evaluate the limit.
lim _ { x
ightarrow - infty } \frac { 2 x + 1 } { - x ^ { 2 } + 8 x + 6 }
simplify any fractions in your answer.

Question

evaluate the limit.
 lim _ { x
ightarrow - infty } \frac { 2 x + 1 } { - x ^ { 2 } + 8 x + 6 }
simplify any fractions in your answer.

Accepted Answer

# Explanation:
## Step1: Divide numerator and denominator by $x^{2}$
Divide each term in the numerator $2x + 1$ and denominator $-x^{2}+8x + 6$ by $x^{2}$.
$$
\begin{align*}
\lim_{x
ightarrow-\infty}\frac{2x + 1}{-x^{2}+8x + 6}&=\lim_{x
ightarrow-\infty}\frac{\frac{2x}{x^{2}}+\frac{1}{x^{2}}}{\frac{-x^{2}}{x^{2}}+\frac{8x}{x^{2}}+\frac{6}{x^{2}}}\
&=\lim_{x
ightarrow-\infty}\frac{\frac{2}{x}+\frac{1}{x^{2}}}{- 1+\frac{8}{x}+\frac{6}{x^{2}}}
\end{align*}
$$
## Step2: Use the limit property $\lim_{x
ightarrow\pm\infty}\frac{c}{x^{n}} = 0$ ($n>0$, $c$ is a constant)
We know that $\lim_{x
ightarrow-\infty}\frac{2}{x}=0$, $\lim_{x
ightarrow-\infty}\frac{1}{x^{2}} = 0$, $\lim_{x
ightarrow-\infty}\frac{8}{x}=0$ and $\lim_{x
ightarrow-\infty}\frac{6}{x^{2}}=0$
$$
\begin{align*}
\lim_{x
ightarrow-\infty}\frac{\frac{2}{x}+\frac{1}{x^{2}}}{-1+\frac{8}{x}+\frac{6}{x^{2}}}&=\frac{\lim_{x
ightarrow-\infty}(\frac{2}{x}+\frac{1}{x^{2}})}{\lim_{x
ightarrow-\infty}(-1+\frac{8}{x}+\frac{6}{x^{2}})}\
&=\frac{0 + 0}{-1+0 + 0}
\end{align*}
$$

# Answer:
$0$

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QUESTION IMAGE

Question

Explanation:

Step1: Divide numerator and denominator by \(x^{2}\)

Step2: Use the limit property \(\lim_{x

Answer: