Proof of sgn function at $x=0$ Is there any proof for sgn function at $x=0$? I am Wondering that how can any one define that $|x|/x$ at  $x=0$ is zero.
 A: The sign function is $-1$ for $x<0$ and $1$ for $x>0$; so no matter what definition we give to $\operatorname{sgn}0$, the function will be discontinuous at $x=0$. This is an example of regularisation. We may as well pick $\operatorname{sgn}0:=0$. It makes $\operatorname{sgn}x$ an odd function. It also gives such nice results as $\int_{\Bbb R}\frac{\sin xy}{y}dy=\pi\operatorname{sgn}x$.
A: The definition of ${\rm sign}(0)$ did not reach a global consensus so as to become a standard.  
The definition to adopt shall be congruent  with that for the [Heaviside step function][1] in $x=0$ $H(0)$, and thus to
the definition of the Dirac delta "function".
See the above Wikipedia link for an analysis of $H(0)$
However the relation with the step function is ambiguous as well. We can in fact define 
$$
{\rm sign}(x) = H(x) - H( - x)
$$
or
$$
{\rm sign}(x) = 2H(x) - 1
$$
So how to specify  ${\rm sign}(0)$ is in practice left to the peculiarities of each application field, where
one or another specification for ${\rm sign}(0)$ may turn out to be "more plain", "more straight"  and finally "more useful".
In the complex field, for instance it is more "natural" to define
$$
{\rm sign}(z)\quad  = {{\left| z \right|} \over z} = e^{\,i\,\,\arg \,(z)\,} 
$$
where $\arg$ is the principal argument, on which there is general consensus to take 
$\arg(0)=0$: this leads to define ${\rm sign}(0)=1$.
A: You can try deriving $ |x| $ as its derivative is the sgn function. Using sub-differentials, you will find that the slope for $ x < 0 $ is $-1$ and for $ x > 0 $ it will be $1$. For $x=0$ you will obtain a set of slopes $(-1,1)$, which you can intuitively regard as the line that connects the 1 and -1 constant lines of the sgn function. Because there are infinitely number of slopes, you can take the average of that set of values, which is zero.
