\infty} \ \frac{3x^3-4x^2+8x-1}{5x^2+8x-3}\\\\
lim_{x->\infty} \ \frac{3x-4+\frac{8}{x}-\frac{1}{x^2}}{5+\frac{8}{x}-\frac{3}{x^2}} = \frac{3x-4}{5}=\frac{3-\frac{4}{x}}{\frac{5}{x}}=+\infty\\\\
" alt="lim_{x->\infty} \ \frac{3x^3-4x^2+8x-1}{5x^2+8x-3}\\\\
lim_{x->\infty} \ \frac{3x-4+\frac{8}{x}-\frac{1}{x^2}}{5+\frac{8}{x}-\frac{3}{x^2}} = \frac{3x-4}{5}=\frac{3-\frac{4}{x}}{\frac{5}{x}}=+\infty\\\\
" align="absmiddle" class="latex-formula">