Proof of Odd functions' Continuity

[bjgawp]

Homework Statement

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 $$\lim_{x \to 0^{-}} g(x)$$ exists and equals to $$\lim_{x \to 0^{+}} g(-x)$$

Homework Equations

The Attempt at a Solution

Suppose $$\lim_{x \to 0^{-}} g(x) = M$$ Let $$\epsilon$$ > 0. We must find $$\delta$$ > 0 such that whenever $$-\delta$$ < x < 0, it follows that |g(x) - M | < $$\epsilon$$. We know that -x < $$\delta$$ but relating this to f(-x) is where I'm stuck. Like the other problem I posted, I can't see what the end result is supposed to be. Any help would be very much appreciated!

 

[qspeechc]

I think your attempted solution takes it too far! For an odd function f(-x) = -f(x), that should be enough to get your proof.

 

[bjgawp]

Hmm I think I do have to incorporate epsilon delta proofs in this one. 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 $$\epsilon>0$$. $$\exists\delta >0$$ such that whenever 0 < |x| < $$\delta$$ it follows that |g(x) - M| < $$\epsilon$$. Since M = 0, then $$|g(x)|<\epsilon$$. (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)| < $$\epsilon$$.

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 ...