\( \lim _{x \rightarrow 0^{-}} \frac{3x^{2}-6x}{3\sqrt [3]{(x^{3}-3x^{2})^{2}}}= \)

To solve the limit \( \lim_{x \rightarrow 0^{-}} \frac{3x^{2}-6x}{3\sqrt[3]{(x^{3}-3x^{2})^{2}}} \), we first simplify the expression. 1. **Numerator**: The numerator is \( 3x^2 - 6x = 3x(x - 2) \). 2. **Denominator**: The denominator is \( 3\sqrt[3]{(x^{3}-3x^{2})^{2}} = 3\sqrt[3]{x^{6} - 6x^{5} + 9x^{4}} \). Now, let's analyze the limit as \( x \) approaches \( 0 \) from the left: As \( x \) approaches \( 0^{-} \): - The numerator \( 3x(x - 2) \) approaches \( 0 \). - The term \( x^3 - 3x^2 \) in the denominator also approaches \( 0 \), leading to a form of \( \frac{0}{0} \), which requires further analysis by substitution or simplification. Since we are mainly interested in the behavior near zero, we can see that the leading terms will dominate. Specifically: - **Numerator**: Near \( x = 0 \), \( 3x(x - 2) \approx -6x \). - **Denominator**: In \( \sqrt[3]{(x^3 - 3x^2)^2} \), \( x^3 - 3x^2 \approx -3x^2 \), leading to \( \sqrt[3]{(-3x^2)^2} = \sqrt[3]{9x^4} = 3^{2/3} |x|^{4/3} \) since we are considering \( x \) approaching from the left. Putting that into the limit gives: \[ \lim_{x \rightarrow 0^{-}} \frac{-6x}{3 \cdot 3^{2/3} |x|^{4/3}} = \lim_{x \rightarrow 0^{-}} \frac{-2}{3^{2/3} |x|^{1/3}}. \] Since \( |x|^{1/3} \to 0 \) as \( x \to 0^{-} \), we can conclude that \[ \lim_{x \rightarrow 0^{-}} \frac{-2}{3^{2/3} |x|^{1/3}} \to -\infty. \] Therefore, the limit is \[ \boxed{-\infty}. \]