TITLE: Is $0$ a removable discontinuity of $\frac{|x-2|-2}{x}$?
QUESTION [6 upvotes]: Given the function $ f(x) = \large\frac{|x-2|-2}{x} $ ,
Is it true to say that the function isn't defined at $x=0$ (because of the denominator!)?
Thus it is a removable discontinuity ? 
The problem is, that if I try to remove the absolute value, I get that in the region $ x<2 $ : $ f(x) = -1$ 
What is the correct logical definition I need to use? 
Thanks

REPLY [5 votes]: Given the function $$ f(x) = \frac{|x-2|-2}{x},$$
  Is it true to say that the function isn't defined at $x=0$ (because of the denominator!)?

Yes, that's correct. As currently defined, the function is undefined at $x = 0$.
But that doesn't mean that the limit of $f(x)$ as $x \to 0$ is undefined. Recall, we are interested in what is happening as $x$ gets very very close to $0$ (not what is happening AT zero).
As $x \to 0, |x - 2| = 2 - x$, so 
$$\lim_{x \to 0}\frac{\left|x-2\right|-2}{x}=\lim_{x \to 0}\frac{2-x-2}{x}=-1$$

Thus it is a removable discontinuity ? 

Yes, indeed, by simply defining  $$f(x) = \begin{cases} \frac{\left|x-2\right|-2}{x} & x\neq 0\\ \\
-1 & x = 0\\ \end{cases}$$
$f(x)$ is then continuous at $x = 0$, hence it is a removable discontinuity.