How to show that for a continuous function on $\mathbb{R}$ that satisfies $-f(-x)=f(x)$ for all $x \neq0$ then $f(0)=0$ How to show that for a continuous function on $\mathbb{R}$ that satisfies $-f(-x)=f(x)$ for all $x \neq0$ then $f(0)=0$
For $-f(-x)=f(x)$ to be true, it appears we must have an odd function without a constant at the end. You can tell this from examination (it's obvious for a linear function, but should also hold true for cubic functions and upward). So it seems obvious to me that the proposal is true, but I don't know of a theorem to prove my observation, or to move from my observation to an actual proof. Any help?
 A: $$f(-x)=-f(x)， \forall x \ne 0$$
$$\lim_{x \to 0}f(-x)=\lim_{x \to 0}-f(x)$$
By continuity,
$$f(\lim_{x \to 0}-x)=-f(\lim_{x \to 0}x)$$
$$f(0)=-f(0)$$
$$2f(0)=0$$
$$f(0)=0$$
A: With an elementary "$\varepsilon$-$\delta$" proof.
By continuity, we have $f(0) = \lim_{x\to 0} f(x)$. We will use the definition of limit. 
Fix any $\varepsilon > 0$; and let $\delta_\varepsilon > 0$ be such that
$$
\forall x \neq 0 \text{ s.t. } |x|\leq \delta_\varepsilon, \, |f(x)-f(0)| \leq \varepsilon. \tag{1}
$$ 
In particular, fix any $x\in (0,\delta_\varepsilon]$. We have, applying the above to both $x$ and $-x$, 
$$
|f(x)-f(0)| \leq \varepsilon \;\text{ and }\; |f(x)+f(0)| = |-f(x)-f(0)| = |f(-x)-f(0)| \leq \varepsilon \tag{2}
$$
so that (using the triangle inequality)
$$
2|f(0)| = |f(0)-f(x)+f(x)+f(0)| \leq |f(0)-f(x)|+|f(x)+f(0)| \leq 2\varepsilon \tag{3}
$$
i.e., $|f(0)| \leq \varepsilon$. Since the above works for every $\varepsilon > 0$, this implies $|f(0)|=0$, i.e., $f(0)=0$.
A: Consider the sequence $$\{1/n\}$$
Due to continuity of $f(x)$ we have $$ \lim _{n\to \infty} f(1/n)
 =f(0)$$
Similarly we have $$ \lim _{n\to \infty} f(-1/n)
 =f(0)$$
Since we have $$f(1/n)=-f(-1/n)$$ We have
$$ f(0)= -f(0)$$
Thus $$f(0)=0$$ 
A: Suppose $f(0)>0$. The case $f(0)<0$ is analogous.
Let $\varepsilon=\dfrac{f(0)}{2}>0$.
How $f$ is continuous at $0$ (in particular), there exists $\delta>0$ such that $|x-0|<\delta \Rightarrow |f(x)-f(0)|<\varepsilon$.
So, $f(x)\in \left( f(0)-\varepsilon, f(0)+\varepsilon \right) = \left( \dfrac{f(0)}{2}, \dfrac{3\cdot f(0)}{2} \right)$
$\Rightarrow f(x)>\dfrac{f(0)}{2}>0$, $\forall x \in (-\delta, \delta)$.
But, if $f(x)>0$ for some $x\in(-\delta,\delta)$, then $f(-x)<0$ with $-x\in(-\delta,\delta)$. Contradiction!
Therefore $f(0)=0$
A: Define $\, g(x) := f(x) + f(-x). \,$ As a sum of continuous functions, $\, g(x) \,$ is continuous  and $\, g(x) = 0 \,$ for all $\, x \neq 0. \,$ By continuity, $\, g(0) = f(0) + f(-0) = 2f(0) = 0. \,$ Now $\, f(0) = 0.$
