1. Problem: Find the limits of the given functions as the variable approaches infinity or negative infinity. 2. Formula and rules: For rational functions, divide numerator and denominator by the highest power of the variable in the denominator. For limits involving roots, factor inside the root to simplify. 3. (a) $$\lim_{x \to +\infty} \frac{2x + 1}{5x - 2}$$ Divide numerator and denominator by $x$: $$\frac{\cancel{x}(2 + \frac{1}{x})}{\cancel{x}(5 - \frac{2}{x})} = \frac{2 + \frac{1}{x}}{5 - \frac{2}{x}}$$ As $x \to +\infty$, $\frac{1}{x} \to 0$, so limit is $\frac{2}{5}$. 4. (b) $$\lim_{s \to +\infty} \frac{4s^2 + 3}{2s^2 - 1}$$ Divide numerator and denominator by $s^2$: $$\frac{\cancel{s^2}(4 + \frac{3}{s^2})}{\cancel{s^2}(2 - \frac{1}{s^2})} = \frac{4 + \frac{3}{s^2}}{2 - \frac{1}{s^2}}$$ As $s \to +\infty$, $\frac{3}{s^2} \to 0$, $\frac{1}{s^2} \to 0$, so limit is $\frac{4}{2} = 2$. 5. (c) $$\lim_{x \to +\infty} \frac{x + 4}{3x^2 - 5}$$ Divide numerator and denominator by $x^2$: $$\frac{\frac{x}{x^2} + \frac{4}{x^2}}{3 - \frac{5}{x^2}} = \frac{\frac{1}{x} + \frac{4}{x^2}}{3 - \frac{5}{x^2}}$$ As $x \to +\infty$, numerator tends to 0, denominator tends to 3, so limit is 0. 6. (d) $$\lim_{x \to +\infty} \frac{x^2 - 2x + 5}{7x + 1}$$ Divide numerator and denominator by $x^2$: $$\frac{1 - \frac{2}{x} + \frac{5}{x^2}}{\frac{7}{x} + \frac{1}{x^2}}$$ As $x \to +\infty$, numerator tends to 1, denominator tends to 0, so limit tends to $+\infty$. 7. (e) $$\lim_{y \to +\infty} \frac{\sqrt{y^2 + 4}}{y + 4}$$ Rewrite numerator: $$\sqrt{y^2 + 4} = y\sqrt{1 + \frac{4}{y^2}}$$ Divide numerator and denominator by $y$: $$\frac{y\sqrt{1 + \frac{4}{y^2}}}{y(1 + \frac{4}{y})} = \frac{\sqrt{1 + \frac{4}{y^2}}}{1 + \frac{4}{y}}$$ As $y \to +\infty$, numerator tends to 1, denominator tends to 1, so limit is 1. 8. (f) $$\lim_{x \to -\infty} \frac{\sqrt{x^2 + 4}}{x + 4}$$ Rewrite numerator: $$\sqrt{x^2 + 4} = |x|\sqrt{1 + \frac{4}{x^2}} = -x\sqrt{1 + \frac{4}{x^2}}$$ since $x$ is negative. Divide numerator and denominator by $x$: $$\frac{-x\sqrt{1 + \frac{4}{x^2}}}{x(1 + \frac{4}{x})} = \frac{-\sqrt{1 + \frac{4}{x^2}}}{1 + \frac{4}{x}}$$ As $x \to -\infty$, numerator tends to $-1$, denominator tends to 1, so limit is $-1$. 9. (g) $$\lim_{x \to -\infty} \frac{4x^3 + 2x^2 - 5}{3x^3 + x + 2}$$ Divide numerator and denominator by $x^3$: $$\frac{4 + \frac{2}{x} - \frac{5}{x^3}}{3 + \frac{1}{x^2} + \frac{2}{x^3}}$$ As $x \to -\infty$, terms with $\frac{1}{x}$ vanish, so limit is $\frac{4}{3}$. 10. (h) $$\lim_{x \to +\infty} (\sqrt{x^2 + 1} - x)$$ Multiply by conjugate: $$\frac{(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)}{\sqrt{x^2 + 1} + x} = \frac{x^2 + 1 - x^2}{\sqrt{x^2 + 1} + x} = \frac{1}{\sqrt{x^2 + 1} + x}$$ As $x \to +\infty$, denominator tends to $+\infty$, so limit is 0. 11. (i) $$\lim_{x \to \infty} (\sqrt[3]{x^3 + x} - \sqrt[3]{x^3 + 1})$$ Rewrite as: $$\sqrt[3]{x^3(1 + \frac{1}{x^2})} - \sqrt[3]{x^3(1 + \frac{1}{x^3})} = x\left(\sqrt[3]{1 + \frac{1}{x^2}} - \sqrt[3]{1 + \frac{1}{x^3}}\right)$$ Use binomial approximation for large $x$: $$\approx x\left(1 + \frac{1}{3x^2} - 1 - \frac{1}{3x^3}\right) = x\left(\frac{1}{3x^2} - \frac{1}{3x^3}\right) = \frac{1}{3x} - \frac{1}{3x^2}$$ As $x \to \infty$, this tends to 0. Final answers: (a) $\frac{2}{5}$ (b) 2 (c) 0 (d) $+\infty$ (e) 1 (f) $-1$ (g) $\frac{4}{3}$ (h) 0 (i) 0