Answer
a) -$\infty$
b) -4
c) 0
Work Step by Step
a) $\lim\limits_{x \to \infty}$$\frac{-4x^2+2x-5}{x}$ = $\lim\limits_{x \to \infty}$-4x + 2 -$\frac{5}{x}$ = $\lim\limits_{x \to \infty}$-4x + 2 - 0 = -4($\infty$)+2 = -$\infty$
b) $\lim\limits_{x \to \infty}$$\frac{-4x^2+2x-5}{x^2}$ = $\lim\limits_{x \to \infty}$-4+$\frac{2}{x}$-$\frac{5}{x^2}$ = -4+0-0 = -4
c) $\lim\limits_{x \to \infty}$$\frac{-4x^2+2x-5}{x^3}$ = $\lim\limits_{x \to \infty}$-$\frac{4}{x}$+$\frac{2}{x^2}$-$\frac{5}{x^3}$ = -0+0-0 = 0
