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I have a problem with the two last questions of this exercise.

1) At which points is the absolute value function f(x) = |x| continuous?

$\forall x \in \mathbb{R}$ f is continuous

2) What about $f(x)=\frac{|x|}{x}$

$\forall x \in \mathbb{R}-{0}$, f is continuous

3) And what about $f(x)=\begin{cases} \frac{|x|}{x} &\text{if }x \neq 0\\0 &\text{if }x=0\end{cases}$

4) What type of discontinuities appear if any at all?

Thank you in advance

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2 Answers 2

Hint: For the third note that if $x>0$ then $|x|=x$ and then $f(x)=1$, however if $x<0$ then $f(x)=-1$ by similar consideration.

Note that $f$ is a bounded and non-decreasing as well.

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So I can say that $\lim_{x \to 0^{-}} f(x)=-1$ and $\lim_{x \to 0^{+}}=1$ –  user43669 Oct 4 '12 at 23:59
    
@user43669: Exactly. And $f(0)$ is neither. –  Asaf Karagila Oct 4 '12 at 23:59
    
So in what points is this function continuous ? –  user43669 Oct 5 '12 at 0:02
    
@user43669: If the function is constant on all positive numbers; and constant on all negative numbers... –  Asaf Karagila Oct 5 '12 at 0:02
    
$\mathbb{R}-{0}$ –  user43669 Oct 5 '12 at 0:03

The function $f(x)$ defined in three is called the signum or sign function. If you look at a graph of it then you can clearly see it's behavior. If we approach the $0$ from the right, then $\frac{x}{|x|} = 1$ for all $x>0$. Similarly, if we approach from the left then $\frac{x}{|x|} = -1$ for all $x<0$. What does continuity require the limits to be?

As for the type of discontinuity, there are only three general scenarios in this case. Either the discontinuity is removable or infinite or it is a jump discontinuity. Can this discontinuity possibly be removed by changing the value of $f(0)$? Why or why not? Is the discontinuity infinite?

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The limits must be equal. And they are not. –  user43669 Oct 5 '12 at 0:00

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