Evaluating limits involving absolute values: $\lim_{x\to\pm6} \frac{2x+12}{|x+6|}$

Question

I have two questions regarding limits involving absolute values. How do I evaluate the following:

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

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

To handle the first problem, I considered two cases: $x\gt -6$ and $x\lt -6$ and thus get $2$ and $-2$ respectively for the right and left hand limits.
However, I am bit confused regarding the second problem. How do I handle that?

Answer 1

For the second problem, when $x$ approaches $6$, $x+6$ will be positive — the absolute values are unecessary, they're just here for show.

(more generally, if $a\neq 0$ and $x_n\operatorname*{\to}_{n\to\infty} a$, then there exists $N \geq 0$ such that $\forall n\geq N$ $x$ and $a$ have same sign)

Answer 2

If $x\rightarrow 6$, eventually it has to be greater than $-6$ and so, we will have that $|x+6| = x+6$. You can then just do regular division and you'll get that the answer is 2.

Answer 3

As $\ x \to 6^+, \ |x+6| \to 12^+ > 0$, so we can replace $|x + 6|$ with $ x+ 6$

As $\ x\to 6^-,\ |x+6| \to 12^- > 0,$ so we can replace $|x+ 6|$ with $x + 6$:

$$ \lim_{\large x\to 6^+} \ \frac{2x+12}{|x+6|} \quad =\quad \lim_{\large x \to 6^-} \ \frac{2x+12}{|x + 6|} \quad= \quad \frac{2(x+6)}{x+ 6} \quad= \quad2$$