Find the limit with greatest integer function












0












$begingroup$


Find the following limit



$$lim_{x to 0}frac{[x]}{x}$$



Where $[x]$ is the greatest integer function.



I tried using the squeeze theorem on this but couldn’t come up with the appropriate functions, can someone help me out?



Edit: I can now see that this limit will not exist.



But what can we say about



$$lim_{x to 0}x[1/x]$$



Where again [x] is the greatest integer function, how can i use squeeze theorem to find this?






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
    $endgroup$
    – Yanko
    Jan 15 at 13:55






  • 1




    $begingroup$
    For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
    $endgroup$
    – Yanko
    Jan 15 at 14:01












  • $begingroup$
    Are you sure you didn't mean least integer function?
    $endgroup$
    – Cuhrazatee
    Jan 15 at 14:36
















0












$begingroup$


Find the following limit



$$lim_{x to 0}frac{[x]}{x}$$



Where $[x]$ is the greatest integer function.



I tried using the squeeze theorem on this but couldn’t come up with the appropriate functions, can someone help me out?



Edit: I can now see that this limit will not exist.



But what can we say about



$$lim_{x to 0}x[1/x]$$



Where again [x] is the greatest integer function, how can i use squeeze theorem to find this?






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
    $endgroup$
    – Yanko
    Jan 15 at 13:55






  • 1




    $begingroup$
    For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
    $endgroup$
    – Yanko
    Jan 15 at 14:01












  • $begingroup$
    Are you sure you didn't mean least integer function?
    $endgroup$
    – Cuhrazatee
    Jan 15 at 14:36














0












0








0





$begingroup$


Find the following limit



$$lim_{x to 0}frac{[x]}{x}$$



Where $[x]$ is the greatest integer function.



I tried using the squeeze theorem on this but couldn’t come up with the appropriate functions, can someone help me out?



Edit: I can now see that this limit will not exist.



But what can we say about



$$lim_{x to 0}x[1/x]$$



Where again [x] is the greatest integer function, how can i use squeeze theorem to find this?






calculus limits







share|cite|improve this question











$endgroup$




Find the following limit



$$lim_{x to 0}frac{[x]}{x}$$



Where $[x]$ is the greatest integer function.



I tried using the squeeze theorem on this but couldn’t come up with the appropriate functions, can someone help me out?



Edit: I can now see that this limit will not exist.



But what can we say about



$$lim_{x to 0}x[1/x]$$



Where again [x] is the greatest integer function, how can i use squeeze theorem to find this?






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 13:59







user601297

















asked Jan 15 at 13:54









user601297user601297

41719




41719












  • $begingroup$
    There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
    $endgroup$
    – Yanko
    Jan 15 at 13:55






  • 1




    $begingroup$
    For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
    $endgroup$
    – Yanko
    Jan 15 at 14:01












  • $begingroup$
    Are you sure you didn't mean least integer function?
    $endgroup$
    – Cuhrazatee
    Jan 15 at 14:36


















  • $begingroup$
    There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
    $endgroup$
    – Yanko
    Jan 15 at 13:55






  • 1




    $begingroup$
    For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
    $endgroup$
    – Yanko
    Jan 15 at 14:01












  • $begingroup$
    Are you sure you didn't mean least integer function?
    $endgroup$
    – Cuhrazatee
    Jan 15 at 14:36
















$begingroup$
There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
$endgroup$
– Yanko
Jan 15 at 13:55




$begingroup$
There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
$endgroup$
– Yanko
Jan 15 at 13:55




1




1




$begingroup$
For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
$endgroup$
– Yanko
Jan 15 at 14:01






$begingroup$
For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
$endgroup$
– Yanko
Jan 15 at 14:01














$begingroup$
Are you sure you didn't mean least integer function?
$endgroup$
– Cuhrazatee
Jan 15 at 14:36




$begingroup$
Are you sure you didn't mean least integer function?
$endgroup$
– Cuhrazatee
Jan 15 at 14:36










2 Answers
2






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oldest

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0












$begingroup$

if the GIF function contains a value that tends to infinity then the gif function can be removed.
hence [1/x] in the second expression will simply become 1/x.
now if the GIF contains some other expression then we must understand that the limit is applied on value given by GIF and not on the expression in the GIF.
Right Hand Limit And Left hand limit must be found(bcoz in negative inputs it gives further negative number ie [-0.2] = -1) and then solved.
Ex. $ lim_{x->0^+}[sinx/x] $
now since sine of a value is always less than the actual value hence sinx/x will be little less than 1 so GIF will return zero so
= lim_{x->0}0=0






