Question

Find $\begin{array}{c}lim\\ x\to \infty \end{array}\frac{11x^3-3x+4}{13x^3-5x^2-7}$

Answer 1

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

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

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

We know that as $x\to \infty ,\frac{1}{x},\frac{1}{x^2},\frac{1}{x^3}\to 0$

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

$=\frac{11}{13}$