Why $lim_{xto0^+}{frac bxleft[frac xaright]}=0$?, where $[x] = sup{n in N, n leq x}$












0












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I ask this question here, and I was told that using squezze theorem I could solve it, but using this idea in this limit I end with.



$frac{b}{a} leq lim_{xto0^+}{frac bxleft[frac xaright]} leq infty$



Why?










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  • $begingroup$
    Been attempting to fix your title, I see you've been helping. Seems better to me now.
    $endgroup$
    – Matt Samuel
    Dec 14 '18 at 16:00










  • $begingroup$
    Yeah, first edit don't work out, but I appreciate your help.
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 16:01
















0












$begingroup$


I ask this question here, and I was told that using squezze theorem I could solve it, but using this idea in this limit I end with.



$frac{b}{a} leq lim_{xto0^+}{frac bxleft[frac xaright]} leq infty$



Why?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Been attempting to fix your title, I see you've been helping. Seems better to me now.
    $endgroup$
    – Matt Samuel
    Dec 14 '18 at 16:00










  • $begingroup$
    Yeah, first edit don't work out, but I appreciate your help.
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 16:01














0












0








0





$begingroup$


I ask this question here, and I was told that using squezze theorem I could solve it, but using this idea in this limit I end with.



$frac{b}{a} leq lim_{xto0^+}{frac bxleft[frac xaright]} leq infty$



Why?










share|cite|improve this question











$endgroup$




I ask this question here, and I was told that using squezze theorem I could solve it, but using this idea in this limit I end with.



$frac{b}{a} leq lim_{xto0^+}{frac bxleft[frac xaright]} leq infty$



Why?







calculus limits fractions






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share|cite|improve this question













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share|cite|improve this question








edited Dec 14 '18 at 16:00









Matt Samuel

38.5k63768




38.5k63768










asked Dec 14 '18 at 15:55









Maria GuthierMaria Guthier

887




887












  • $begingroup$
    Been attempting to fix your title, I see you've been helping. Seems better to me now.
    $endgroup$
    – Matt Samuel
    Dec 14 '18 at 16:00










  • $begingroup$
    Yeah, first edit don't work out, but I appreciate your help.
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 16:01


















  • $begingroup$
    Been attempting to fix your title, I see you've been helping. Seems better to me now.
    $endgroup$
    – Matt Samuel
    Dec 14 '18 at 16:00










  • $begingroup$
    Yeah, first edit don't work out, but I appreciate your help.
    $endgroup$
    – Maria Guthier
    Dec 14 '18 at 16:01
















$begingroup$
Been attempting to fix your title, I see you've been helping. Seems better to me now.
$endgroup$
– Matt Samuel
Dec 14 '18 at 16:00




$begingroup$
Been attempting to fix your title, I see you've been helping. Seems better to me now.
$endgroup$
– Matt Samuel
Dec 14 '18 at 16:00












$begingroup$
Yeah, first edit don't work out, but I appreciate your help.
$endgroup$
– Maria Guthier
Dec 14 '18 at 16:01




$begingroup$
Yeah, first edit don't work out, but I appreciate your help.
$endgroup$
– Maria Guthier
Dec 14 '18 at 16:01










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Hint: $x-1leq left [ x right ]leq x$






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    1 Answer
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    Hint: $x-1leq left [ x right ]leq x$






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      Hint: $x-1leq left [ x right ]leq x$






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        Hint: $x-1leq left [ x right ]leq x$







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        answered Dec 14 '18 at 15:58









        John11John11

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