Question

Hello, I am in need of help!For the following problem I need to figure out: (1.) IF THE LIMIT EXISTS, (2.) IF IT INDICATES THE EXISTENCE OF HORIZONTAL OR VERTICAL ASYMPTOTES, (3.) GIVE THE EQUATION OF SAID ASYMPTOTE, Any help is appreciated! (This is a Calculus problem by the way)

Answer 1

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:

`\begin{gathered} \lim _(x\rightarrow\infty)(6x^3-3x+1)/(-2x^2+4x+7)=\lim _(x\rightarrow\infty)((6x^3-3x+1)/(x^(^2)))/((-2x^2+4x+7)/(x^3)) \\ \lim _(x\rightarrow\infty)\frac{6x-\frac{3}{x^{}}+(1)/(x^(^2))}{-2+\frac{4}{x^{}}+(7)/(x^(^2))} \end{gathered}`

Now, since:

`6x-(3)/(x)+(1)/(x^2)\rightarrow\infty`

and

`-2+(4)/(x)-(7)/(x^2)\rightarrow-2`

we conclude that:

`\lim _(x\rightarrow\infty)(6x^3-3x+1)/(-2x^2+4x+7)=-\infty`

(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

`\begin{gathered} \lim _(x\rightarrow\infty)f(x)=L\text{ } \\ or \\ \lim _(x\rightarrow-\infty)f(x)=L\text{ } \end{gathered}`

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:

`\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}`

Since this is not the case we conclude that his is not a vertical asymptote.