Question Number 105036 by bobhans last updated on 25/Jul/20
![lim_(x→0) x.[ (1/x) ] ? note [ ] = greatest integer function](https://www.tinkutara.com/question/Q105036.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}.\left[\:\frac{\mathrm{1}}{{x}}\:\right]\:? \\ $$$${note}\:\left[\:\right]\:=\:{greatest}\:{integer}\:{function} \\ $$
Answered by john santu last updated on 25/Jul/20
$${the}\:{limit}\:{as}\:{x}\rightarrow\mathrm{0}^{+} \:{can}\:{be} \\ $$$${transformed}\:{into}\:\underset{{p}\rightarrow\infty} {\mathrm{lim}}\frac{\lfloor\:{p}\:\rfloor}{{p}}\: \\ $$$${set}\:\left\{\:{p}\:\right\}\:=\:{p}\:−\:\lfloor\:{p}\:\rfloor\: \\ $$$${we}\:{have}\:\frac{\lfloor\:{p}\:\rfloor}{{p}}\:=\:\mathrm{1}−\:\frac{\left\{{p}\right\}}{{p}}\:,\:{since} \\ $$$$\mathrm{0}\leqslant\:\left\{{p}\right\}\:<\:\mathrm{1}\:.\:{so}\:\underset{{p}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\left\{{p}\right\}}{{p}}\right)= \\ $$$$\mathrm{1}−\mathrm{0}\:=\:\mathrm{1}\:.{similarly}\:{for} \\ $$$${the}\:{limit}\:{as}\:{x}\rightarrow\mathrm{0}^{−} \: \\ $$$$\left({JS}\:\spadesuit\blacklozenge\right)\: \\ $$
Answered by mathmax by abdo last updated on 25/Jul/20
![we have [(1/x)] ≤(1/x)<[(1/x)] +1 ⇒ for x>0 ⇒x[(1/x)]≤1<x[(1/x)]+x ⇒ { ((x[(1/x)]≤1 ⇒ 1−x <x[(1/x)]≤1 we passe to limit (x→o^+ ) ⇒)),((1−x<x[(1/x)])) :} lim_(x→0^+ ) x[(1/x)] =1](https://www.tinkutara.com/question/Q105050.png)
$$\mathrm{we}\:\mathrm{have}\:\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\:\leqslant\frac{\mathrm{1}}{\mathrm{x}}<\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\:+\mathrm{1}\:\Rightarrow\:\mathrm{for}\:\mathrm{x}>\mathrm{0}\:\Rightarrow\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\leqslant\mathrm{1}<\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]+\mathrm{x}\:\Rightarrow \\ $$$$\begin{cases}{\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\leqslant\mathrm{1}\:\:\Rightarrow\:\:\:\mathrm{1}−\mathrm{x}\:<\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\leqslant\mathrm{1}\:\:\mathrm{we}\:\mathrm{passe}\:\mathrm{to}\:\mathrm{limit}\:\:\left(\mathrm{x}\rightarrow\mathrm{o}^{+} \right)\:\Rightarrow}\\{\mathrm{1}−\mathrm{x}<\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]}\end{cases} \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{+} } \:\:\:\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\:=\mathrm{1} \\ $$