To find the limit given we need to divide the numerator and denominator by the highest power of x in the denominator. We notice that the highest power is x squared, then we divide by this power:
[tex]\begin{gathered} \lim _{x\rightarrow\infty}\frac{6x^3-3x+1}{-2x^2+4x+7}=\lim _{x\rightarrow\infty}\frac{\frac{6x^3-3x+1}{x^{^2}}}{\frac{-2x^2+4x+7}{x^3}} \\ \lim _{x\rightarrow\infty}\frac{6x-\frac{3}{x^{}}+\frac{1}{x^{^2}}}{-2+\frac{4}{x^{}}+\frac{7}{x^{^2}}} \end{gathered}[/tex]Now, since:
[tex]6x-\frac{3}{x}+\frac{1}{x^2}\rightarrow\infty[/tex]and
[tex]-2+\frac{4}{x}-\frac{7}{x^2}\rightarrow-2[/tex]we conclude that:
[tex]\lim _{x\rightarrow\infty}\frac{6x^3-3x+1}{-2x^2+4x+7}=-\infty[/tex](Remember that this does not mean that the limit exist, this just means that the values of the function become small as x becomes large. This is a short way to express that)
Now, remember that an horizontal asymptote is defined as:
The line y=L is called a horizontal asymptote of the curve y=f(x) if either
[tex]\begin{gathered} \lim _{x\rightarrow\infty}f(x)=L\text{ } \\ or \\ \lim _{x\rightarrow-\infty}f(x)=L\text{ } \end{gathered}[/tex]Since this is not the case we conclude that his is not a horizontal asymptote.
Now a vertical asymptote is defined as:
The vertical line x=a is called a vertical asymptote of the curve y=f(x) if at least one of the following statements is true:
[tex]\begin{gathered} \lim _{x\rightarrow a}f(x)=\infty \\ \lim _{x\rightarrow a^-}f(x)=\infty \\ \lim _{x\rightarrow a^+}f(x)=\infty \\ \lim _{x\rightarrow a}f(x)=-\infty \\ \lim _{x\rightarrow a^-}f(x)=-\infty \\ \lim _{x\rightarrow a^+}f(x)=-\infty \end{gathered}[/tex]Since this is not the case we conclude that his is not a vertical asymptote.
