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hw6 limits at infinity and asymptotes (target l3; §2.2,4.6)
score: 4/6 answered: 4/6
question 5
evaluate the limit
lim_{x
ightarrowinfty}\frac{sqrt{7 + 2x^{3}}}{9+11x}

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 $x$.
\[

$$\begin{align*} \lim_{x ightarrow+\infty}\frac{\sqrt{7 + 2x^{3}}}{9 + 11x}&=\lim_{x ightarrow+\infty}\frac{\sqrt{\frac{7}{x^{2}}+2x}}{\frac{9}{x}+11} \end{align*}$$

\]

Step2: Evaluate the limit of each term

We know that $\lim_{x
ightarrow+\infty}\frac{7}{x^{2}} = 0$ and $\lim_{x
ightarrow+\infty}\frac{9}{x}=0$. Also, as $x
ightarrow+\infty$, $\sqrt{\frac{7}{x^{2}}+2x}
ightarrow+\infty$.
\[

$$\begin{align*} \lim_{x ightarrow+\infty}\frac{\sqrt{\frac{7}{x^{2}}+2x}}{\frac{9}{x}+11}&=\frac{\lim_{x ightarrow+\infty}\sqrt{\frac{7}{x^{2}}+2x}}{\lim_{x ightarrow+\infty}(\frac{9}{x}+11)}\\ &=\frac{+\infty}{0 + 11}=+\infty \end{align*}$$

\]

Answer:

$+\infty$