How to show that an odd function always goes through zero? I have the standard definition of an odd-function from wikipedia:


  Again, let f(x) be a real-valued function of a real variable. Then f
  is odd if the following equation holds for all x and -x in the domain
  of
  $-f(x) = f(-x)$


Can anyone help me how to do this?
Do I have to show, that it converges to zero?
If $-f(x)=f(-x)$ for all $x$ in the domain for which $-x$ is in the domain, then if $0$ is in the domain
$$-f(0)=f(-0)=f(0),$$
which means $2f(0)=0$, and so $f(0)=0$.
The way I think of this property is through the intermediate value theorem. Assume that the odd function $f$, is continuous. Then choose some point $ a &gt; 0 $. Suppose without loss of generality that $ f(a) &gt; 0 $. Then since $ f(-a) = -f(a) &lt;0 $, by the IVT, there must be a point $ b $ in $ (-a,a) $ such that $ f(b) = 0 $. Note that this only works if $ f $ is continuous.