1. **State the problem:** Find the limits as $x \to +\infty$ for the given rational functions. 2. **Recall the rule for limits at infinity of rational functions:** When $x \to +\infty$, the behavior of a rational function $\frac{P(x)}{Q(x)}$ is dominated by the highest degree terms in numerator and denominator. 3. **First limit:** $$\lim_{x \to +\infty} \frac{x^6 + 3x^3}{x^3 + 3}$$ The highest degree term in numerator is $x^6$, in denominator is $x^3$. 4. **Divide numerator and denominator by $x^3$ (the highest power in denominator):** $$\lim_{x \to +\infty} \frac{\frac{x^6}{x^3} + \frac{3x^3}{x^3}}{\frac{x^3}{x^3} + \frac{3}{x^3}} = \lim_{x \to +\infty} \frac{x^3 + 3}{1 + \frac{3}{x^3}}$$ 5. As $x \to +\infty$, $\frac{3}{x^3} \to 0$, so the limit becomes: $$\lim_{x \to +\infty} (x^3 + 3) = +\infty$$ 6. **Second limit:** $$\lim_{x \to +\infty} \frac{4x^2}{5x^2 + 3x}$$ 7. Divide numerator and denominator by $x^2$: $$\lim_{x \to +\infty} \frac{4}{5 + \frac{3}{x}}$$ 8. As $x \to +\infty$, $\frac{3}{x} \to 0$, so the limit is: $$\frac{4}{5}$$ 9. **Third limit:** $$\lim_{x \to +\infty} \frac{2x + 10}{3x^2 + 3.24x + 70}$$ 10. Divide numerator and denominator by $x^2$ (highest power in denominator): $$\lim_{x \to +\infty} \frac{\frac{2x}{x^2} + \frac{10}{x^2}}{3 + \frac{3.24x}{x^2} + \frac{70}{x^2}} = \lim_{x \to +\infty} \frac{\frac{2}{x} + \frac{10}{x^2}}{3 + \frac{3.24}{x} + \frac{70}{x^2}}$$ 11. As $x \to +\infty$, all terms with $\frac{1}{x}$ or $\frac{1}{x^2}$ go to 0, so the limit is: $$\frac{0 + 0}{3 + 0 + 0} = 0$$ **Final answers:** - $\lim_{x \to +\infty} \frac{x^6 + 3x^3}{x^3 + 3} = +\infty$ - $\lim_{x \to +\infty} \frac{4x^2}{5x^2 + 3x} = \frac{4}{5}$ - $\lim_{x \to +\infty} \frac{2x + 10}{3x^2 + 3.24x + 70} = 0$