1. **Problem 1:** Find $\lim_{x \to +\infty} f(x)$ and $\lim_{x \to -\infty} f(x)$ for the piecewise function $$f(x) = \begin{cases} \frac{\sqrt{x^2 + x + 2} - 2}{x^2 -1}, & x > 1, \\ \frac{4x^2 + x - 2}{x^2 + 7}, & x \leq 1. \end{cases}$$ 2. For $x > 1$, analyze the limit as $x \to +\infty$: $$\lim_{x \to +\infty} \frac{\sqrt{x^2 + x + 2} - 2}{x^2 -1}$$ 3. Simplify the numerator for large $x$: $$\sqrt{x^2 + x + 2} = x \sqrt{1 + \frac{1}{x} + \frac{2}{x^2}} \approx x \left(1 + \frac{1}{2x}\right) = x + \frac{1}{2}$$ 4. So numerator $\approx (x + \frac{1}{2}) - 2 = x - \frac{3}{2}$, denominator $\approx x^2$. 5. Therefore, $$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \frac{x - \frac{3}{2}}{x^2} = \lim_{x \to +\infty} \frac{1 - \frac{3}{2x}}{x} = 0$$ 6. For $x \leq 1$, analyze $\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{4x^2 + x - 2}{x^2 + 7}$. 7. Divide numerator and denominator by $x^2$: $$\frac{4 + \frac{1}{x} - \frac{2}{x^2}}{1 + \frac{7}{x^2}} \to 4$$ 8. So, $$\lim_{x \to -\infty} f(x) = 4$$ --- 9. **Problem 2:** For $$f(x) = x - 4 \sqrt{x-2} - 1,$$ find $\lim_{x \to +\infty} f(x)$ and $\lim_{x \to +\infty} \frac{f(x)}{x}$. 10. As $x \to +\infty$, $\sqrt{x-2} \sim \sqrt{x}$, so $$f(x) \approx x - 4 \sqrt{x} - 1$$ 11. Since $x$ grows faster than $\sqrt{x}$, $f(x) \to +\infty$. 12. Now, $$\frac{f(x)}{x} = \frac{x - 4 \sqrt{x-2} - 1}{x} = 1 - 4 \frac{\sqrt{x-2}}{x} - \frac{1}{x}$$ 13. Note $\frac{\sqrt{x-2}}{x} = \frac{\sqrt{x-2}}{x} = \frac{\sqrt{x-2}}{\sqrt{x} \sqrt{x}} = \frac{1}{\sqrt{x}} \to 0$. 14. So, $$\lim_{x \to +\infty} \frac{f(x)}{x} = 1 - 0 - 0 = 1$$ --- 15. **Problem 3:** Given $$g(x) = x^3 + 3x + 8,$$ no limit requested, so no further steps. --- 16. **Problem 4:** For $$f(x) = \frac{x^3 - 4}{x^2 + 1},$$ find $\lim_{x \to +\infty} f(x)$ and $\lim_{x \to -\infty} f(x)$. 17. Divide numerator and denominator by $x^2$: $$f(x) = \frac{x^3 - 4}{x^2 + 1} = \frac{x - \frac{4}{x^2}}{1 + \frac{1}{x^2}}$$ 18. As $x \to +\infty$, $f(x) \approx x$ which tends to $+\infty$. 19. As $x \to -\infty$, $f(x) \approx x$ which tends to $-\infty$. --- 20. Additional given: $$f(\alpha) = \frac{3 \alpha}{2}$$ and $$f'(x) = \frac{x g(x)}{(x^2 + 1)^2}$$ where $g(x) = x^3 + 3x + 8$. --- **Final answers:** - $\lim_{x \to +\infty} f(x)$ for piecewise function = $0$ - $\lim_{x \to -\infty} f(x)$ for piecewise function = $4$ - $\lim_{x \to +\infty} f(x) = +\infty$ for $f(x) = x - 4 \sqrt{x-2} - 1$ - $\lim_{x \to +\infty} \frac{f(x)}{x} = 1$ - $\lim_{x \to +\infty} \frac{x^3 - 4}{x^2 + 1} = +\infty$ - $\lim_{x \to -\infty} \frac{x^3 - 4}{x^2 + 1} = -\infty$