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Defining $u={1over x}$ we must obtain two limits$$lim_{uto infty}{lfloor urfloorover u}$$and $$lim_{uto -infty}{lfloor urfloorover u}$$for $uto infty$ we can write$${lfloor urfloorover lfloor urfloor+1}< {lfloor urfloorover u}le1$$and for $uto -infty$$$1le {lfloor urfloorover u}<{lfloor urfloorover lfloor urfloor+1}$$by Squeeze theorem, both tend to $1$ and so does $lim_{xto 0}xlfloor{1over x}rfloor$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      if the GIF function contains a value that tends to infinity then the gif function can be removed.
      hence [1/x] in the second expression will simply become 1/x.
      now if the GIF contains some other expression then we must understand that the limit is applied on value given by GIF and not on the expression in the GIF.
      Right Hand Limit And Left hand limit must be found(bcoz in negative inputs it gives further negative number ie [-0.2] = -1) and then solved.
      Ex. $ lim_{x->0^+}[sinx/x] $
      now since sine of a value is always less than the actual value hence sinx/x will be little less than 1 so GIF will return zero so
      = lim_{x->0}0=0






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        if the GIF function contains a value that tends to infinity then the gif function can be removed.
        hence [1/x] in the second expression will simply become 1/x.
        now if the GIF contains some other expression then we must understand that the limit is applied on value given by GIF and not on the expression in the GIF.
        Right Hand Limit And Left hand limit must be found(bcoz in negative inputs it gives further negative number ie [-0.2] = -1) and then solved.
        Ex. $ lim_{x->0^+}[sinx/x] $
        now since sine of a value is always less than the actual value hence sinx/x will be little less than 1 so GIF will return zero so
        = lim_{x->0}0=0






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          if the GIF function contains a value that tends to infinity then the gif function can be removed.
          hence [1/x] in the second expression will simply become 1/x.
          now if the GIF contains some other expression then we must understand that the limit is applied on value given by GIF and not on the expression in the GIF.
          Right Hand Limit And Left hand limit must be found(bcoz in negative inputs it gives further negative number ie [-0.2] = -1) and then solved.
          Ex. $ lim_{x->0^+}[sinx/x] $
          now since sine of a value is always less than the actual value hence sinx/x will be little less than 1 so GIF will return zero so
          = lim_{x->0}0=0






          share|cite|improve this answer









          $endgroup$



          if the GIF function contains a value that tends to infinity then the gif function can be removed.
          hence [1/x] in the second expression will simply become 1/x.
          now if the GIF contains some other expression then we must understand that the limit is applied on value given by GIF and not on the expression in the GIF.
          Right Hand Limit And Left hand limit must be found(bcoz in negative inputs it gives further negative number ie [-0.2] = -1) and then solved.
          Ex. $ lim_{x->0^+}[sinx/x] $
          now since sine of a value is always less than the actual value hence sinx/x will be little less than 1 so GIF will return zero so
          = lim_{x->0}0=0







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 15 at 14:45









          Harsh WasnikHarsh Wasnik

          365




          365























              0












              $begingroup$

              Defining $u={1over x}$ we must obtain two limits$$lim_{uto infty}{lfloor urfloorover u}$$and $$lim_{uto -infty}{lfloor urfloorover u}$$for $uto infty$ we can write$${lfloor urfloorover lfloor urfloor+1}< {lfloor urfloorover u}le1$$and for $uto -infty$$$1le {lfloor urfloorover u}<{lfloor urfloorover lfloor urfloor+1}$$by Squeeze theorem, both tend to $1$ and so does $lim_{xto 0}xlfloor{1over x}rfloor$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Defining $u={1over x}$ we must obtain two limits$$lim_{uto infty}{lfloor urfloorover u}$$and $$lim_{uto -infty}{lfloor urfloorover u}$$for $uto infty$ we can write$${lfloor urfloorover lfloor urfloor+1}< {lfloor urfloorover u}le1$$and for $uto -infty$$$1le {lfloor urfloorover u}<{lfloor urfloorover lfloor urfloor+1}$$by Squeeze theorem, both tend to $1$ and so does $lim_{xto 0}xlfloor{1over x}rfloor$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Defining $u={1over x}$ we must obtain two limits$$lim_{uto infty}{lfloor urfloorover u}$$and $$lim_{uto -infty}{lfloor urfloorover u}$$for $uto infty$ we can write$${lfloor urfloorover lfloor urfloor+1}< {lfloor urfloorover u}le1$$and for $uto -infty$$$1le {lfloor urfloorover u}<{lfloor urfloorover lfloor urfloor+1}$$by Squeeze theorem, both tend to $1$ and so does $lim_{xto 0}xlfloor{1over x}rfloor$






                  share|cite|improve this answer









                  $endgroup$



                  Defining $u={1over x}$ we must obtain two limits$$lim_{uto infty}{lfloor urfloorover u}$$and $$lim_{uto -infty}{lfloor urfloorover u}$$for $uto infty$ we can write$${lfloor urfloorover lfloor urfloor+1}< {lfloor urfloorover u}le1$$and for $uto -infty$$$1le {lfloor urfloorover u}<{lfloor urfloorover lfloor urfloor+1}$$by Squeeze theorem, both tend to $1$ and so does $lim_{xto 0}xlfloor{1over x}rfloor$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 15 at 15:10









                  Mostafa AyazMostafa Ayaz

                  18.1k31040




                  18.1k31040






























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