## Calculus: Early Transcendentals 8th Edition

$\frac{2}{5}$
$\lim\limits_{x \to \infty}\frac{2x^{2}-7}{5^{2}+x-3}$ Divide both the numerator and the denominator by $x^{2}$. $\lim\limits_{x \to \infty}\frac{\frac{2x^{2}}{x^{2}}\frac{7}{x^{2}}}{\frac{5x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{3}{x^{2}}} =$ $\lim\limits_{x \to \infty} \frac{2 - \frac{7}{x^{2}}}{5 + \frac{1}{x} - \frac{3}{x^{2}}} =$ Now use the formula of: $\lim\limits_{x \to a} \frac{p(x)}{q(x)} = \frac{\lim\limits_{x \to a} p(x)}{\lim\limits_{x \to a} q(x)}$ $\lim\limits_{x \to \infty} \frac{\lim\limits_{x \to \infty} 2 - \frac{7}{x^{2}}}{\lim\limits_{x \to \infty} 5 + \frac{1}{x} - \frac{3}{x^{2}}} =$ Now use the formula of: $\lim\limits_{x \to a} p(x) + q(x) = \lim\limits_{x \to a} p(x) + \lim\limits_{x \to a} q(x)$. $\lim\limits_{x \to \infty} \frac{\lim\limits_{x \to \infty} 2 - \lim\limits_{x \to \infty} \frac{7}{x^{2}}}{\lim\limits_{x \to \infty} 5 + \lim\limits_{x \to \infty} \frac{1}{x} - \lim\limits_{x \to \infty} \frac{3}{x^{2}}} =$ By theorem: $\lim\limits_{x \to \infty} \frac{1}{x^r} = 0$ $\lim\limits_{x \to \infty} = \frac{2}{5 + 0 - 0} = \frac{2}{5}$