Proof of an odd function's continuity

[o_O]

Here's the problem I'm trying to solve:

If an odd function g(x) is right-continuous at x = 0, show that it is continuous at x = 0 and that g(0) = 0. Hint: Prove first that $\displaystyle \lim_{x \to 0^{-}} g(x)$ exists and equals to $\displaystyle \lim_{x \to 0^{+}} g(-x)$

Here's my attempt:
I was thinking that since -g(x) = g(-x) then |-g(x)| = |g(-x)|. And since we're proving that the limit of odd continuous functions have a limit at 0 (I think that's a safe assumption) then:
Given $\displaystyle \epsilon>0$. $\displaystyle \exists\delta >0$ such that whenever 0 < |x| < $\displaystyle \delta$ it follows that |g(x) - M| < $\displaystyle \epsilon$. Since M = 0, then $\displaystyle |g(x)|<\epsilon$. (All this is the function approaching 0 from the left)

Then looking at g(-x), for the same epsilon and delta (since the definition of odd functions implies symmetry about the origin) we arrive at the same conclusion |g(x)| < $\displaystyle \epsilon$.

Do I seem to be on the right track or is this just dead wrong? It's difficult making the transition from concrete, repetitive epsilon-delta proofs to more general and abstract ones. The methods don't quite seem to be the same ...

Thanks a lot in advance! Any help would be deeply appreciated : )

 

[pka]

\(\displaystyle \begin{array}{l}
g\left( { - x} \right) = - g\left( x \right) \\
g( - 0) = g(0) = - g(0) \\
g(0) = 0 \\
\end{array}\)
As you can see, any odd function has the property g(0)=0.
You assumed that did you not?
Otherwise, your work is right.

 

[o_O]

Ooh yes yes. Thanks a lot! If you don't mind me asking, do you have any suggestions on where to practice doing epsilon-delta proofs? I'm having trouble with more abstract ones (such as this one) and my textbook (Stewart) just provides with rather simplistic examples and questions. Anywho, thanks a lot again!