1. evaluate the limits at infinity\n$limlimits_{x\\to -\\infty}\\frac{3x^{5}+x^{4}-7x + 1}{2x^{4}+x^{3}-x +…

1. evaluate the limits at infinity\n$limlimits_{x\\to -\\infty}\\frac{3x^{5}+x^{4}-7x + 1}{2x^{4}+x^{3}-x + 1}$\n$limlimits_{x\\to -\\infty}\\frac{3x^{4}+x^{4}-7x + 1}{x - x^{2}}$\n$limlimits_{x\\to\\infty}\\frac{3x^{3}+x^{2}-7x + 1}{2x^{4}-3x + 5}$\n$limlimits_{x\\to -\\infty}\\sqrt{x^{2}+2x - 3}$\n2. find the horizontal asymptote(s) of

1. evaluate the limits at infinity\n$limlimits_{x\\to -\\infty}\\frac{3x^{5}+x^{4}-7x + 1}{2x^{4}+x^{3}-x + 1}$\n$limlimits_{x\\to -\\infty}\\frac{3x^{4}+x^{4}-7x + 1}{x - x^{2}}$\n$limlimits_{x\\to\\infty}\\frac{3x^{3}+x^{2}-7x + 1}{2x^{4}-3x + 5}$\n$limlimits_{x\\to -\\infty}\\sqrt{x^{2}+2x - 3}$\n2. find the horizontal asymptote(s) of

Answer

Explanation:

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

For (\lim_{x\rightarrow-\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\rightarrow-\infty}\frac{3x^{5}+x^{4}-7x + 1}{2x^{4}+x^{3}-x + 1}&=\lim_{x\rightarrow-\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\rightarrow-\infty), (\frac{1}{x}\rightarrow0), (\frac{1}{x^{3}}\rightarrow0), (\frac{1}{x^{4}}\rightarrow0). [ \begin{align*} \lim_{x\rightarrow-\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\rightarrow-\infty}\frac{3x+1}{2}\ \end{align*} ] Since (x\rightarrow-\infty), (3x\rightarrow-\infty) and (\lim_{x\rightarrow-\infty}\frac{3x + 1}{2}=-\infty).

For (\lim_{x\rightarrow\infty}\frac{3x^{3}+x^{2}-7x + 1}{2x^{4}-3x + 5}), divide numerator and denominator by (x^{4}). [ \begin{align*} \lim_{x\rightarrow\infty}\frac{3x^{3}+x^{2}-7x + 1}{2x^{4}-3x + 5}&=\lim_{x\rightarrow\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\rightarrow\infty), (\frac{1}{x}\rightarrow0), (\frac{1}{x^{2}}\rightarrow0), (\frac{1}{x^{3}}\rightarrow0), (\frac{1}{x^{4}}\rightarrow0). [ \begin{align*} \lim_{x\rightarrow\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\rightarrow-\infty}\frac{3x^{4}+x^{4}-7x + 1}{x - x^{2}}), divide numerator and denominator by (x^{2}). [ \begin{align*} \lim_{x\rightarrow-\infty}\frac{3x^{4}+x^{4}-7x + 1}{x - x^{2}}&=\lim_{x\rightarrow-\infty}\frac{4x^{2}-\frac{7}{x}+\frac{1}{x^{2}}}{\frac{1}{x}-1}\ \end{align*} ] As (x\rightarrow-\infty), (\frac{1}{x}\rightarrow0), (\frac{1}{x^{2}}\rightarrow0). [ \begin{align*} \lim_{x\rightarrow-\infty}\frac{4x^{2}-\frac{7}{x}+\frac{1}{x^{2}}}{\frac{1}{x}-1}&=\lim_{x\rightarrow-\infty}\frac{4x^{2}}{-1}=-\infty\ \end{align*} ]

For (\lim_{x\rightarrow-\infty}\frac{\sqrt{x^{2}+2x - 3}}{x}), when (x\rightarrow-\infty), (\sqrt{x^{2}}=-x) (since (x<0)). [ \begin{align*} \lim_{x\rightarrow-\infty}\frac{\sqrt{x^{2}+2x - 3}}{x}&=\lim_{x\rightarrow-\infty}\frac{\sqrt{x^{2}(1+\frac{2}{x}-\frac{3}{x^{2}})}}{x}\ &=\lim_{x\rightarrow-\infty}\frac{-x\sqrt{1+\frac{2}{x}-\frac{3}{x^{2}}}}{x}\ &=\lim_{x\rightarrow-\infty}-\sqrt{1+\frac{2}{x}-\frac{3}{x^{2}}}\ \end{align*} ] As (x\rightarrow-\infty), (\frac{2}{x}\rightarrow0), (\frac{3}{x^{2}}\rightarrow0). [ \begin{align*} \lim_{x\rightarrow-\infty}-\sqrt{1+\frac{2}{x}-\frac{3}{x^{2}}}&=- 1\ \end{align*} ]

Answer:

(\lim_{x\rightarrow-\infty}\frac{3x^{5}+x^{4}-7x + 1}{2x^{4}+x^{3}-x + 1}=-\infty), (\lim_{x\rightarrow\infty}\frac{3x^{3}+x^{2}-7x + 1}{2x^{4}-3x + 5}=0), (\lim_{x\rightarrow-\infty}\frac{3x^{4}+x^{4}-7x + 1}{x - x^{2}}=-\infty), (\lim_{x\rightarrow-\infty}\frac{\sqrt{x^{2}+2x - 3}}{x}=-1)