6
$\begingroup$

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

  • 0
    I changed the title because it made no sense and seemed to have no $r$elevance to the question. Hope you like it!2012-12-21

2 Answers 2

9

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 $0$.

  • 0
    @BabakSorouh : great, thanks!2012-12-21
5

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.

  • 0
    @Sami Me too: via googling some question I had, three of the first hits turned up math.stackexchange.com!2014-07-15