For each $x \in \mathbb R$ let $\lfloor x \rfloor := \max\{n ∈ \mathbb Z : n ≤ x\}$ and consider $f: \mathbb R → \mathbb R$ such that $$f(x) = \begin{cases} 0 & \textrm{when } x=0, \\ x^2\left\lfloor \tfrac1x \right\rfloor & \textrm{otherwise}. \end{cases}$$
Prove that $f$ is discontinuous at all points of the form $x_0 = \tfrac1n$, with $n \in \mathbb Z \setminus \{0\}$.
If possible, can you please help solve it in epsilon-delta criterion? Thank you very much.
hint
Assume $ n>1$.
$$\begin{align*} x\in\left(\frac1n,\frac1{n-1}\right]&\implies n-1\le\frac1x<n\\ &\implies f(x)=x^2(n-1)\\ &\implies\lim_{x\to\frac1n^+}f(x)=\frac{n-1}{n^2} \end{align*}$$
$$x\in\left(\frac{1}{n+1},\frac 1n\right]\implies $$ $$n\le\frac 1x<n+1 \implies$$ $$f(x)=x^2n\implies$$ $$\lim_{x\to\frac 1n^-}f(x)=\frac 1n$$
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