Say we have $\lim_{x\to 2}\ \frac{x-2}{|x-2|}$
Now, being that an absolute value is used, this becomes a piecewise function.
$f(x) = \begin{cases}
x-2, & \text{$x$ > 2} \\
-(x-2), & \text{$x$ < 2}
\end{cases}$
When solving the left side limit we can do:
$\lim_{x\to 2}-\ \frac{x-2}{-(x-2)}$
$\lim_{x\to 2}-\ = -1$
Why is $|x-2|$ equal to $-(x-2)$ when $x<2$ ? aren't we dealing with an absolute value so the range is always $>=0$ ?
Best Answer
The range is $\geq 0$, but if $x < 2$, then $x-2 \boxed{\color{red}{\leq}} 0$ so we need to multiply it by $-1$ to make it positive again.That is, $-(x-2) = -1 \times (x-2) \boxed{\color{blue}{\geq}} 0$ , so indeed $|x-2| = -1 \times (x-2)$.