1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{x+2}{x^2 \ln x}$$. 2. **Analyze the expression:** The function involves $x^2$ and $\ln x$. Note that $\ln x$ is only defined for $x>0$, so we consider the limit as $x \to 0^+$. 3. **Behavior of components:** As $x \to 0^+$, $x+2 \to 2$, $x^2 \to 0$, and $\ln x \to -\infty$. 4. **Rewrite the expression:** $$\frac{x+2}{x^2 \ln x} \approx \frac{2}{x^2 \ln x}$$ 5. **Consider the denominator:** $x^2$ approaches 0, and $\ln x$ approaches $-\infty$, so $x^2 \ln x$ approaches 0 from the negative side (since $x^2$ is positive and $\ln x$ is negative). 6. **Evaluate the limit:** The denominator tends to 0 negatively, numerator tends to 2, so the fraction tends to $-\infty$. **Final answer:** $$\lim_{x \to 0^+} \frac{x+2}{x^2 \ln x} = -\infty$$