evaluate the limit.
lim_{x o -infty}\frac{-8x^{2}+5x - 4}{7x - 1}
simplify any fractions in your answer.

Question

evaluate the limit.
 lim_{x	o -infty}\frac{-8x^{2}+5x - 4}{7x - 1}
simplify any fractions in your answer.

Accepted Answer

# Explanation:
## Step1: Divide numerator and denominator by highest power of $x$ in denominator
The highest power of $x$ in the denominator $7x - 1$ is $x$.
Divide each term in $\frac{-8x^{2}+5x - 4}{7x - 1}$ by $x$:
$$
\begin{align*}
\frac{-8x^{2}+5x - 4}{7x - 1}&=\frac{\frac{-8x^{2}}{x}+\frac{5x}{x}-\frac{4}{x}}{\frac{7x}{x}-\frac{1}{x}}\
&=\frac{-8x + 5-\frac{4}{x}}{7-\frac{1}{x}}
\end{align*}
$$
## Step2: Apply limit as $x\to-\infty$
We know that $\lim_{x\to-\infty}\frac{1}{x}=0$.
$$
\begin{align*}
\lim_{x\to-\infty}\frac{-8x + 5-\frac{4}{x}}{7-\frac{1}{x}}&=\frac{\lim_{x\to-\infty}(-8x)+\lim_{x\to-\infty}5-\lim_{x\to-\infty}\frac{4}{x}}{\lim_{x\to-\infty}7-\lim_{x\to-\infty}\frac{1}{x}}\
\end{align*}
$$
Since $\lim_{x\to-\infty}\frac{4}{x}=0$, $\lim_{x\to-\infty}\frac{1}{x}=0$ and $\lim_{x\to-\infty}5 = 5$, $\lim_{x\to-\infty}7=7$
$$
\begin{align*}
&=\frac{-\infty+5 - 0}{7-0}\
&=-\infty
\end{align*}
$$

# Answer:
$-\infty$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Divide numerator and denominator by highest power of \(x\) in denominator

Step2: Apply limit as \(x\to-\infty\)

Answer: