# 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?

• write the equation for $x=0$.
– vnd
Commented Dec 14, 2015 at 16:17
• It depends a little on fine details of the definition. If $0$ is in the domain of $f$, we are quickly finished, But one might consider $1/x$ to be an odd function. Commented Dec 14, 2015 at 16:21

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 $f(0)=0$.
HINT: Substitute $x=0$ into the equation defining oddness of $f$: $-f(0)=f(-0)$.
Nothing can be concluded about convergence. But to answer the question in the title, if $0$ is in the domain then $-f(0)=f(0)$ so it must be $0.$ If $0$ is not in the domain then the answer is that it doesn't necessarily do so.
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 $a > 0$. Suppose without loss of generality that $f(a) > 0$. Then since $f(-a) = -f(a) <0$, 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.