\( \lim 9 ( x ^ { 2 } - x ) ( \frac { \sqrt { x } + x ^ { 2 } + 3 } { 2 x + \log x } ) = \)

To evaluate the limit \[ \lim_{x \to 0} 9 (x^2 - x) \left( \frac{\sqrt{x} + x^2 + 3}{2x + \log x} \right), \] we first analyze the behavior of each component as \( x \to 0 \). 1. **Analyzing \( x^2 - x \)**: \[ x^2 - x = x(x - 1). \] As \( x \to 0 \), \( x^2 - x \to 0 \). 2. **Analyzing \( \sqrt{x} + x^2 + 3 \)**: \[ \sqrt{x} + x^2 + 3 \to 0 + 0 + 3 = 3 \quad \text{as } x \to 0. \] 3. **Analyzing \( 2x + \log x \)**: As \( x \to 0 \), \( \log x \to -\infty \). Therefore, \[ 2x + \log x \to -\infty. \] Now, we can combine these results into the limit expression: \[ \lim_{x \to 0} 9 (x^2 - x) \left( \frac{\sqrt{x} + x^2 + 3}{2x + \log x} \right) = \lim_{x \to 0} 9 (x^2 - x) \left( \frac{3}{2x + \log x} \right). \] Next, we substitute the limits we found: - The numerator approaches \( 3 \) as \( x \to 0 \). - The denominator approaches \( -\infty \). Thus, we have: \[ \lim_{x \to 0} 9 (x^2 - x) \left( \frac{3}{2x + \log x} \right) = \lim_{x \to 0} 9 (x^2 - x) \cdot \frac{3}{-\infty} = 0. \] Since \( x^2 - x \to 0 \) and the fraction approaches \( 0 \) as well, we conclude that the entire limit evaluates to: \[ \boxed{0}. \]