To do this all we need to do is factor out the largest power of \(x\) that is in the denominator from both the denominator and the numerator. Then all we need to do is use basic limit properties along with Fact 1 from this section to evaluate the limit.

For the denominator we need to be a little careful. The power of \(x\) in the denominator needs to be outside of the root so it can cancel against the \(x\)’s in the numerator. The largest power of \(x\) outside of the root that we can get (and leave something we can deal with in the root) will be just \({x^2}\). We get this by factoring an \({x^4}\) out of the root.

So, let’s do the first couple of steps in this process to get us started.

\[\begin{align*}\mathop {\lim }\limits_{x \to \, - \infty } \frac{{8 + x - 4{x^2}}}{{\sqrt {6 + {x^2} + 7{x^4}} }} & = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}\left( {\frac{8}{{{x^2}}} + \frac{1}{x} - 4} \right)}}{{\sqrt {{x^4}\left( {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} \right)} }}\\ & = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}\left( {\frac{8}{{{x^2}}} + \frac{1}{x} - 4} \right)}}{{\sqrt {{x^4}} \sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}\left( {\frac{8}{{{x^2}}} + \frac{1}{x} - 4} \right)}}{{\left| {{x^2}} \right|\sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }}\end{align*}\]

Recall from the discussion in this section that.

\[\sqrt {{x^2}} = \left| x \right|\]

So, in this case we’ll have.

\[\sqrt {{x^4}} = \left| {{x^2}} \right| = {x^2}\]

and note that we can get rid of the absolute value bars because we know that \({x^2} \ge 0\). So, let’s finish the limit up.

\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{8 + x - 4{x^2}}}{{\sqrt {6 + {x^2} + 7{x^4}} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}\left( {\frac{8}{{{x^2}}} + \frac{1}{x} - 4} \right)}}{{{x^2}\sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{\frac{8}{{{x^2}}} + \frac{1}{x} - 4}}{{\sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{ - 4}}{{\sqrt 7 }}}}\]

Unlike the previous two problems with roots in them all of the mathematical manipulations in this case did not depend upon the actual limit because we were factoring an \({x^2}\) out which will always be positive and so there will be no reason to redo all of that work.

Here is this limit (with most of the work excluded).

For this part all of the mathematical manipulations we did in the first part up to dealing with the absolute value did not depend upon the limit itself and so don’t really need to be redone here. So, up to that part we have.

\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{8 + x - 4{x^2}}}{{\sqrt {6 + {x^2} + 7{x^4}} }} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{{x^2}\left( {\frac{8}{{{x^2}}} + \frac{1}{x} - 4} \right)}}{{{x^2}\sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{\frac{8}{{{x^2}}} + \frac{1}{x} - 4}}{{\sqrt {\frac{6}{{{x^4}}} + \frac{1}{{{x^2}}} + 7} }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{ - 4}}{{\sqrt 7 }}}}\]

We know that there will be a horizontal asymptote for \(x \to - \infty \) if \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\) exists and is a finite number. Likewise, we’ll have a horizontal asymptote for \(x \to \infty \) if \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\) exists and is a finite number.

Therefore, from the first two parts, we can see that we will get the horizontal asymptote.

\[y = - \frac{4}{{\sqrt 7 }}\]

For both \(x \to - \infty \) and \(x \to \infty \).