My attempt: I found the left hand limit and right hand limit at $x=1$ and equated them From that it can be written that $\forall x,y\in \mathbb R, f(x)$ is continuous I don't see how $f$ can be discontinous at $x=0$

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  • what about $f(x)=x^2$ ?? – giannispapav Dec 6 at 16:30
  • If $f$ is the constant one function, $f(xy) = f(x)f(y)$ and $f$ is continuous everywhere, so you cannot prove what is stated in your title. – Mees de Vries Dec 6 at 16:30
  • Is this true though, how about $f(x) = 1$ – caverac Dec 6 at 16:31
  • $f(x)=0\forall x$ is a counterexample. Do you mean $f(xy) = f(x) + f(y)$ ("$+$" instead of "$\cdot$") ? – Caroline Dec 6 at 16:38
  • @Caroline. No, same counter example. – William Elliot Dec 7 at 3:38