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$f:\mathbb{R} \to \mathbb{R}$ is a bounded odd function. What can be said about $\lim_{x \to 0}f(x)$?

  1. Must be $0$
  2. If exists then $0$
  3. Must exist but may not be $0$

I am stuck at this question. I know, for continuous odd function we must have value $0$ at $x=0$. But we don't know if this function is continuous or not. How to proceed? Thanks in advance.

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By odd parity,

$$\lim_{x\to0^-}f(x)=-\lim_{x\to0^+}f(x).$$

If those limits exist, they must be equal for the ordinary limit to exist, hence 2.

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  • $\begingroup$ Why are you saying that the limit may not exist? I cannot think of any counterexample here $\endgroup$ – Prof.Shanku Jul 18 '19 at 12:22
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    $\begingroup$ @user587126 Take the function which is $0$ at $x = 0$, $1$ on $\mathbb R^+$, and $-1$ on $\mathbb R^-$. $\endgroup$ – Alex Provost Jul 18 '19 at 12:29
  • $\begingroup$ @AlexProvost Now i get it. Thanks man! $\endgroup$ – Prof.Shanku Jul 18 '19 at 12:32
  • $\begingroup$ @user587126: because limits are not "obliged" to exist ! $\endgroup$ – Yves Daoust Jul 18 '19 at 12:45
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    $\begingroup$ @user587126 Consider $$f(x) = \begin{cases} 1, &\text{ if $x\notin \mathbb{Q}, x > 0$}\\ -1, &\text{ if $x\notin \mathbb{Q}, x < 0$}\\ 0, &\text{ if $x\in \mathbb{Q}$}\\ \end{cases}$$ Then $\lim_{x\to 0^+} f(x)$ and $\lim_{x\to 0^-} f(x)$ both don't exist. $\endgroup$ – mechanodroid Jul 18 '19 at 12:59

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