## Calculus: Early Transcendentals 8th Edition

$3$
$\lim\limits_{x \to \infty}\sqrt {\frac{9x^{3}+8x-4}{3-5x+x^{3}}}$ Divide both the numerator and the denominator by $x^{3}$. $\lim\limits_{x \to \infty}\sqrt \frac{\frac{9x^{3}}{x^{3}}+\frac{8}{x^{3}}-\frac{x}{x^{3}}}{\frac{3}{x^{3}}+\frac{5x}{x^{3}}-\frac{x^{3}}{x^{3}}} =$ $\lim\limits_{x \to \infty} \sqrt \frac{9 - \frac{8}{x^{3}}-\frac{1}{x^{2}}}{\frac{3}{x^{3}} + \frac{5}{x^{2}} - 1} =$ Now use the formula of: $\lim\limits_{x \to a} \frac{p(x)}{q(x)} = \frac{\lim\limits_{x \to a} p(x)}{\lim\limits_{x \to a} q(x)}$ $\lim\limits_{x \to \infty} \sqrt \frac{\lim\limits_{x \to \infty} 9 - \frac{8}{x^{3}} - \frac{1}{x^{3}}}{\lim\limits_{x \to \infty} \frac{3}{x^{3}} + \frac{5}{x^{3}} -1}$ Now use the formula of: $\lim\limits_{x \to a} p(x) + q(x) = \lim\limits_{x \to a} p(x) + \lim\limits_{x \to a} q(x)$. $\lim\limits_{x \to \infty} \sqrt \frac{\lim\limits_{x \to \infty} 9 - \lim\limits_{x \to \infty} \frac{8}{x^{3}} - \lim\limits_{x \to \infty}\frac{1}{x^{3}}}{\lim\limits_{x \to \infty} \frac{3}{x^{3}} + \lim\limits_{x \to \infty} \frac{5}{x^{3}} \lim\limits_{x \to \infty} -1}$ By theorem: $\lim\limits_{x \to \infty} \frac{1}{x^r} = 0$ $\lim\limits_{x \to \infty} = \sqrt \frac{9 - 0 - 0}{0 + 0 - 1} = \sqrt \frac{9}{1} = \sqrt 9 = 3$