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$f(x)$ is given by: $f(x)=1/(1/[x]),\quad 0\leq x\leq 1,$ where $[x]$ represents the largest integer less than or equal to $x$. How to prove that $f(x)$ is discontinuous at infinitely many points on $(0,1)$?

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I assume you mean $f(x) = \dfrac1{\lfloor 1/x \rfloor}$ For $x \in \left(\dfrac1{n+1}, \dfrac1n \right]$, we have $\dfrac1x \in \left[n,n+1 \right)$. This means $f(x) = \dfrac1n$.

Do you now see the points of discontinuity of $f(x)$?