Answer
(a) = 0
(b) = 1
(c) = $\infty$
Work Step by Step
(a) $\lim\limits_{x \to \infty}$$\frac{x^2+2}{x^3-1}$ = $\lim\limits_{x \to \infty}$$\frac{\frac{x^2+2}{x^3}}{\frac{x^3-1}{x^3}}$ = $\lim\limits_{x \to \infty}$$\frac{\frac{1}{x}+\frac{2}{x^3}}{1-\frac{1}{x^3}}$ = $\frac{0 + 0}{1-0}$ + $\frac{0}{1}$ = 0
(b) $\lim\limits_{x \to \infty}$$\frac{x^2+2}{x^2-1}$ = $\lim\limits_{x \to \infty}$$\frac{\frac{x^2+2}{x^2}}{\frac{x^2-1}{x^2}}$ = $\lim\limits_{x \to \infty}$$\frac{1+\frac{2}{x^2}}{1-\frac{1}{x^2}}$ = $\frac{1+0}{1-0}$ = $\frac{1}{1}$ = 1
(c) $\lim\limits_{x \to \infty}$$\frac{x^2+2}{x-1}$ = $\lim\limits_{x \to \infty}$$\frac{\frac{x^2+2}{x}}{\frac{x-1}{x}}$ = $\lim\limits_{x \to \infty}$$\frac{x+\frac{2}{x}}{1-\frac{1}{x}}$ = $\lim\limits_{x \to \infty}$$\frac{x+0}{1-0}$ = $\lim\limits_{x \to \infty}$$\frac{x}{1}$ = $\lim\limits_{x \to \infty}x$ = $\infty$
