\infty} \frac{9x^3+7x}{9x^3-4x^2+5}=lim_{x->\infty} \frac{9+\frac{7}{x^2}}{9-\frac{4}{x^2}+\frac{5}{x^3}}=\frac{9+0}{9-0+0}=1" alt="lim_{x->\infty} \frac{9x^3+7x}{9x^3-4x^2+5}=lim_{x->\infty} \frac{9+\frac{7}{x^2}}{9-\frac{4}{x^2}+\frac{5}{x^3}}=\frac{9+0}{9-0+0}=1" align="absmiddle" class="latex-formula">