Question

5. Calcule os seguintes limites (do tipo ∞ / ∞):

a) $\lim_{x \to \infty}$ $\frac{2 x^{2} - 4x - 25}{18 x^{3} - 9 x^{2} }$
[o infinito é positivo!]

b) $\lim_{x \to \infty}$ $\frac{ x(x - 3)(2x + 5)}{(x - 1)(3x + 4)(2 - x)}$
[ o infinito é negativo! ]

Answer 1

$\lim_{x \to \infty} \frac{2x^2-4x-25}{18x^3-9x^2}= \lim_{x \to \infty} \frac{2x^2-4x-25}{18x^3-9x^2}* \frac{ \frac{1}{x^3} }{ \frac{1}{x^3} }=\lim_{x \to \infty} \frac{\frac2x-\frac{4}{x^2}-\frac{25}{x^3}}{18-\frac9x}=$
$\frac{\frac{2}{\infty}-\frac{4}{\infty}-\frac{25}{\infty}}{18-\frac{9}{\infty}}=\frac{0-0-0}{18-0}=\frac{0}{18}=\boxed{0}$

$\lim_{x \to -\infty} \frac{x(x-3)(2x+5)}{(x-1)(3x+4)(2-x)}= \lim_{x \to -\infty} \frac{-2x^3+x^2+15x}{3x^3-5x^2-6x+8}=$
$\lim_{x \to -\infty} \frac{-2x^3+x^2+15x}{3x^3-5x^2-6x+8}*\frac{\frac{1}{x^3}}{\frac{1}{x^3}}=\lim_{x \to -\infty} \frac{-2+\frac{1}{x}+\frac{15}{x^2}}{3-\frac{5}{x}-\frac{6}{x^2}+\frac{8}{x^3}}= \frac{-2+\frac{1}{-\infty}+\frac{15}{\infty}}{3-\frac{5}{-\infty}-\frac{6}{\infty}+\frac{8}{-\infty}}=$
$\frac{-2-0+0}{3+0-0-0}=\boxed{\frac{-2}{3}}$