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 ext{ (simplify your answer.)}

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	ext{ (simplify your answer.)}

Accepted 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:
$0$

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QUESTION IMAGE

Question

Explanation:

Step1: Identify highest power

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

Step3: Use limit rules for negative - power functions

Answer: