Question

divide the numerator and the denominator by the highest power of x in the denominator and proceed from there. find the limit. write ∞ or - ∞ where appropriate.
lim_{x
ightarrow-infty}\frac{sqrt3{x}-5x + 1}{\frac{2}{6x + x^{3}-5}}
lim_{x
ightarrow-infty}\frac{sqrt3{x}-5x + 1}{\frac{2}{6x + x^{3}-5}}=square\text{ (simplify your answer.)}

Explanation:

Step1: Identify highest power

The highest - power of \(x\) in the denominator is \(x^{3}\).

Step2: Divide numerator and denominator by \(x^{3}\)

\[

$$\begin{align*} \lim_{x ightarrow-\infty}\frac{\sqrt[3]{x}-5x + 1}{6x+x^{2}-5}&=\lim_{x ightarrow-\infty}\frac{\frac{\sqrt[3]{x}}{x^{3}}-\frac{5x}{x^{3}}+\frac{1}{x^{3}}}{\frac{6x}{x^{3}}+\frac{x^{2}}{x^{3}}-\frac{5}{x^{3}}}\\ &=\lim_{x ightarrow-\infty}\frac{x^{\frac{1}{3}-3}-5x^{1 - 3}+x^{-3}}{6x^{1-3}+x^{2 - 3}-5x^{-3}}\\ &=\lim_{x ightarrow-\infty}\frac{x^{-\frac{8}{3}}-5x^{-2}+x^{-3}}{6x^{-2}+x^{-1}-5x^{-3}} \end{align*}$$

\]

Step3: Use limit rules for negative - power functions

As \(x
ightarrow-\infty\), for \(n>0\), \(\lim_{x
ightarrow-\infty}x^{-n}=0\).
\[

$$\begin{align*} \lim_{x ightarrow-\infty}\frac{x^{-\frac{8}{3}}-5x^{-2}+x^{-3}}{6x^{-2}+x^{-1}-5x^{-3}}&=\frac{\lim_{x ightarrow-\infty}x^{-\frac{8}{3}}-5\lim_{x ightarrow-\infty}x^{-2}+\lim_{x ightarrow-\infty}x^{-3}}{6\lim_{x ightarrow-\infty}x^{-2}+\lim_{x ightarrow-\infty}x^{-1}-5\lim_{x ightarrow-\infty}x^{-3}}\\ &=\frac{0 - 0+0}{0 + 0-0}=0 \end{align*}$$

\]

Answer:

\(0\)