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) $