1. **Problem:** Find $$\lim_{x \to \infty} \frac{8x^3 + 6x^2 - 10}{5x^3 + 7x + 8}$$. 2. **Formula and rule:** For limits at infinity of rational functions, divide numerator and denominator by the highest power of $x$ in the denominator. 3. Divide numerator and denominator by $x^3$: $$\frac{\frac{8x^3}{x^3} + \frac{6x^2}{x^3} - \frac{10}{x^3}}{\frac{5x^3}{x^3} + \frac{7x}{x^3} + \frac{8}{x^3}} = \frac{8 + \frac{6}{x} - \frac{10}{x^3}}{5 + \frac{7}{x^2} + \frac{8}{x^3}}$$ 4. As $x \to \infty$, terms with $\frac{1}{x^n} \to 0$, so limit becomes: $$\frac{8 + 0 - 0}{5 + 0 + 0} = \frac{8}{5}$$ --- 5. **Problem:** Find $$\lim_{x \to \infty} \frac{27x^8 - 7x + 9}{9x^8 - 8x^4 - 6}$$. 6. Divide numerator and denominator by $x^8$: $$\frac{\frac{27x^8}{x^8} - \frac{7x}{x^8} + \frac{9}{x^8}}{\frac{9x^8}{x^8} - \frac{8x^4}{x^8} - \frac{6}{x^8}} = \frac{27 - \frac{7}{x^7} + \frac{9}{x^8}}{9 - \frac{8}{x^4} - \frac{6}{x^8}}$$ 7. As $x \to \infty$, terms with $\frac{1}{x^n} \to 0$, so limit is: $$\frac{27}{9} = 3$$ --- 8. **Problem:** Find $$\lim_{x \to \infty} \frac{6x^2 + 2}{18x^2 - 10x - 9}$$. 9. Divide numerator and denominator by $x^2$: $$\frac{\frac{6x^2}{x^2} + \frac{2}{x^2}}{\frac{18x^2}{x^2} - \frac{10x}{x^2} - \frac{9}{x^2}} = \frac{6 + \frac{2}{x^2}}{18 - \frac{10}{x} - \frac{9}{x^2}}$$ 10. As $x \to \infty$, terms with $\frac{1}{x^n} \to 0$, so limit is: $$\frac{6}{18} = \frac{1}{3}$$