Question

enter an exact answer. provide your answer below: $lim_{x
ightarrowinfty}\frac{x^{2}+6x - 8}{2x^{2}-13x - 11}=$ $lim_{x
ightarrowinfty}\frac{x^{2}+6x - 8}{2x^{2}-13x - 11}$

Explanation:

Step1: Divide numerator and denominator by $x^{2}$

$\lim_{x
ightarrow\infty}\frac{x^{2}+6x - 8}{2x^{2}-13x - 11}=\lim_{x
ightarrow\infty}\frac{\frac{x^{2}}{x^{2}}+\frac{6x}{x^{2}}-\frac{8}{x^{2}}}{\frac{2x^{2}}{x^{2}}-\frac{13x}{x^{2}}-\frac{11}{x^{2}}}=\lim_{x
ightarrow\infty}\frac{1 + \frac{6}{x}-\frac{8}{x^{2}}}{2-\frac{13}{x}-\frac{11}{x^{2}}}$

Step2: Use limit properties

As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{1}{x}=0$ and $\lim_{x
ightarrow\infty}\frac{1}{x^{2}} = 0$. So $\lim_{x
ightarrow\infty}\frac{1+\frac{6}{x}-\frac{8}{x^{2}}}{2-\frac{13}{x}-\frac{11}{x^{2}}}=\frac{1 + 0-0}{2-0 - 0}$

Answer:

$\frac{1}{2}$