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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?


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What's wrong having with $f(x)=-1$ in the region $x<2$ ? – Ted Dec 21 '12 at 17:34
I changed the title because it made no sense and seemed to have no relevance to the question. Hope you like it! – rschwieb Dec 21 '12 at 17:51
up vote 9 down vote accepted

As you said, $f$ isn't defined at $0$. However, $$\lim_{x\to 0}\frac{\left|x-2\right|-2}{x}=\lim_{x\to 0}\frac{2-x-2}{x}=-1$$ and so if we define $f(0)=-1$, $f$ becomes continuous at $0$.

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Thanks. That's indeed what I thought. – joshua Dec 21 '12 at 17:34
@joshua: Note that when $x\to 0$, we always consider $x\neq 0$. – Babak S. Dec 21 '12 at 17:34
@BabakSorouh : great, thanks! – joshua Dec 21 '12 at 17:36

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.

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@Sami Me too: via googling some question I had, three of the first hits turned up! – amWhy Jul 15 '14 at 11:45

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