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 \(x\). We get this by factoring an \({x^2}\) out of the root.

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

\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{x + 8}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{\sqrt {{x^2}\left( {2 + \frac{3}{{{x^2}}}} \right)} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{\sqrt {{x^2}} \sqrt {2 + \frac{3}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{\left| x \right|\sqrt {2 + \frac{3}{{{x^2}}}} }}\]

Recall from the discussion in this section that.

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

and we do need to be careful with that.

Now, because we are looking at the limit \(x \to - \infty \) it is safe to assume that \(x < 0\). Therefore, from the definition of the absolute value we get.

\[\left| x \right| = - x\]

and the limit is then.

\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{x + 8}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{ - x\sqrt {2 + \frac{3}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{1 + \frac{8}{x}}}{{ - \sqrt {2 + \frac{3}{{{x^2}}}} }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{ - \sqrt 2 }}}}\]

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.

\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{x + 8}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{\left| x \right|\sqrt {2 + \frac{3}{{{x^2}}}} }}\]

In this part we are looking at the limit \(x \to \infty \) and so it will be safe to assume in this part that \(x > 0\). Therefore, from the definition of the absolute value we get.

\[\left| x \right| = x\]

and the limit is then.

\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{x + 8}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{x\left( {1 + \frac{8}{x}} \right)}}{{x\sqrt {2 + \frac{3}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to \,\infty } \frac{{1 + \frac{8}{x}}}{{\sqrt {2 + \frac{3}{{{x^2}}}} }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{\sqrt 2 }}}}\]

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.

\[y = - \frac{1}{{\sqrt 2 }}\]

for \(x \to - \infty \) and we have the horizontal asymptote.

\[y = \frac{1}{{\sqrt 2 }}\]

for \(x \to \infty \).