Question Number 3655 by 123456 last updated on 17/Dec/15
$${f}\left({x}\right)=\begin{cases}{{p}+{q}}&{{x}=\frac{{p}}{{q}},{p}\in\mathbb{Z},{q}\in\mathbb{Z},{q}\neq\mathrm{0},\left({p},{q}\right)=\mathrm{1}}\\{\lfloor{x}\rfloor+\lfloor\mathrm{10}{x}\rfloor}&{\mathrm{overtise}}\end{cases} \\ $$$$\mathrm{does}\:{f}\:\mathrm{is}\:\mathrm{continuous}? \\ $$
Commented by prakash jain last updated on 18/Dec/15
$${choose}\:{p}\notin\mathbb{Q},\:\mathrm{say}\:{p}=\pi \\ $$$$\mathrm{Checking}\:\mathrm{for}\:\mathrm{definition} \\ $$$$\forall\epsilon>\mathrm{0},\exists\delta>\mathrm{0}\:\mathrm{that}\:\mathrm{if}\:\mid{p}−{x}\mid<\epsilon\:\mathrm{then}\:\mid{f}\left({p}\right)−{f}\left({x}\right)\mid<\delta \\ $$$${f}\left({p}\right)=\mathrm{3}+\mathrm{31}=\mathrm{34} \\ $$$$\mathrm{Choose}\:\epsilon=\mathrm{1},\:\mathrm{for}\:\mathrm{any}\:\delta\:\mathrm{we}\:\mathrm{can}\:\mathrm{choose}\:{x} \\ $$$$\mathrm{3}+\frac{\mathrm{1}}{\mathrm{34}+\lfloor\delta+\mathrm{1}\rfloor}\Rightarrow\mid{f}\left({p}\right)−{f}\left({x}\right)\mid>\delta \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{continous}. \\ $$