{dx}(x^x) = x^x (\ln(x) + 1)$ (using logarithmic differentiation).
   • $\frac{d}{dx}(b^b) = 0$ (since it’s a constant).

   So, the derivative of the denominator is: $x^x (\ln(x) + 1).$

5. Evaluate the limit:

   Now, the limit becomes:

   $\lim_{x \to b} \frac{b \cdot x^{b-1} - b^x \ln(b)}{x^x (\ln(x) + 1)}.$

   Substituting $x = b$ into this expression:

   • Numerator: $b \cdot b^{b-1} - b^b \ln(b) = b^b - b^b \ln(b) = b^b (1 - \ln(b))$.
   • Denominator: $b^b (\ln(b) + 1)$.

   So the limit becomes:

   $\frac{b^b (1 - \ln(b))}{b^b (\ln(b) + 1)} = \frac{1 - \ln(b)}{\ln(b) + 1}.$

6. Final Answer:

   $\lim_{x \to b} \frac{x^b - b^x}{x^x - b^b} = \frac{1 - \ln(b)}{\ln(b) + 1}.$

Let me know if you want further clarifications or have any questions!

Follow-up Questions:

1. What is L'Hôpital's Rule, and when can it be applied?
2. How does logarithmic differentiation work in cases like $x^x$?
3. What happens to the limit if $b = e$ (the base of natural logarithms)?
4. Can the limit be solved by series expansion around $x = b$?
5. How does the indeterminate form $\frac{0}{0}$ appear in other contexts?

Tip: When dealing with limits involving exponential functions, checking for $0/0$ or $\infty/\infty$ forms can guide you toward using L'Hôpital's Rule effectively.