Suppose $f(x)=\frac {3x-1}{\lfloor3x\rfloor}$ how can prove that $f$ is continuous in $D_f$ where $D_f=\mathopen]-\infty;0\mathclose[\cup[\frac{1}{3}; +\infty\mathclose[$.
continuity of a floor function in $D_f$
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limits
continuity
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0yeap i$t$'s discon$t$inuous in 1/3Z and Z but how can I prove that using math – 2012-11-06
1 Answers
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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 $x=1/3$.