1. **Evaluate the limits at infinity and zero:** - $L_1 = \lim_{x \to \pm \infty} \frac{x^2 - 8}{3x^2 - 2}$ - $L_2 = \lim_{x \to \pm \infty} \frac{2x^2 + x + 4}{3x^2 + x}$ - $L_3 = \lim_{x \to \pm \infty} \left(1 - \frac{1}{x^2}\right)$ - $L_4 = \lim_{x \to 0} \frac{\sin(3x)}{\tan(2x)}$ - $L_5 = \lim_{x \to 0} \frac{\sqrt{3 + x} - \sqrt{5}}{2x}$ - $L_6 = \lim_{x \to 0} \frac{\sqrt{20x^2 + x} - 3}{2x^2 - 1}$ 2. **Order the square roots:** $\sqrt{16}, \sqrt{3}, \sqrt{2}, \sqrt{4}$ 3. **Simplify the expression:** $A = \frac{a^3 \sqrt{b}}{\sqrt{b} \sqrt{a}}$ 4. **Solve the equation:** $x^8 - 3x^4 - 4 = 0$ 5. **Solve the inequality:** $\sqrt[3]{7x + 11} < 2$ 6. **Analyze the piecewise function:** $$g(x) = \begin{cases} \frac{\sqrt{x^2 - 3}}{x - 7}, & x > 7 \\ x^2 - x + 14, & x \leq 7 \end{cases}$$ 7. **Intervals:** - $I = [0, 1]$ - $J = ]-\infty, 0]$ - $K = [1, +\infty[$ - $L = (0, 2)$ 8. **Functions:** - $h(x) = 2x^3 - 3x^2 - 1$ - $f(x) = x \sqrt{x^2 - 1}$ on $I = [1, +\infty[$ --- ### Step-by-step solutions: 1. **Limits at infinity:** - For rational functions, divide numerator and denominator by highest power of $x$ in denominator. $L_1 = \lim_{x \to \pm \infty} \frac{x^2 - 8}{3x^2 - 2} = \lim_{x \to \pm \infty} \frac{1 - \frac{8}{x^2}}{3 - \frac{2}{x^2}} = \frac{1}{3}$ - Similarly for $L_2$: $L_2 = \lim_{x \to \pm \infty} \frac{2x^2 + x + 4}{3x^2 + x} = \lim_{x \to \pm \infty} \frac{2 + \frac{1}{x} + \frac{4}{x^2}}{3 + \frac{1}{x}} = \frac{2}{3}$ - For $L_3$: $L_3 = \lim_{x \to \pm \infty} \left(1 - \frac{1}{x^2}\right) = 1$ 2. **Limit involving trigonometric functions at 0:** Use small angle approximations $\sin(kx) \approx kx$, $\tan(kx) \approx kx$ as $x \to 0$: $L_4 = \lim_{x \to 0} \frac{\sin(3x)}{\tan(2x)} \approx \lim_{x \to 0} \frac{3x}{2x} = \frac{3}{2}$ 3. **Limit with square roots at 0:** For $L_5$: Multiply numerator and denominator by conjugate: $$L_5 = \lim_{x \to 0} \frac{\sqrt{3 + x} - \sqrt{5}}{2x} \times \frac{\sqrt{3 + x} + \sqrt{5}}{\sqrt{3 + x} + \sqrt{5}} = \lim_{x \to 0} \frac{(3 + x) - 5}{2x (\sqrt{3 + x} + \sqrt{5})} = \lim_{x \to 0} \frac{x - 2}{2x (\sqrt{3 + x} + \sqrt{5})}$$ Split numerator: $$= \lim_{x \to 0} \frac{x}{2x (\sqrt{3 + x} + \sqrt{5})} - \lim_{x \to 0} \frac{2}{2x (\sqrt{3 + x} + \sqrt{5})}$$ The second term diverges, so re-check approach: Actually numerator is $x - 2$, but original numerator is $(3 + x) - 5 = x - 2$, so at $x \to 0$ numerator $-2$, denominator $0$, limit diverges to $-\infty$ or $+\infty$ depending on sign. Re-examining, original numerator is $\sqrt{3 + x} - \sqrt{5}$, at $x=0$ numerator $\sqrt{3} - \sqrt{5} < 0$, denominator $0$, so limit tends to $-\infty$. 4. **Limit $L_6$ at 0:** $$L_6 = \lim_{x \to 0} \frac{\sqrt{20x^2 + x} - 3}{2x^2 - 1}$$ At $x=0$, numerator $\sqrt{0} - 3 = -3$, denominator $-1$, so limit is $\frac{-3}{-1} = 3$ 5. **Order square roots:** Calculate approximate values: - $\sqrt{16} = 4$ - $\sqrt{3} \approx 1.732$ - $\sqrt{2} \approx 1.414$ - $\sqrt{4} = 2$ Order ascending: $\sqrt{2} < \sqrt{3} < \sqrt{4} < \sqrt{16}$ 6. **Simplify $A$:** $$A = \frac{a^3 \sqrt{b}}{\sqrt{b} \sqrt{a}} = \frac{a^3 \sqrt{b}}{\sqrt{b} a^{1/2}} = a^{3 - 1/2} = a^{5/2}$$ 7. **Solve equation:** $$x^8 - 3x^4 - 4 = 0$$ Let $y = x^4$, then: $$y^2 - 3y - 4 = 0$$ Solve quadratic: $$y = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2}$$ So $y_1 = 4$, $y_2 = -1$ (discard negative for $x^4$) Then $x^4 = 4 \Rightarrow x = \pm \sqrt{2}$ 8. **Solve inequality:** $$\sqrt[3]{7x + 11} < 2$$ Cube both sides (monotonic): $$7x + 11 < 8 \Rightarrow 7x < -3 \Rightarrow x < -\frac{3}{7}$$ 9. **Piecewise function $g(x)$:** - For $x > 7$, domain requires $x^2 - 3 \geq 0$, true for $x > 7$ - For $x \leq 7$, polynomial defined everywhere 10. **Intervals:** - $I = [0,1]$ closed interval - $J = ]-\infty, 0]$ left-unbounded, right-closed - $K = [1, +\infty[$ left-closed, right-unbounded - $L = (0,2)$ open interval 11. **Functions:** - $h(x) = 2x^3 - 3x^2 - 1$ - $f(x) = x \sqrt{x^2 - 1}$ on $[1, +\infty[$ domain valid since $x^2 - 1 \geq 0$ for $x \geq 1$ --- **Final answers:** - $L_1 = \frac{1}{3}$ - $L_2 = \frac{2}{3}$ - $L_3 = 1$ - $L_4 = \frac{3}{2}$ - $L_5$ diverges to $-\infty$ - $L_6 = 3$ - Order: $\sqrt{2} < \sqrt{3} < \sqrt{4} < \sqrt{16}$ - $A = a^{5/2}$ - Equation roots: $x = \pm \sqrt{2}$ - Inequality solution: $x < -\frac{3}{7}$