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→ - ∞) (∛x - 5x + 1)/(6x + x^(2/3) - 5)
lim(x→ - ∞) (∛x - 5x + 1)/(6x + x^(2/3) - 5)=0 (simplify your answer)

Explanation:

Step1: Identify highest - power in denominator

The highest power of $x$ in the denominator $6x + x^{\frac{2}{3}}-5$ is $x$.

Step2: Divide numerator and denominator by $x$

\[

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

\]

Step3: Evaluate the limit of each term

As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}x^{-\frac{2}{3}} = 0$, $\lim_{x
ightarrow-\infty}\frac{1}{x}=0$, $\lim_{x
ightarrow-\infty}x^{-\frac{1}{3}} = 0$.
\[

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

\]

Answer:

$-\frac{5}{6}$