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Suppose a function f which maps a real no to a real no is an odd function satisfying $\lim_{x\to 0^+} f(x) = f(0)$

Show that $f(0)=0$ and f is continuous at $x=0$

I am stuck at proving the continuity part.

Heres my partial solution

Since f is an odd function, $f(-x)=-f(x)$ Let $x=0$, $f(-0) = -f(0)$ The only way this can hold true is for $f(0)=0$

To prove continuity, i need to show $\lim_{x\to 0^-} f(x) =\lim_{x\to 0^+} f(x)$. I am not sure how to find $\lim_{x\to 0^-} f(x) $

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up vote 3 down vote accepted

$\lim_{x\to 0^-}f(x)=\lim_{h\to 0}f(0-h)=\lim_{h\to 0}f(-h)=-\lim_{h\to 0}f(h)=-\lim_{x\to 0^+}f(x)=0$

Here, i have used $\lim_{h\to 0}f(-h)=-\lim_{h\to 0}f(h)$ (since $f$ is odd so $f(-h)=-f(h)$)

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Note that $$\lim_{x\to 0^-}f(x)=\lim_{y\to 0^+}f(-y)=-\lim_{y\to 0^+}f(y)=-f(0)$$

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