details\nno additional details were added for this assignment.\nhw6 limits at infinity and asymptotes…

Explanation:

Step1: Divide numerator and denominator by highest - power of x in denominator

Divide both $\sqrt{7 + 2x^{3}}$ and $9+11x$ by $x$. Since $x=\sqrt{x^{2}}$ for $x>0$, we divide $\sqrt{7 + 2x^{3}}$ by $\sqrt{x^{2}}$ and divide $9 + 11x$ by $x$. [ \begin{align*} \lim_{x\rightarrow+\infty}\frac{\sqrt{7 + 2x^{3}}}{9 + 11x}&=\lim_{x\rightarrow+\infty}\frac{\sqrt{\frac{7}{x^{2}}+2x}}{\frac{9}{x}+11} \end{align*} ]

Step2: Evaluate the limit of each term

We know that $\lim_{x\rightarrow+\infty}\frac{7}{x^{2}} = 0$ and $\lim_{x\rightarrow+\infty}\frac{9}{x}=0$. Also, as $x\rightarrow+\infty$, $\sqrt{\frac{7}{x^{2}}+2x}\rightarrow+\infty$. [ \begin{align*} \lim_{x\rightarrow+\infty}\frac{\sqrt{\frac{7}{x^{2}}+2x}}{\frac{9}{x}+11}&=\frac{\lim_{x\rightarrow+\infty}\sqrt{\frac{7}{x^{2}}+2x}}{\lim_{x\rightarrow+\infty}(\frac{9}{x}+11)}\ &=\frac{+\infty}{0 + 11}=+\infty \end{align*} ]

Answer:

$+\infty$