Find the value of `|(lim_( x to 1) (x^x-1)/(x log x )) /( lim_( xto0) (log (1-3x))/x)|`

Answer

To solve the problem, we need to evaluate the expression:

\[ \left| \frac{\lim_{x \to 1} \frac{x^x - 1}{x \log x}}{\lim_{x \to 0} \frac{\log(1 - 3x)}{x}} \right| \]

We will break this down into two parts: the numerator and the denominator.

### Step 1: Evaluate the numerator \( i = \lim_{x \to 1} \frac{x^x - 1}{x \log x} \)

1. **Substituting \( x = 1 \)**:
\[ x^x = 1^1 = 1 \quad \text{and} \quad \log(1) = 0 \]
Thus, we have:
\[ i = \frac{1 - 1}{1 \cdot 0} = \frac{0}{0} \]
This is an indeterminate form, so we apply L'Hôpital's Rule.

2. **Differentiate the numerator and denominator**:
- The numerator: \( f(x) = x^x - 1 \)
- Using logarithmic differentiation:
\[ y = x^x \implies \log y = x \log x \implies \frac{dy}{dx} = y \left( \log x + 1 \right) = x^x \left( \log x + 1 \right) \]
- The denominator: \( g(x) = x \log x \)
- Using the product rule:
\[ g'(x) = \log x + 1 \]

3. **Apply L'Hôpital's Rule**:
\[ i = \lim_{x \to 1} \frac{x^x (\log x + 1)}{\log x + 1} \]
When \( x \to 1 \):
\[ i = \frac{1 \cdot (0 + 1)}{0 + 1} = 1 \]

### Step 2: Evaluate the denominator \( j = \lim_{x \to 0} \frac{\log(1 - 3x)}{x} \)

1. **Substituting \( x = 0 \)**:
\[ \log(1 - 3 \cdot 0) = \log(1) = 0 \]
Thus, we have:
\[ j = \frac{0}{0} \]
This is also an indeterminate form, so we apply L'Hôpital's Rule.

2. **Differentiate the numerator and denominator**:
- The numerator: \( f(x) = \log(1 - 3x) \)
- Using the chain rule:
\[ f'(x) = \frac{-3}{1 - 3x} \]
- The denominator: \( g(x) = x \)
- The derivative is simply \( g'(x) = 1 \).

3. **Apply L'Hôpital's Rule**:
\[ j = \lim_{x \to 0} \frac{-3}{1 - 3x} = \frac{-3}{1} = -3 \]

### Step 3: Combine results to find the final value

Now we have:
\[ i = 1 \quad \text{and} \quad j = -3 \]

Thus,
\[ \left| \frac{i}{j} \right| = \left| \frac{1}{-3} \right| = \frac{1}{3} \]

### Final Answer
The value of the expression is:
\[ \frac{1}{3} \]