divide numerator and denominator by the highest power of x in the denominator and proceed from there. find $lim_{x
ightarrowinfty}\frac{4sqrt{x}+x^{-7}}{3x - 2}$. write $infty$ or $-infty$ where appropriate.
$lim_{x
ightarrowinfty}\frac{4sqrt{x}+x^{-7}}{3x - 2}=square$ (simplify your answer.)

Question

Accepted Answer

# Explanation:
## Step1: Identify highest - power of x
The highest power of x in the denominator $3x - 2$ is $x^1$.
## Step2: Divide numerator and denominator by x
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{4\sqrt{x}+x^{-7}}{3x - 2}&=\lim_{x
ightarrow\infty}\frac{\frac{4\sqrt{x}}{x}+\frac{x^{-7}}{x}}{\frac{3x}{x}-\frac{2}{x}}\
&=\lim_{x
ightarrow\infty}\frac{4x^{-\frac{1}{2}}+x^{-8}}{3 - 2x^{-1}}
\end{align*}
$$
## Step3: Use limit rules
We know that $\lim_{x
ightarrow\infty}x^{-n}=0$ for $n>0$. So $\lim_{x
ightarrow\infty}4x^{-\frac{1}{2}} = 0$, $\lim_{x
ightarrow\infty}x^{-8}=0$ and $\lim_{x
ightarrow\infty}2x^{-1}=0$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{4x^{-\frac{1}{2}}+x^{-8}}{3 - 2x^{-1}}&=\frac{\lim_{x
ightarrow\infty}(4x^{-\frac{1}{2}}+x^{-8})}{\lim_{x
ightarrow\infty}(3 - 2x^{-1})}\
&=\frac{0 + 0}{3-0}
\end{align*}
$$
# Answer:
0

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Question

Explanation:

Step1: Identify highest - power of x

Step2: Divide numerator and denominator by x

Step3: Use limit rules

Answer: