Limits of the function $[x] = y$, where $y$ is the bigger integer smaller or equal $x$?












1












$begingroup$


I am reading a notes in Analysis and I find this limit ($a$ and $b$ are positive):



$$lim_{xto0^+}{frac xaleft[frac bxright]}=frac baqquadlim_{xto0^+}{frac bxleft[frac xaright]}=0\
lim_{xto0^-}{frac xaleft[frac bxright]}=frac baqquadlim_{xto0^-}{frac bxleft[frac xaright]}=infty$$



(Image that replaced math).



However I do not understand why that is the case.






real-analysis calculus limits







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$endgroup$












  • $begingroup$
    Try to use $y leqslant [y] < y+1$ and squeeze theorem.
    $endgroup$
    – xbh
    Dec 14 '18 at 2:46










  • $begingroup$
    I still don't see it
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 3:00
















1












$begingroup$


I am reading a notes in Analysis and I find this limit ($a$ and $b$ are positive):



$$lim_{xto0^+}{frac xaleft[frac bxright]}=frac baqquadlim_{xto0^+}{frac bxleft[frac xaright]}=0\
lim_{xto0^-}{frac xaleft[frac bxright]}=frac baqquadlim_{xto0^-}{frac bxleft[frac xaright]}=infty$$



(Image that replaced math).



However I do not understand why that is the case.






real-analysis calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try to use $y leqslant [y] < y+1$ and squeeze theorem.
    $endgroup$
    – xbh
    Dec 14 '18 at 2:46










  • $begingroup$
    I still don't see it
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 3:00














1












1








1


1



$begingroup$


I am reading a notes in Analysis and I find this limit ($a$ and $b$ are positive):



$$lim_{xto0^+}{frac xaleft[frac bxright]}=frac baqquadlim_{xto0^+}{frac bxleft[frac xaright]}=0\
lim_{xto0^-}{frac xaleft[frac bxright]}=frac baqquadlim_{xto0^-}{frac bxleft[frac xaright]}=infty$$



(Image that replaced math).



However I do not understand why that is the case.






real-analysis calculus limits







share|cite|improve this question











$endgroup$




I am reading a notes in Analysis and I find this limit ($a$ and $b$ are positive):



$$lim_{xto0^+}{frac xaleft[frac bxright]}=frac baqquadlim_{xto0^+}{frac bxleft[frac xaright]}=0\
lim_{xto0^-}{frac xaleft[frac bxright]}=frac baqquadlim_{xto0^-}{frac bxleft[frac xaright]}=infty$$



(Image that replaced math).



However I do not understand why that is the case.






real-analysis calculus limits




real-analysis calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 7:16









manooooh

6271517




6271517










asked Dec 14 '18 at 2:45









Maria GuthierMaria Guthier

887




887












  • $begingroup$
    Try to use $y leqslant [y] < y+1$ and squeeze theorem.
    $endgroup$
    – xbh
    Dec 14 '18 at 2:46










  • $begingroup$
    I still don't see it
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 3:00


















  • $begingroup$
    Try to use $y leqslant [y] < y+1$ and squeeze theorem.
    $endgroup$
    – xbh
    Dec 14 '18 at 2:46










  • $begingroup$
    I still don't see it
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 3:00
















$begingroup$
Try to use $y leqslant [y] < y+1$ and squeeze theorem.
$endgroup$
– xbh
Dec 14 '18 at 2:46




$begingroup$
Try to use $y leqslant [y] < y+1$ and squeeze theorem.
$endgroup$
– xbh
Dec 14 '18 at 2:46












$begingroup$
I still don't see it
$endgroup$
– Maria Guthier
Dec 14 '18 at 3:00




$begingroup$
I still don't see it
$endgroup$
– Maria Guthier
Dec 14 '18 at 3:00










1 Answer
1






active

oldest

votes


















2












$begingroup$

Hint: As the comment above suggests you can use the fact that for all $cinBbb R$, $cleqslant[c]lt c+1$ hence $$frac{b}{x}leqslantleft[frac{b}{x}right]lt frac{b}{x}+1.$$ When we're approaching $0$ from the right we have $xgt 0$ hence multiplying both sides by $frac xa$ yields, $$frac{x}{a}frac{b}{x}leqslantfrac{x}{a}left[frac{b}{x}right]lt frac{x}{a}left(dfrac{b}{x}+1right).$$ After simplifying use the squeeze theorem. Do the same for the other cases.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But in the second case I end with $frac{a}{b} leq frac{b}{x} [frac{x}{a}] < infty$
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 15:34












  • $begingroup$
    @MariaGuthier So it seems you already got your answer in math.stackexchange.com/questions/3039534/… right :) ?
    $endgroup$
    – Stupid Questions Inc
    Dec 15 '18 at 7:28










  • $begingroup$
    @StupidQuestionsInc Somewhat off-topic: where did you find your profile picture ?
    $endgroup$
    – Gabriel Romon
    Dec 22 '18 at 20:10










  • $begingroup$
    @GabrielRomon somewhere on twitter
    $endgroup$
    – Stupid Questions Inc
    Dec 23 '18 at 7:49











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Hint: As the comment above suggests you can use the fact that for all $cinBbb R$, $cleqslant[c]lt c+1$ hence $$frac{b}{x}leqslantleft[frac{b}{x}right]lt frac{b}{x}+1.$$ When we're approaching $0$ from the right we have $xgt 0$ hence multiplying both sides by $frac xa$ yields, $$frac{x}{a}frac{b}{x}leqslantfrac{x}{a}left[frac{b}{x}right]lt frac{x}{a}left(dfrac{b}{x}+1right).$$ After simplifying use the squeeze theorem. Do the same for the other cases.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But in the second case I end with $frac{a}{b} leq frac{b}{x} [frac{x}{a}] < infty$
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 15:34












  • $begingroup$
    @MariaGuthier So it seems you already got your answer in math.stackexchange.com/questions/3039534/… right :) ?
    $endgroup$
    – Stupid Questions Inc
    Dec 15 '18 at 7:28










  • $begingroup$
    @StupidQuestionsInc Somewhat off-topic: where did you find your profile picture ?
    $endgroup$
    – Gabriel Romon
    Dec 22 '18 at 20:10










  • $begingroup$
    @GabrielRomon somewhere on twitter
    $endgroup$
    – Stupid Questions Inc
    Dec 23 '18 at 7:49
















2












$begingroup$

Hint: As the comment above suggests you can use the fact that for all $cinBbb R$, $cleqslant[c]lt c+1$ hence $$frac{b}{x}leqslantleft[frac{b}{x}right]lt frac{b}{x}+1.$$ When we're approaching $0$ from the right we have $xgt 0$ hence multiplying both sides by $frac xa$ yields, $$frac{x}{a}frac{b}{x}leqslantfrac{x}{a}left[frac{b}{x}right]lt frac{x}{a}left(dfrac{b}{x}+1right).$$ After simplifying use the squeeze theorem. Do the same for the other cases.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But in the second case I end with $frac{a}{b} leq frac{b}{x} [frac{x}{a}] < infty$
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 15:34












  • $begingroup$
    @MariaGuthier So it seems you already got your answer in math.stackexchange.com/questions/3039534/… right :) ?
    $endgroup$
    – Stupid Questions Inc
    Dec 15 '18 at 7:28










  • $begingroup$
    @StupidQuestionsInc Somewhat off-topic: where did you find your profile picture ?
    $endgroup$
    – Gabriel Romon
    Dec 22 '18 at 20:10










  • $begingroup$
    @GabrielRomon somewhere on twitter
    $endgroup$
    – Stupid Questions Inc
    Dec 23 '18 at 7:49














