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### Section 2.6 : Infinite Limits

8. Find all the vertical asymptotes of $$\displaystyle g\left( x \right) = \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}}$$.

Hint : Remember how vertical asymptotes are defined and use the examples above to help determine where they are liable to be for the given function. Once you have the locations for the possible vertical asymptotes verify that they are in fact vertical asymptotes.
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Recall that vertical asymptotes will occur at $$x = a$$ if any of the limits (one-sided or overall limit) at $$x = a$$ are plus or minus infinity.

From previous examples we can see that for rational expressions vertical asymptotes will occur where there is division by zero. Therefore, it looks like we will have possible vertical asymptote at $$x = - 5$$ and $$x = 9$$.

Now let’s verify that these are in fact vertical asymptotes by evaluating the two one-sided limits for each of them.

Let’s start with $$x = - 5$$. We’ll need to evaluate,

$\mathop {\lim }\limits_{x \to \, - {5^{\, - }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}}\hspace{0.25in}\,\,\,\,{\rm{and}}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{x \to \, - {5^{\, + }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}}$

In either case as $$x \to - 5$$ (from both left and right) the numerator is a constant -8.

For the one-sided limits we have the following information,

\begin{align*}x \to - {5^ - }\hspace{0.25in} \Rightarrow \hspace{0.25in}x < - 5\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x + 5 < 0\\ x \to - {5^ + }\hspace{0.25in} \Rightarrow \hspace{0.25in}x > - 5\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x + 5 > 0\end{align*}

Also, note that for $$x$$’s close enough to -5 (which because we’re looking at $$x \to - 5$$ is safe enough to assume), we will have $$x - 9 < 0$$.

So, in the left-hand limit, the numerator is a fixed negative number and the denominator is positive (a product of two negative numbers) and increasingly small. Likewise, for the right-hand limit, the denominator is negative (a product of a positive and negative number) and increasingly small. Therefore, we will have,

$\mathop {\lim }\limits_{x \to \, - {5^{\, - }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = - \infty \hspace{0.25in}\,\,\,\,{\rm{and}}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{x \to \, - {5^{\, + }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = \infty$

Now for $$x = 9$$. Again, the numerator is a constant -8. We also have,

\begin{align*}x \to {9^ - }\hspace{0.25in} \Rightarrow \hspace{0.25in}x < 9\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x - 9 < 0\\ x \to {9^ + }\hspace{0.25in} \Rightarrow \hspace{0.25in}x > 9\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x - 9 > 0\end{align*}

Finally, for $$x$$’s close enough to 9 (which because we’re looking at $$x \to 9$$ is safe enough to assume), we will have $$x + 5 > 0$$.

So, in the left-hand limit, the numerator is a fixed negative number and the denominator is negative (a product of a positive and negative number) and increasingly small. Likewise, for the right-hand limit, the denominator is positive (a product of two positive numbers) and increasingly small. Therefore, we will have,

$\mathop {\lim }\limits_{x \to \,{9^{\, - }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = \infty \hspace{0.25in}\,\,\,\,{\rm{and}}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{x \to \,{9^{\, + }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = - \infty$

So, as all of these limits show we do in fact have vertical asymptotes at $$x = - 5$$ and$$x = 9$$.