We are tasked with finding the limit:

\[
\lim_{x \to 1} \left( \frac{x^x - x}{x - 1} - \log x \right)
\]

Let’s proceed step by step.

### Step 1: Analyze \( x^x \) as \( x \to 1 \)
Recall that:
\[
x^x = e^{x \log x}
\]
At \( x = 1 \):
\[
x^x = e^{1 \cdot \log 1} = e^0 = 1
\]
Thus, \( x^x - x = 1 - 1 = 0 \) when \( x = 1 \), which suggests that the expression \( \frac{x^x - x}{x - 1} \) might have an indeterminate form. We'll need to apply L'Hopital's Rule to this part.

### Step 2: Apply L'Hopital’s Rule to \( \frac{x^x - x}{x - 1} \)
To apply L'Hopital's Rule, we differentiate the numerator and denominator.

1. **Numerator derivative \( f(x) = x^x - x \):**
   - Derivative of \( x^x \):
     \[
     \frac{d}{dx} (x^x) = x^x (\log x + 1)
     \]
   - Derivative of \( -x \) is \( -1 \).

   So the derivative of the numerator is:
   \[
   \frac{d}{dx} (x^x - x) = x^x (\log x + 1) - 1
   \]

2. **Denominator derivative \( g(x) = x - 1 \):**
   The derivative of \( x - 1 \) is simply 1.

Therefore, applying L'Hopital's Rule:
\[
\lim_{x \to 1} \frac{x^x - x}{x - 1} = \lim_{x \to 1} \left( x^x (\log x + 1) - 1 \right)
\]
At \( x = 1 \), this becomes:
\[
1^1 (\log 1 + 1) - 1 = 1 \times (0 + 1) - 1 = 1 - 1 = 0
\]

### Step 3: Handle the logarithmic term
Now, we need to handle the \( -\log x \) term in the limit. As \( x \to 1 \), \( \log x \to 0 \).

### Step 4: Combine the results
We found that both the terms \( \frac{x^x - x}{x - 1} \) and \( \log x \) tend to 0 as \( x \to 1 \). Therefore, the overall limit is:
\[
\lim_{x \to 1} \left( \frac{x^x - x}{x - 1} - \log x \right) = 0 - 0 = 0
\]

### Final Answer:
\[
\lim_{x \to 1} \left( \frac{x^x - x}{x - 1} - \log x \right) = 0
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How can we prove \( x^x \) is differentiable at all points?
2. What are some other ways to compute limits involving indeterminate forms?
3. How does L'Hopital’s Rule work for higher-order indeterminate forms?
4. Can we apply Taylor series expansion around \( x = 1 \) to approximate the limit?
5. How does the behavior of logarithmic functions compare to power functions near singularities?

**Tip:** L'Hopital’s Rule is useful when both the numerator and denominator tend to 0 or infinity; always check if these conditions hold before applying.

Explore how to solve the limit problem involving exponentials and logarithms: lim x tends to 1 (x^x - x)/(x - 1) - log(x). This step-by-step solution uses L'Hopital's Rule and logarithmic functions, making it ideal for advanced high school students in Grades 11-12.

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