Using Epsilon Delta to prove that for a continuous odd function, f(0)=0

Question

Suppose f(x) is a continuous odd function defined on [−∞,∞].

Prove that f(0) = 0.

I believe the answer requires using the definition of continuity/limits in order to prove that for an odd function, f(0)=0

Answer 1

1.) f is odd $\Rightarrow f(-x)=-f(x)$ for all $x \in \mathbb{R}$

2.) f is continous in $\mathbb{R}$, this implies that f is continous in zero. Therefore $$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) =f(0)$$

Now, the first lateral limit can be rewritten $$\lim_{x \rightarrow 0^-} f(x)= \lim_{x \rightarrow 0^+} f(-x)$$

and f is odd, so $$\lim_{x \rightarrow 0^-} f(x)= \lim_{x \rightarrow 0^+} f(-x)=-\lim_{x \rightarrow 0^+} f(x)$$

By continuity $$ \left\lbrace \begin{array}{l} \lim\limits_{x \rightarrow 0^+} f(x)=-f(0) \\ \lim\limits_{x \rightarrow 0^+} f(x)=f(0) \\ \end{array}\right\rbrace. \Rightarrow f(0)=-f(0) $$

And the only solution for $f(0)=-f(0)$ is $f(0)=0$.