Respuesta :
[tex]f(x)= \frac{9x^2-3x-8}{4x^2-5x+3}
[/tex]
Horizontal asymptote = [tex] \lim_{n \to \infty} f(x)[/tex]
[tex]\lim_{n \to \infty} \frac{9x^2-3x-8}{4x^2-5x+3} =\lim_{n \to \infty} \frac{\frac{9x^2}{x^2}-\frac{3x}{x^2}-\frac{8}{x^2}}{\frac{4x^2}{x^2}-\frac{5x}{x^2}+\frac{3}{x^2}} =\lim_{n \to \infty} \frac{9-\frac{3}{x}-\frac{8}{x^2}}{4-\frac{5}{x}+\frac{3}{x^2}} = \frac{9-0-0}{4-0+0}=\frac{9}{4} [/tex]
Horizontal asymptote = [tex] \lim_{n \to \infty} f(x)[/tex]
[tex]\lim_{n \to \infty} \frac{9x^2-3x-8}{4x^2-5x+3} =\lim_{n \to \infty} \frac{\frac{9x^2}{x^2}-\frac{3x}{x^2}-\frac{8}{x^2}}{\frac{4x^2}{x^2}-\frac{5x}{x^2}+\frac{3}{x^2}} =\lim_{n \to \infty} \frac{9-\frac{3}{x}-\frac{8}{x^2}}{4-\frac{5}{x}+\frac{3}{x^2}} = \frac{9-0-0}{4-0+0}=\frac{9}{4} [/tex]
