Determine the value, when defined The Next CEO of Stack OverflowHow can calculate this limit?Finding the value of one-sided limits and greatest integer function.Find the limit $lim_ nrightarrow infty fracleftlfloorfrac3n10rightrfloorn$What is the value of $lim _xto 0leftlfloorfractan x sin xx^2rightrfloor$Finding a delta for the greatest integer function given an epsilon = 1/2If $lfloorcdotrfloor$ denotes the greatest integer function $ninmathbb N$, what is $lim_xto 0leftlfloorfracnsin(x) x rightrfloor$?Find the value of $lim_xto 0 fracsinxx$Can Greatest integer function and limit be InterchangedEvaluate $ lim_x to -0.5^- leftlfloorfrac1x leftlfloor frac-1x rightrfloorrightrfloor$Let $x>0$ , $lfloor xrfloor$ denotes the greatest integer less than or equal to $x$. Then find limit

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Determine the value, when defined



The Next CEO of Stack OverflowHow can calculate this limit?Finding the value of one-sided limits and greatest integer function.Find the limit $lim_ nrightarrow infty fracleftlfloorfrac3n10rightrfloorn$What is the value of $lim _xto 0leftlfloorfractan x sin xx^2rightrfloor$Finding a delta for the greatest integer function given an epsilon = 1/2If $lfloorcdotrfloor$ denotes the greatest integer function $ninmathbb N$, what is $lim_xto 0leftlfloorfracnsin(x) x rightrfloor$?Find the value of $lim_xto 0 fracsinxx$Can Greatest integer function and limit be InterchangedEvaluate $ lim_x to -0.5^- leftlfloorfrac1x leftlfloor frac-1x rightrfloorrightrfloor$Let $x>0$ , $lfloor xrfloor$ denotes the greatest integer less than or equal to $x$. Then find limit










0












$begingroup$


Determine the value, when defined, of $$lim_xtoinfty frac1x^2-lfloor xrfloor$$ where $lfloor xrfloor$ is the greatest integer.



I know $lim_xtoinfty lfloor xrfloor = infty$ and $lim_xtoinfty x^2$=$infty$ but how do I combine them in order to get $$lim_xtoinfty frac1x^2-[x]$$






limits analysis







share|cite|improve this question











$endgroup$











  • $begingroup$
    What do you mean by $[x]$?
    $endgroup$
    – copper.hat
    2 days ago










  • $begingroup$
    $lim_xtoinfty lfloor xrfloor =infty,$ not $-infty.$
    $endgroup$
    – Thomas Andrews
    2 days ago










  • $begingroup$
    copper.hat greatest integer
    $endgroup$
    – user597188
    2 days ago















0












$begingroup$


Determine the value, when defined, of $$lim_xtoinfty frac1x^2-lfloor xrfloor$$ where $lfloor xrfloor$ is the greatest integer.



I know $lim_xtoinfty lfloor xrfloor = infty$ and $lim_xtoinfty x^2$=$infty$ but how do I combine them in order to get $$lim_xtoinfty frac1x^2-[x]$$






limits analysis







share|cite|improve this question











$endgroup$











  • $begingroup$
    What do you mean by $[x]$?
    $endgroup$
    – copper.hat
    2 days ago










  • $begingroup$
    $lim_xtoinfty lfloor xrfloor =infty,$ not $-infty.$
    $endgroup$
    – Thomas Andrews
    2 days ago










  • $begingroup$
    copper.hat greatest integer
    $endgroup$
    – user597188
    2 days ago













0












0








0





$begingroup$


Determine the value, when defined, of $$lim_xtoinfty frac1x^2-lfloor xrfloor$$ where $lfloor xrfloor$ is the greatest integer.



I know $lim_xtoinfty lfloor xrfloor = infty$ and $lim_xtoinfty x^2$=$infty$ but how do I combine them in order to get $$lim_xtoinfty frac1x^2-[x]$$






limits analysis







share|cite|improve this question











$endgroup$




Determine the value, when defined, of $$lim_xtoinfty frac1x^2-lfloor xrfloor$$ where $lfloor xrfloor$ is the greatest integer.



