Let $x>0$ , $lfloor xrfloor$ denotes the greatest integer less than or equal to $x$. Then find limit












1












$begingroup$



For $x>0$, let $lfloor xrfloor$ denote the greatest integer less than or equal to $x$. Then
$$
lim_{xto0^+}xleft(leftlfloorfrac{1}{x}rightrfloor
+leftlfloorfrac{2}{x}rightrfloor+dots
+leftlfloorfrac{10}{x}rightrfloorright)=___
$$




I know , $lfloor xrfloor= x $ if $xin Z$ &
$n$ if $xnotin Z$ and $n in Z$ and $ n<x<n+1$.



I don't know how to proceed further !!






real-analysis calculus algebra-precalculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    The way you defined $lfloor x rfloor$ in the title does not agree with your interpretation in the third line.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 9:17










  • $begingroup$
    note: if $x$ is not an integer $lfloor xrfloor ne x-1$ . However, $lfloor xrfloor = x$ "wihout the decimal part after the comma". Eg. $lfloor 2.568rfloor = 2$
    $endgroup$
    – TheD0ubleT
    Jan 10 at 9:17






  • 3




    $begingroup$
    Hint : Use $x-1le lfloor x rfloorle x$
    $endgroup$
    – Peter
    Jan 10 at 9:22






  • 2




    $begingroup$
    Hint: change the limit to $lim_{ytoinfty}frac{1}{y}(lfloor yrfloor+lfloor 2yrfloor+dots+lfloor 10yrfloor)$.
    $endgroup$
    – egreg
    Jan 10 at 9:32
















1












$begingroup$



For $x>0$, let $lfloor xrfloor$ denote the greatest integer less than or equal to $x$. Then
$$
lim_{xto0^+}xleft(leftlfloorfrac{1}{x}rightrfloor
+leftlfloorfrac{2}{x}rightrfloor+dots
+leftlfloorfrac{10}{x}rightrfloorright)=___
$$




I know , $lfloor xrfloor= x $ if $xin Z$ &
$n$ if $xnotin Z$ and $n in Z$ and $ n<x<n+1$.



I don't know how to proceed further !!






real-analysis calculus algebra-precalculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    The way you defined $lfloor x rfloor$ in the title does not agree with your interpretation in the third line.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 9:17










  • $begingroup$
    note: if $x$ is not an integer $lfloor xrfloor ne x-1$ . However, $lfloor xrfloor = x$ "wihout the decimal part after the comma". Eg. $lfloor 2.568rfloor = 2$
    $endgroup$
    – TheD0ubleT
    Jan 10 at 9:17






  • 3




    $begingroup$
    Hint : Use $x-1le lfloor x rfloorle x$
    $endgroup$
    – Peter
    Jan 10 at 9:22






  • 2




    $begingroup$
    Hint: change the limit to $lim_{ytoinfty}frac{1}{y}(lfloor yrfloor+lfloor 2yrfloor+dots+lfloor 10yrfloor)$.
    $endgroup$
    – egreg
    Jan 10 at 9:32














1












1








1





$begingroup$



For $x>0$, let $lfloor xrfloor$ denote the greatest integer less than or equal to $x$. Then
$$
lim_{xto0^+}xleft(leftlfloorfrac{1}{x}rightrfloor
+leftlfloorfrac{2}{x}rightrfloor+dots
+leftlfloorfrac{10}{x}rightrfloorright)=___
$$




I know , $lfloor xrfloor= x $ if $xin Z$ &
$n$ if $xnotin Z$ and $n in Z$ and $ n<x<n+1$.



I don't know how to proceed further !!






real-analysis calculus algebra-precalculus limits







share|cite|improve this question











$endgroup$





For $x>0$, let $lfloor xrfloor$ denote the greatest integer less than or equal to $x$. Then
$$
lim_{xto0^+}xleft(leftlfloorfrac{1}{x}rightrfloor
+leftlfloorfrac{2}{x}rightrfloor+dots
+leftlfloorfrac{10}{x}rightrfloorright)=___
$$




I know , $lfloor xrfloor= x $ if $xin Z$ &
$n$ if $xnotin Z$ and $n in Z$ and $ n<x<n+1$.



I don't know how to proceed further !!






real-analysis calculus algebra-precalculus limits




real-analysis calculus algebra-precalculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 9:29









egreg

183k1486205




183k1486205










asked Jan 10 at 9:14









sejysejy

1579




1579












  • $begingroup$
    The way you defined $lfloor x rfloor$ in the title does not agree with your interpretation in the third line.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 9:17










  • $begingroup$
    note: if $x$ is not an integer $lfloor xrfloor ne x-1$ . However, $lfloor xrfloor = x$ "wihout the decimal part after the comma". Eg. $lfloor 2.568rfloor = 2$
    $endgroup$
    – TheD0ubleT
    Jan 10 at 9:17






  • 3




    $begingroup$
    Hint : Use $x-1le lfloor x rfloorle x$
    $endgroup$
    – Peter
    Jan 10 at 9:22






  • 2




    $begingroup$
    Hint: change the limit to $lim_{ytoinfty}frac{1}{y}(lfloor yrfloor+lfloor 2yrfloor+dots+lfloor 10yrfloor)$.
    $endgroup$
    – egreg
    Jan 10 at 9:32


















  • $begingroup$
    The way you defined $lfloor x rfloor$ in the title does not agree with your interpretation in the third line.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 9:17










  • $begingroup$
    note: if $x$ is not an integer $lfloor xrfloor ne x-1$ . However, $lfloor xrfloor = x$ "wihout the decimal part after the comma". Eg. $lfloor 2.568rfloor = 2$
    $endgroup$
    – TheD0ubleT
    Jan 10 at 9:17






  • 3




    $begingroup$
    Hint : Use $x-1le lfloor x rfloorle x$
    $endgroup$
    – Peter
    Jan 10 at 9:22






  • 2




    $begingroup$
    Hint: change the limit to $lim_{ytoinfty}frac{1}{y}(lfloor yrfloor+lfloor 2yrfloor+dots+lfloor 10yrfloor)$.
    $endgroup$
    – egreg
    Jan 10 at 9:32
















$begingroup$
The way you defined $lfloor x rfloor$ in the title does not agree with your interpretation in the third line.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 9:17




$begingroup$
The way you defined $lfloor x rfloor$ in the title does not agree with your interpretation in the third line.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 9:17












$begingroup$
note: if $x$ is not an integer $lfloor xrfloor ne x-1$ . However, $lfloor xrfloor = x$ "wihout the decimal part after the comma". Eg. $lfloor 2.568rfloor = 2$
$endgroup$
– TheD0ubleT
Jan 10 at 9:17




$begingroup$
note: if $x$ is not an integer $lfloor xrfloor ne x-1$ . However, $lfloor xrfloor = x$ "wihout the decimal part after the comma". Eg. $lfloor 2.568rfloor = 2$
$endgroup$
– TheD0ubleT
Jan 10 at 9:17




3




3




$begingroup$
Hint : Use $x-1le lfloor x rfloorle x$
$endgroup$
– Peter
Jan 10 at 9:22




$begingroup$
Hint : Use $x-1le lfloor x rfloorle x$
$endgroup$
– Peter
Jan 10 at 9:22




2




2




$begingroup$
Hint: change the limit to $lim_{ytoinfty}frac{1}{y}(lfloor yrfloor+lfloor 2yrfloor+dots+lfloor 10yrfloor)$.
$endgroup$
– egreg
Jan 10 at 9:32




$begingroup$
Hint: change the limit to $lim_{ytoinfty}frac{1}{y}(lfloor yrfloor+lfloor 2yrfloor+dots+lfloor 10yrfloor)$.
$endgroup$
– egreg
Jan 10 at 9:32










1 Answer
1






active

oldest

votes


















0












$begingroup$

$frac{1}{x} -1< lfloor frac{1}{x}rfloor<frac{1}{x}$



multiplying by x we get
$implies$x[$frac{1}{x} -1]< xlfloor frac{1}{x}rfloor<x[frac{1}{x}]$



this gives $1-x< lfloor frac{1}{x}rfloor<1$
taking limit x $to 0^{+}$ by squeez theorem ,we get
$ lim_{xto 0^{+}} xlfloor frac{1}{x}rfloor=1$



similarly,



$frac{2}{x} -1< lfloor frac{2}{x}rfloor<frac{2}{x}$



multiplying by x we get
$implies$x[$frac{2}{x} -1]< xlfloor frac{2}{x}rfloor<x[frac{2}{x}]$



this gives $2-x< lfloor frac{2}{x}rfloor<2$



taking limit x $to 0^{+}$ by squeez theorem ,we get
$ lim_{xto 0^{+}} xlfloor frac{2}{x}rfloor=2$



so this give value of complete expression =55.






