$\underset{x\to \mathrm{\infty }}{lim} \frac{11x^3-3x+4}{13x^3-5x^2-7}$

$\underset{x\to \mathrm{\infty }}{lim} \frac{11x^3-3x+4}{13x^3-5x^2-7}=\underset{x\to \mathrm{\infty }}{lim} \frac{x^3\left(11-\frac{3}{x^2}+\frac{4}{x^3}\right)}{x^3\left(13-\frac{5}{x}-\frac{7}{x^3}\right)}$

$\text{ As }x\to \mathrm{\infty }\Rightarrow \frac{1}{x}\to 0,\frac{1}{x^2}\to 0,\frac{1}{x^3}\to 0$

$=\frac{11-0+0}{13-0-0}=\frac{11}{13}$