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Proof of Odd functions' Continuity

by bjgawp
Tags: continuity, functions, proof
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Oct18-07, 10:03 PM
P: 84
1. The problem statement, all variables and given/known data
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 [tex]\lim_{x \to 0^{-}} g(x) [/tex] exists and equals to [tex]\lim_{x \to 0^{+}} g(-x)[/tex]

2. Relevant equations

3. The attempt at a solution
Suppose [tex]\lim_{x \to 0^{-}} g(x) = M[/tex] Let [tex]\epsilon[/tex] > 0. We must find [tex]\delta[/tex] > 0 such that whenever [tex]-\delta[/tex] < x < 0, it follows that |g(x) - M | < [tex]\epsilon[/tex]. We know that -x < [tex]\delta[/tex] 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!
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Oct19-07, 05:12 AM
P: 792
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.
Oct19-07, 10:00 PM
P: 84
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 [tex]\epsilon>0[/tex]. [tex]\exists\delta >0[/tex] such that whenever 0 < |x| < [tex]\delta[/tex] it follows that |g(x) - M| < [tex]\epsilon[/tex]. Since M = 0, then [tex]|g(x)|<\epsilon[/tex]. (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)| < [tex]\epsilon[/tex].

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 ...

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