# Absolute value and limit reasoning.

I am trying to develop my reasoning ability with absolute value. So, I wanted to know if the following reasoning is correct:

Find $\lim_{x \to -6}\dfrac{2x+12}{|x+6|}$

By definition of absolute value we have $|x| = x$ when $x > 0$ and $|x| = -x$ when $x<0$

So for the above limit we can consider the limit from the left and the limit from the right: $(x+6)<0$ and $(x+6)>0$:

Case $(x+6)<0$:

$\dfrac{2x+12}{-(x+6)} = \dfrac{2(x+6)}{-(x+6)} = -2$

Case $(x+6) > 0$:

$\dfrac{2x+12}{(x+6)} = \dfrac{2(x+6)}{(x+6)} = 2$

Hence, the limits from the left and right are not and equal and we conclude that the limit does not exist.

1. If we were not considering the limit at $-6$ we could just evaluate the function at any point since the function is continuous everywhere else
2. I graphed this function and I see that I have a vertical asymptote at $x=-6$. What do these left and right limits evaluating to $-2$ and $2$ mean then?
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Your reasoning is absolutely correct. –  M. Strochyk Oct 28 '12 at 19:18

As echoed by others, your reasoning is correct.

Your graph should consist of 2 horizontal lines:

$y=2$ if $x>6$

$y=-2$ if $x<6$

The function is undefined at $x=6$.

Approaching $x=-6$ from the left, you will approach -2, even though the function is undefined at $x=6$. Likewise, approaching $x=-6$ from the right, you will approach 2.

The meaning of "approaching" in this case is pretty intuitive, as the function doesn't change value on each side of $x=6$. For a more rigorous definition of what "approaching" means in the general case, read about the epsilon-delta definition of limit.

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