Lim-x-4x-3-2x-2-5x-4-9x-3-4x-2-9-

Question Number 188985 by normans last updated on 10/Mar/23

{Lim}_{x \to\sim} \frac{4 x^{3} - 2 x^{2} - 5 x + 4}{9 x^{3} - 4 x^{2} + 9} = ? ?

Commented by MJS_new last updated on 10/Mar/23

\lim_{x \to \pm \infty} \frac{{a x}^{n} + c_{1} x^{n - 1} + \dots + c_{n} x^{0}}{{b x}^{n} + d_{1} x^{n - 1} + \dots d_{n} x^{0}} = \frac{a}{b}

Answered by cortano12 last updated on 10/Mar/23

\lim_{x \to \infty} \frac{{4 x}^{3} - {2 x}^{2} - 5 x + 4}{{9 x}^{3} - {4 x}^{2} + 9}

= \lim_{x \to \infty} \frac{x^{3} (4 - \frac{2}{x} - \frac{5}{x^{2}} + \frac{4}{x^{3}})}{x^{3} (9 - \frac{4}{x} + \frac{9}{x^{3}})}

= \frac{4}{9}