# Thread: Limits of rational functions with absolute values in the denominator

1. ## Limits of rational functions with absolute values in the denominator

Consider $\lim_{x \to -6} \frac{2x+12}{|x+6|}$.

I thought the best way to approach this was to consider the two sides of the limit seperately.

$\lim_{x \to -6^+} \frac{2x+12}{|x+6|}$

$= \lim_{x \to -6^+} \frac{2x+12}{6-x}$ because x < 0.

Now the limit will evaluate to 0. I had to check numerically to discover the limits on both sides of -6 are -2 and 2 respectively, so there is no limit.

I had a hunch from the fact that the limit seemed to be equal to 0.

But how can I, without tedious checking values,

a) Establish whether there is a limit or not
b) The value of the limit if it exists?

If I carried on applying algebraic rules and limit laws to this problem I would get the answer 0, and be none the wiser.

2. $\frac{{2x + 12}}
{{\left| {x + 6} \right|}} = \left\{ \begin{gathered}
2,\;x > - 6 \hfill \\
- 2,\;x < - 6 \hfill \\
\end{gathered} \right.$

3. Originally Posted by Plato
$\frac{{2x + 12}}
{{\left| {x + 6} \right|}} = \left\{ \begin{gathered}
2,\;x > - 6 \hfill \\
- 2,\;x < - 6 \hfill \\
\end{gathered} \right.$
Can you explain how you came to that conclusion? I did it by plugging in values on either side, as I mentioned. But I really don't want to plug in values if there's perhaps a nicer way to evaluate this.

4. Originally Posted by rowe
Can you explain how you came to that conclusion? I did it by plugging in values on either side, as I mentioned. But I really don't want to plug in values if there's perhaps a nicer way to evaluate this.
It is truly elementary.
Note that $2x+12=2(x+6)$

Then note that $\left| {x + 6} \right| = \left\{ {\begin{array}{rl}
{(x + 6),} & {x > - 6} \\
{ - (x + 6),} & {x < - 6} \\ \end{array} } \right.$

5. Just a bit more detail: For x> -6, the is $\frac{2x+ 12}{x+ 6}= \frac{2(x+6)}{x+6}$ and, since x is not equal to -6, we can cancel to get 2. If x< -6, this is $\frac{2x+12}{-(x+6)}= \frac{2(x+6)}{-(x+6)}= -2$.

Of course, you then use the basic property: "If f(x)= g(x) for all x except, possibly, x= 1, then $\lim_{x\to a^+} f(x)= \lim_{x\to a^+} g(x)$ and $\lim_{x\to a^-} f(x)= \lim_{x\to a^-} g(x)$".