1. **Problem a:** Evaluate $$\lim_{x \to \infty} \frac{x^4 - 2x^3 - 1}{x^2 - x}$$ 2. **Step 1:** Identify the highest powers of $x$ in numerator and denominator. - Numerator highest power: $x^4$ - Denominator highest power: $x^2$ 3. **Step 2:** Divide numerator and denominator by $x^2$ (the highest power in denominator) to simplify the expression: $$\lim_{x \to \infty} \frac{\frac{x^4}{x^2} - \frac{2x^3}{x^2} - \frac{1}{x^2}}{\frac{x^2}{x^2} - \frac{x}{x^2}} = \lim_{x \to \infty} \frac{x^2 - 2x - \frac{1}{x^2}}{1 - \frac{1}{x}}$$ 4. **Step 3:** As $x \to \infty$, terms with $\frac{1}{x}$ or $\frac{1}{x^2}$ approach 0: $$\lim_{x \to \infty} \frac{x^2 - 2x - 0}{1 - 0} = \lim_{x \to \infty} (x^2 - 2x)$$ 5. **Step 4:** Since $x^2$ dominates $-2x$ for large $x$, the limit grows without bound: $$\lim_{x \to \infty} (x^2 - 2x) = \infty$$ **Answer for a:** The limit diverges to infinity. --- 6. **Problem b:** Evaluate $$\lim_{x \to 0} \left( \ln \frac{1}{x} \right)^x$$ 7. **Step 1:** Rewrite the expression inside the limit: $$\left( \ln \frac{1}{x} \right)^x = ( -\ln x )^x$$ 8. **Step 2:** Use the exponential and logarithm relationship: $$ ( -\ln x )^x = e^{x \ln(-\ln x)} $$ 9. **Step 3:** Analyze the exponent as $x \to 0^+$: - $\ln x \to -\infty$ so $-\ln x \to \infty$ - $\ln(-\ln x) \to \infty$ but slowly 10. **Step 4:** Consider the limit of the exponent: $$ \lim_{x \to 0^+} x \ln(-\ln x) $$ 11. **Step 5:** Since $x \to 0$ and $\ln(-\ln x) \to \infty$ slowly, the product tends to 0 because $x$ goes to zero faster than $\ln(-\ln x)$ grows. 12. **Step 6:** Therefore, $$ \lim_{x \to 0^+} e^{x \ln(-\ln x)} = e^0 = 1 $$ **Answer for b:** The limit is 1.