# Proof for discontinuity in $f(x)$ at $x=0$ if $f(xy)=f(x)f(y)$ and $f(x)$ is continuous at $x=1$ [closed]

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 $$x=0$$

## closed as off-topic by amWhy, José Carlos Santos, user10354138, KReiser, RebellosDec 7 at 10:57

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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