# continuity of a floor function in $D_f$

Suppose $f(x)=\frac {3x-1}{\lfloor3x\rfloor}$ how can prove that $f$ is continuous in $D_f$ where $D_f=\mathopen]-\infty;0\mathclose[\cup[\frac{1}{3}; +\infty\mathclose[$.

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$f$ will not be continuous. Try to think why this is so. –  Pedro Tamaroff Nov 5 '12 at 14:06
I draw the graph but I didn't understand if not how can I prove it's discontinuous? –  pourjour Nov 5 '12 at 14:29
Find a value where you think a discontinuity might occur (namely $x\in\frac{1}{3}\Bbb Z$ in the domain), then show that the left and right-sided limits disagree with each other, or alternately that the limit is not equal to the actual value, at your particular choice of $x$. –  anon Nov 6 '12 at 13:45
yeap it's discontinuous in 1/3Z and Z but how can I prove that using math –  pourjour Nov 6 '12 at 19:25

Theorem Let $f$ be continuous and $g$ discontinuous on $E$ then $h(x) = f(x)/g(x)$ is discontinuous on $E \setminus f^{-1}(0)$.
proof Suppose $h$ was continuous at those points, then $g(x) = f(x)/h(x)$ is the quotient of two continuous functions so it's continuous at those points too. Contradiction.
Therefore $\frac {3x-1}{\lfloor3x\rfloor}$ is discontinuous where $\lfloor3x\rfloor$ is, except $x=1/3$.