1. evaluate the limits at infinity
$limlimits_{x\to -\infty}\frac{3x^{5}+x^{4}-7x + 1}{2x^{4}+x^{3}-x + 1}$
$limlimits_{x\to -\infty}\frac{3x^{4}+x^{4}-7x + 1}{x - x^{2}}$
$limlimits_{x\to\infty}\frac{3x^{3}+x^{2}-7x + 1}{2x^{4}-3x + 5}$
$limlimits_{x\to -\infty}\sqrt{x^{2}+2x - 3}$
2. find the horizontal asymptote(s) of

Question

Accepted Answer

# Explanation:
## Step1: Divide numerator and denominator by highest - power of $x$ in denominator
For $\lim_{x
ightarrow-\infty}\frac{3x^{5}+x^{4}-7x + 1}{2x^{4}+x^{3}-x + 1}$, divide numerator and denominator by $x^{4}$.
$$
\begin{align*}
\lim_{x
ightarrow-\infty}\frac{3x^{5}+x^{4}-7x + 1}{2x^{4}+x^{3}-x + 1}&=\lim_{x
ightarrow-\infty}\frac{3x + 1-\frac{7}{x^{3}}+\frac{1}{x^{4}}}{2+\frac{1}{x}-\frac{1}{x^{3}}+\frac{1}{x^{4}}}\
\end{align*}
$$
As $x
ightarrow-\infty$, $\frac{1}{x}
ightarrow0$, $\frac{1}{x^{3}}
ightarrow0$, $\frac{1}{x^{4}}
ightarrow0$.
$$
\begin{align*}
\lim_{x
ightarrow-\infty}\frac{3x + 1-\frac{7}{x^{3}}+\frac{1}{x^{4}}}{2+\frac{1}{x}-\frac{1}{x^{3}}+\frac{1}{x^{4}}}&=\lim_{x
ightarrow-\infty}\frac{3x+1}{2}\
\end{align*}
$$
Since $x
ightarrow-\infty$, $3x
ightarrow-\infty$ and $\lim_{x
ightarrow-\infty}\frac{3x + 1}{2}=-\infty$.

For $\lim_{x
ightarrow\infty}\frac{3x^{3}+x^{2}-7x + 1}{2x^{4}-3x + 5}$, divide numerator and denominator by $x^{4}$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{3x^{3}+x^{2}-7x + 1}{2x^{4}-3x + 5}&=\lim_{x
ightarrow\infty}\frac{\frac{3}{x}+\frac{1}{x^{2}}-\frac{7}{x^{3}}+\frac{1}{x^{4}}}{2-\frac{3}{x^{3}}+\frac{5}{x^{4}}}\
\end{align*}
$$
As $x
ightarrow\infty$, $\frac{1}{x}
ightarrow0$, $\frac{1}{x^{2}}
ightarrow0$, $\frac{1}{x^{3}}
ightarrow0$, $\frac{1}{x^{4}}
ightarrow0$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{\frac{3}{x}+\frac{1}{x^{2}}-\frac{7}{x^{3}}+\frac{1}{x^{4}}}{2-\frac{3}{x^{3}}+\frac{5}{x^{4}}}&=\frac{0 + 0-0 + 0}{2-0 + 0}=0\
\end{align*}
$$

For $\lim_{x
ightarrow-\infty}\frac{3x^{4}+x^{4}-7x + 1}{x - x^{2}}$, divide numerator and denominator by $x^{2}$.
$$
\begin{align*}
\lim_{x
ightarrow-\infty}\frac{3x^{4}+x^{4}-7x + 1}{x - x^{2}}&=\lim_{x
ightarrow-\infty}\frac{4x^{2}-\frac{7}{x}+\frac{1}{x^{2}}}{\frac{1}{x}-1}\
\end{align*}
$$
As $x
ightarrow-\infty$, $\frac{1}{x}
ightarrow0$, $\frac{1}{x^{2}}
ightarrow0$.
$$
\begin{align*}
\lim_{x
ightarrow-\infty}\frac{4x^{2}-\frac{7}{x}+\frac{1}{x^{2}}}{\frac{1}{x}-1}&=\lim_{x
ightarrow-\infty}\frac{4x^{2}}{-1}=-\infty\
\end{align*}
$$

For $\lim_{x
ightarrow-\infty}\frac{\sqrt{x^{2}+2x - 3}}{x}$, when $x
ightarrow-\infty$, $\sqrt{x^{2}}=-x$ (since $x<0$).
$$
\begin{align*}
\lim_{x
ightarrow-\infty}\frac{\sqrt{x^{2}+2x - 3}}{x}&=\lim_{x
ightarrow-\infty}\frac{\sqrt{x^{2}(1+\frac{2}{x}-\frac{3}{x^{2}})}}{x}\
&=\lim_{x
ightarrow-\infty}\frac{-x\sqrt{1+\frac{2}{x}-\frac{3}{x^{2}}}}{x}\
&=\lim_{x
ightarrow-\infty}-\sqrt{1+\frac{2}{x}-\frac{3}{x^{2}}}\
\end{align*}
$$
As $x
ightarrow-\infty$, $\frac{2}{x}
ightarrow0$, $\frac{3}{x^{2}}
ightarrow0$.
$$
\begin{align*}
\lim_{x
ightarrow-\infty}-\sqrt{1+\frac{2}{x}-\frac{3}{x^{2}}}&=- 1\
\end{align*}
$$

# Answer:
$\lim_{x
ightarrow-\infty}\frac{3x^{5}+x^{4}-7x + 1}{2x^{4}+x^{3}-x + 1}=-\infty$, $\lim_{x
ightarrow\infty}\frac{3x^{3}+x^{2}-7x + 1}{2x^{4}-3x + 5}=0$, $\lim_{x
ightarrow-\infty}\frac{3x^{4}+x^{4}-7x + 1}{x - x^{2}}=-\infty$, $\lim_{x
ightarrow-\infty}\frac{\sqrt{x^{2}+2x - 3}}{x}=-1$

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QUESTION IMAGE

Question

Explanation:

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

Answer: