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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:**

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

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

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### 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.

Evaluate lim_{x → b} (x^b - b^x) / (x^x - b^b).

Step-by-step solution to evaluate the limit lim_{x → b} (x^b - b^x) / (x^x - b^b) using L'Hôpital's Rule. Learn about logarithmic differentiation, exponential functions, and how to handle indeterminate forms. Suitable for advanced calculus students.

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