hw6 limits at infinity and asymptotes (target l3; §2.2,4.6)\nscore: 4/6 answered: 5/6\nquestion 5\nscore on…

Explanation:

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

Divide $\frac{\sqrt{7 + 2x^{3}}}{9+11x}$ by $x$. Since $x=\sqrt{x^{2}}$ for $x>0$ and the highest - power of $x$ in the denominator is $x$, we rewrite the expression as $\lim_{x\rightarrow\infty}\frac{\sqrt{\frac{7}{x^{2}}+2x}}{\frac{9}{x}+11}$.

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, $\lim_{x\rightarrow\infty}\sqrt{\frac{7}{x^{2}}+2x}=\infty$ because as $x\rightarrow\infty$, the term $2x$ dominates in $\frac{7}{x^{2}}+2x$. And $\lim_{x\rightarrow\infty}(\frac{9}{x}+11)=11$.

Step3: Determine the overall limit

Since $\lim_{x\rightarrow\infty}\sqrt{\frac{7}{x^{2}}+2x}=\infty$ and $\lim_{x\rightarrow\infty}(\frac{9}{x}+11)=11$, then $\lim_{x\rightarrow\infty}\frac{\sqrt{\frac{7}{x^{2}}+2x}}{\frac{9}{x}+11}=\infty$.

Answer:

$\infty$