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Question Number 105036 by bobhans last updated on 25/Jul/20

lim_(x→0) x.[ (1/x) ] ?  note [ ] = greatest integer function

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}.\left[\:\frac{\mathrm{1}}{{x}}\:\right]\:? \\ $$$${note}\:\left[\:\right]\:=\:{greatest}\:{integer}\:{function} \\ $$

Answered by john santu last updated on 25/Jul/20

the limit as x→0^+  can be  transformed into lim_(p→∞) ((⌊ p ⌋)/p)   set { p } = p − ⌊ p ⌋   we have ((⌊ p ⌋)/p) = 1− (({p})/p) , since  0≤ {p} < 1 . so lim_(p→∞) (1−(({p})/p))=  1−0 = 1 .similarly for  the limit as x→0^−    (JS ♠⧫)

$${the}\:{limit}\:{as}\:{x}\rightarrow\mathrm{0}^{+} \:{can}\:{be} \\ $$$${transformed}\:{into}\:\underset{{p}\rightarrow\infty} {\mathrm{lim}}\frac{\lfloor\:{p}\:\rfloor}{{p}}\: \\ $$$${set}\:\left\{\:{p}\:\right\}\:=\:{p}\:−\:\lfloor\:{p}\:\rfloor\: \\ $$$${we}\:{have}\:\frac{\lfloor\:{p}\:\rfloor}{{p}}\:=\:\mathrm{1}−\:\frac{\left\{{p}\right\}}{{p}}\:,\:{since} \\ $$$$\mathrm{0}\leqslant\:\left\{{p}\right\}\:<\:\mathrm{1}\:.\:{so}\:\underset{{p}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\left\{{p}\right\}}{{p}}\right)= \\ $$$$\mathrm{1}−\mathrm{0}\:=\:\mathrm{1}\:.{similarly}\:{for} \\ $$$${the}\:{limit}\:{as}\:{x}\rightarrow\mathrm{0}^{−} \: \\ $$$$\left({JS}\:\spadesuit\blacklozenge\right)\: \\ $$

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

$$\mathrm{we}\:\mathrm{have}\:\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\:\leqslant\frac{\mathrm{1}}{\mathrm{x}}<\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\:+\mathrm{1}\:\Rightarrow\:\mathrm{for}\:\mathrm{x}>\mathrm{0}\:\Rightarrow\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\leqslant\mathrm{1}<\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]+\mathrm{x}\:\Rightarrow \\ $$$$\begin{cases}{\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\leqslant\mathrm{1}\:\:\Rightarrow\:\:\:\mathrm{1}−\mathrm{x}\:<\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\leqslant\mathrm{1}\:\:\mathrm{we}\:\mathrm{passe}\:\mathrm{to}\:\mathrm{limit}\:\:\left(\mathrm{x}\rightarrow\mathrm{o}^{+} \right)\:\Rightarrow}\\{\mathrm{1}−\mathrm{x}<\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]}\end{cases} \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{+} } \:\:\:\mathrm{x}\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\:=\mathrm{1} \\ $$

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