(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Proof of Odd functions' Continuity

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