2












2








2





$begingroup$

Hint: As the comment above suggests you can use the fact that for all $cinBbb R$, $cleqslant[c]lt c+1$ hence $$frac{b}{x}leqslantleft[frac{b}{x}right]lt frac{b}{x}+1.$$ When we're approaching $0$ from the right we have $xgt 0$ hence multiplying both sides by $frac xa$ yields, $$frac{x}{a}frac{b}{x}leqslantfrac{x}{a}left[frac{b}{x}right]lt frac{x}{a}left(dfrac{b}{x}+1right).$$ After simplifying use the squeeze theorem. Do the same for the other cases.






share|cite|improve this answer









$endgroup$



Hint: As the comment above suggests you can use the fact that for all $cinBbb R$, $cleqslant[c]lt c+1$ hence $$frac{b}{x}leqslantleft[frac{b}{x}right]lt frac{b}{x}+1.$$ When we're approaching $0$ from the right we have $xgt 0$ hence multiplying both sides by $frac xa$ yields, $$frac{x}{a}frac{b}{x}leqslantfrac{x}{a}left[frac{b}{x}right]lt frac{x}{a}left(dfrac{b}{x}+1right).$$ After simplifying use the squeeze theorem. Do the same for the other cases.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 6:57









Stupid Questions IncStupid Questions Inc

7010




7010












  • $begingroup$
    But in the second case I end with $frac{a}{b} leq frac{b}{x} [frac{x}{a}] < infty$
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 15:34












  • $begingroup$
    @MariaGuthier So it seems you already got your answer in math.stackexchange.com/questions/3039534/… right :) ?
    $endgroup$
    – Stupid Questions Inc
    Dec 15 '18 at 7:28










  • $begingroup$
    @StupidQuestionsInc Somewhat off-topic: where did you find your profile picture ?
    $endgroup$
    – Gabriel Romon
    Dec 22 '18 at 20:10










  • $begingroup$
    @GabrielRomon somewhere on twitter
    $endgroup$
    – Stupid Questions Inc
    Dec 23 '18 at 7:49


















  • $begingroup$
    But in the second case I end with $frac{a}{b} leq frac{b}{x} [frac{x}{a}] < infty$
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 15:34












  • $begingroup$
    @MariaGuthier So it seems you already got your answer in math.stackexchange.com/questions/3039534/… right :) ?
    $endgroup$
    – Stupid Questions Inc
    Dec 15 '18 at 7:28










  • $begingroup$
    @StupidQuestionsInc Somewhat off-topic: where did you find your profile picture ?
    $endgroup$
    – Gabriel Romon
    Dec 22 '18 at 20:10










  • $begingroup$
    @GabrielRomon somewhere on twitter
    $endgroup$
    – Stupid Questions Inc
    Dec 23 '18 at 7:49
















$begingroup$
But in the second case I end with $frac{a}{b} leq frac{b}{x} [frac{x}{a}] < infty$
$endgroup$
– Maria Guthier
Dec 14 '18 at 15:34






$begingroup$
But in the second case I end with $frac{a}{b} leq frac{b}{x} [frac{x}{a}] < infty$
$endgroup$
– Maria Guthier
Dec 14 '18 at 15:34














$begingroup$
@MariaGuthier So it seems you already got your answer in math.stackexchange.com/questions/3039534/… right :) ?
$endgroup$
– Stupid Questions Inc
Dec 15 '18 at 7:28




$begingroup$
@MariaGuthier So it seems you already got your answer in math.stackexchange.com/questions/3039534/… right :) ?
$endgroup$
– Stupid Questions Inc
Dec 15 '18 at 7:28












$begingroup$
@StupidQuestionsInc Somewhat off-topic: where did you find your profile picture ?
$endgroup$
– Gabriel Romon
Dec 22 '18 at 20:10




$begingroup$
@StupidQuestionsInc Somewhat off-topic: where did you find your profile picture ?
$endgroup$
– Gabriel Romon
Dec 22 '18 at 20:10












$begingroup$
@GabrielRomon somewhere on twitter
$endgroup$
– Stupid Questions Inc
Dec 23 '18 at 7:49




$begingroup$
@GabrielRomon somewhere on twitter
$endgroup$
– Stupid Questions Inc
Dec 23 '18 at 7:49


















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