$f:\mathbb{R} \to \mathbb{R}$ 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.


By odd parity,


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

  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ @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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