Question

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.)

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