1. Problem: Find the limits of the following expressions 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 to simplify. For limits involving roots, factor inside the root or use conjugates or dominant terms. (a) $$\lim_{x \to +\infty} \frac{2x + 1}{5x - 2}$$ Divide numerator and denominator by $x$: $$\lim_{x \to +\infty} \frac{\frac{2x}{x} + \frac{1}{x}}{\frac{5x}{x} - \frac{2}{x}} = \lim_{x \to +\infty} \frac{2 + \frac{1}{x}}{5 - \frac{2}{x}}$$ As $x \to +\infty$, $\frac{1}{x} \to 0$, so limit is $\frac{2}{5}$. (b) $$\lim_{s \to +\infty} \frac{4s^2 + 3}{2s^2 - 1}$$ Divide numerator and denominator by $s^2$: $$\lim_{s \to +\infty} \frac{4 + \frac{3}{s^2}}{2 - \frac{1}{s^2}}$$ As $s \to +\infty$, terms with $\frac{1}{s^2} \to 0$, so limit is $\frac{4}{2} = 2$. (c) $$\lim_{x \to +\infty} \frac{x + 4}{3x^2 - 5}$$ Divide numerator and denominator by $x^2$: $$\lim_{x \to +\infty} \frac{\frac{x}{x^2} + \frac{4}{x^2}}{3 - \frac{5}{x^2}} = \lim_{x \to +\infty} \frac{\frac{1}{x} + \frac{4}{x^2}}{3 - \frac{5}{x^2}}$$ As $x \to +\infty$, numerator tends to 0, denominator tends to 3, so limit is 0. (d) $$\lim_{x \to +\infty} \frac{x^2 - 2x + 5}{7x + 1}$$ Divide numerator and denominator by $x$: $$\lim_{x \to +\infty} \frac{\frac{x^2}{x} - \frac{2x}{x} + \frac{5}{x}}{7 + \frac{1}{x}} = \lim_{x \to +\infty} \frac{x - 2 + \frac{5}{x}}{7 + \frac{1}{x}}$$ As $x \to +\infty$, numerator tends to $+\infty$, denominator tends to 7, so limit is $+\infty$. (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 $y$: $$\lim_{y \to +\infty} \frac{y \sqrt{1 + \frac{4}{y^2}}}{y (1 + \frac{4}{y})} = \lim_{y \to +\infty} \frac{\sqrt{1 + \frac{4}{y^2}}}{1 + \frac{4}{y}}$$ As $y \to +\infty$, terms with $\frac{1}{y}$ and $\frac{1}{y^2}$ tend to 0, so limit is $\frac{1}{1} = 1$. (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 $x$: $$\lim_{x \to -\infty} \frac{-x \sqrt{1 + \frac{4}{x^2}}}{x (1 + \frac{4}{x})} = \lim_{x \to -\infty} \frac{-\sqrt{1 + \frac{4}{x^2}}}{1 + \frac{4}{x}}$$ As $x \to -\infty$, terms with $\frac{1}{x}$ and $\frac{1}{x^2}$ tend to 0, so limit is $\frac{-1}{1} = -1$. (g) $$\lim_{x \to -\infty} \frac{4x^3 + 2x^2 - 5}{3x^3 + x + 2}$$ Divide numerator and denominator by $x^3$: $$\lim_{x \to -\infty} \frac{4 + \frac{2}{x} - \frac{5}{x^3}}{3 + \frac{1}{x^2} + \frac{2}{x^3}}$$ As $x \to -\infty$, terms with $\frac{1}{x}$ and higher powers tend to 0, so limit is $\frac{4}{3}$. (h) $$\lim_{x \to +\infty} \sqrt{x^2 + 1 - x}$$ Rewrite inside root: $$x^2 + 1 - x = x^2 - x + 1$$ For large $x$, dominant term is $x^2$, so $$\sqrt{x^2 - x + 1} = |x| \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} = x \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}}$$ since $x > 0$. As $x \to +\infty$, terms with $\frac{1}{x}$ and $\frac{1}{x^2}$ tend to 0, so limit is $+\infty$. (i) $$\lim_{x \to +\infty} \left( \sqrt[3]{x^3 + x} - \sqrt[3]{x^3 + 1} \right)$$ Rewrite each cube root: $$\sqrt[3]{x^3 + x} = x \sqrt[3]{1 + \frac{1}{x^2}}, \quad \sqrt[3]{x^3 + 1} = x \sqrt[3]{1 + \frac{1}{x^3}}$$ Use the difference of cube roots approximation for large $x$: $$\sqrt[3]{a} - \sqrt[3]{b} \approx \frac{a - b}{3 a^{2/3}}$$ Set $a = x^3 + x$, $b = x^3 + 1$, so $$a - b = x - 1$$ For large $x$, $a^{2/3} \approx x^2$, so $$\lim_{x \to +\infty} \left( \sqrt[3]{x^3 + x} - \sqrt[3]{x^3 + 1} \right) \approx \lim_{x \to +\infty} \frac{x - 1}{3 x^2} = \lim_{x \to +\infty} \frac{\cancel{x} - \frac{1}{x}}{3 \cancel{x}^2} = 0$$ Final answers: (a) $\frac{2}{5}$ (b) 2 (c) 0 (d) $+\infty$ (e) 1 (f) $-1$ (g) $\frac{4}{3}$ (h) $+\infty$ (i) 0