share|cite|improve this answer









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    0












    $begingroup$

    $frac{1}{x} -1< lfloor frac{1}{x}rfloor<frac{1}{x}$



    multiplying by x we get
    $implies$x[$frac{1}{x} -1]< xlfloor frac{1}{x}rfloor<x[frac{1}{x}]$



    this gives $1-x< lfloor frac{1}{x}rfloor<1$
    taking limit x $to 0^{+}$ by squeez theorem ,we get
    $ lim_{xto 0^{+}} xlfloor frac{1}{x}rfloor=1$



    similarly,



    $frac{2}{x} -1< lfloor frac{2}{x}rfloor<frac{2}{x}$



    multiplying by x we get
    $implies$x[$frac{2}{x} -1]< xlfloor frac{2}{x}rfloor<x[frac{2}{x}]$



    this gives $2-x< lfloor frac{2}{x}rfloor<2$



    taking limit x $to 0^{+}$ by squeez theorem ,we get
    $ lim_{xto 0^{+}} xlfloor frac{2}{x}rfloor=2$



    so this give value of complete expression =55.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $frac{1}{x} -1< lfloor frac{1}{x}rfloor<frac{1}{x}$



      multiplying by x we get
      $implies$x[$frac{1}{x} -1]< xlfloor frac{1}{x}rfloor<x[frac{1}{x}]$



      this gives $1-x< lfloor frac{1}{x}rfloor<1$
      taking limit x $to 0^{+}$ by squeez theorem ,we get
      $ lim_{xto 0^{+}} xlfloor frac{1}{x}rfloor=1$



      similarly,



      $frac{2}{x} -1< lfloor frac{2}{x}rfloor<frac{2}{x}$



      multiplying by x we get
      $implies$x[$frac{2}{x} -1]< xlfloor frac{2}{x}rfloor<x[frac{2}{x}]$



      this gives $2-x< lfloor frac{2}{x}rfloor<2$



      taking limit x $to 0^{+}$ by squeez theorem ,we get
      $ lim_{xto 0^{+}} xlfloor frac{2}{x}rfloor=2$



      so this give value of complete expression =55.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $frac{1}{x} -1< lfloor frac{1}{x}rfloor<frac{1}{x}$



        multiplying by x we get
        $implies$x[$frac{1}{x} -1]< xlfloor frac{1}{x}rfloor<x[frac{1}{x}]$



        this gives $1-x< lfloor frac{1}{x}rfloor<1$
        taking limit x $to 0^{+}$ by squeez theorem ,we get
        $ lim_{xto 0^{+}} xlfloor frac{1}{x}rfloor=1$



        similarly,



        $frac{2}{x} -1< lfloor frac{2}{x}rfloor<frac{2}{x}$



        multiplying by x we get
        $implies$x[$frac{2}{x} -1]< xlfloor frac{2}{x}rfloor<x[frac{2}{x}]$



        this gives $2-x< lfloor frac{2}{x}rfloor<2$



        taking limit x $to 0^{+}$ by squeez theorem ,we get
        $ lim_{xto 0^{+}} xlfloor frac{2}{x}rfloor=2$



        so this give value of complete expression =55.






        share|cite|improve this answer









        $endgroup$



        $frac{1}{x} -1< lfloor frac{1}{x}rfloor<frac{1}{x}$



        multiplying by x we get
        $implies$x[$frac{1}{x} -1]< xlfloor frac{1}{x}rfloor<x[frac{1}{x}]$



        this gives $1-x< lfloor frac{1}{x}rfloor<1$
        taking limit x $to 0^{+}$ by squeez theorem ,we get
        $ lim_{xto 0^{+}} xlfloor frac{1}{x}rfloor=1$



        similarly,



        $frac{2}{x} -1< lfloor frac{2}{x}rfloor<frac{2}{x}$



        multiplying by x we get
        $implies$x[$frac{2}{x} -1]< xlfloor frac{2}{x}rfloor<x[frac{2}{x}]$



        this gives $2-x< lfloor frac{2}{x}rfloor<2$



        taking limit x $to 0^{+}$ by squeez theorem ,we get
        $ lim_{xto 0^{+}} xlfloor frac{2}{x}rfloor=2$



        so this give value of complete expression =55.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 9:48









        sejysejy

        1579




        1579






























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