I know $lim_xtoinfty lfloor xrfloor = infty$ and $lim_xtoinfty x^2$=$infty$ but how do I combine them in order to get $$lim_xtoinfty frac1x^2-[x]$$






limits analysis




limits analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







user597188

















asked 2 days ago









user597188user597188

235




235











  • $begingroup$
    What do you mean by $[x]$?
    $endgroup$
    – copper.hat
    2 days ago










  • $begingroup$
    $lim_xtoinfty lfloor xrfloor =infty,$ not $-infty.$
    $endgroup$
    – Thomas Andrews
    2 days ago










  • $begingroup$
    copper.hat greatest integer
    $endgroup$
    – user597188
    2 days ago
















  • $begingroup$
    What do you mean by $[x]$?
    $endgroup$
    – copper.hat
    2 days ago










  • $begingroup$
    $lim_xtoinfty lfloor xrfloor =infty,$ not $-infty.$
    $endgroup$
    – Thomas Andrews
    2 days ago










  • $begingroup$
    copper.hat greatest integer
    $endgroup$
    – user597188
    2 days ago















$begingroup$
What do you mean by $[x]$?
$endgroup$
– copper.hat
2 days ago




$begingroup$
What do you mean by $[x]$?
$endgroup$
– copper.hat
2 days ago












$begingroup$
$lim_xtoinfty lfloor xrfloor =infty,$ not $-infty.$
$endgroup$
– Thomas Andrews
2 days ago




$begingroup$
$lim_xtoinfty lfloor xrfloor =infty,$ not $-infty.$
$endgroup$
– Thomas Andrews
2 days ago












$begingroup$
copper.hat greatest integer
$endgroup$
– user597188
2 days ago




$begingroup$
copper.hat greatest integer
$endgroup$
– user597188
2 days ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

Notice that $ x-1 leq [x]leq x$ so:



$$lim_xto inftyfrac1x^2-x+1leqlim_xto infty frac1x^2-[x]leq lim_xto infty frac1x^2-x$$



So:



$$0leqlim_xto infty frac1x^2-[x]leq 0$$



And:
$$lim_xto infty frac1x^2-[x]=0$$



:)






share|cite|improve this answer








New contributor




Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




















    1












    $begingroup$

    We always have $lfloor xrfloorle x$.
    For $x ge 2$ we have $x^2 over 2 ge x$.



    Hence, if $x ge 2$, we have $x^2-lfloor xrfloor ge x^2 -x ge x^2 over 2$ and
    so
    $1 over x^2-lfloor xrfloor le 2 over x^2$.






    share|cite|improve this answer











    $endgroup$




















      0












      $begingroup$

      Is in not obvious that $x^2$ grows proportionally faster than $[x]$ so $x^2 - [x]to infty$?



      To verify:



      $x ge [x] > 0$ so $x^2 ge [x]^2$ and $frac 1 x^2 - [x] le frac 1[x]^2 - [x] = frac 1[x]([x] - 1)< frac 1([x]-1)^2to 0$.



      To formalize:



      If $epsilon > 0$ and we want $[x]- 1 > sqrtfrac 1epsilon$, which we can get if we choose $x > lceil sqrtfrac 1epsilon rceil + 1=N$.



      Now if $x > N =lceil sqrtfrac 1epsilon rceil + 1$ then $[x] ge lceil sqrtfrac 1epsilon rceil + 1 ge sqrtfrac 1epsilon + 1$



      $[x]- 1 ge sqrtfrac 1 epsilon$



      $x^2 - [x]>[x]^2 - [x]=[x]([x]- 1)>([x]-1)^2 ge frac 1epsilon$



      $frac 1x^2 -[x] < epsilon$.



      So $limlimits_xto infty frac 1x^2 -[x] = infty$.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Notice that $ x-1 leq [x]leq x$ so:



        $$lim_xto inftyfrac1x^2-x+1leqlim_xto infty frac1x^2-[x]leq lim_xto infty frac1x^2-x$$



        So:



        $$0leqlim_xto infty frac1x^2-[x]leq 0$$



        And:
        $$lim_xto infty frac1x^2-[x]=0$$



        :)






        share|cite|improve this answer








        New contributor




        Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$

















          3












          $begingroup$

          Notice that $ x-1 leq [x]leq x$ so:



          $$lim_xto inftyfrac1x^2-x+1leqlim_xto infty frac1x^2-[x]leq lim_xto infty frac1x^2-x$$



          So:



          $$0leqlim_xto infty frac1x^2-[x]leq 0$$



          And:
          $$lim_xto infty frac1x^2-[x]=0$$



          :)






          share|cite|improve this answer








          New contributor




          Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$















            3












            3








            3





            $begingroup$

            Notice that $ x-1 leq [x]leq x$ so:



            $$lim_xto inftyfrac1x^2-x+1leqlim_xto infty frac1x^2-[x]leq lim_xto infty frac1x^2-x$$



            So:



            $$0leqlim_xto infty frac1x^2-[x]leq 0$$



            And:
            $$lim_xto infty frac1x^2-[x]=0$$



            :)






            share|cite|improve this answer








            New contributor




            Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$



            Notice that $ x-1 leq [x]leq x$ so:



            $$lim_xto inftyfrac1x^2-x+1leqlim_xto infty frac1x^2-[x]leq lim_xto infty frac1x^2-x$$



            So:



            $$0leqlim_xto infty frac1x^2-[x]leq 0$$



            And:
            $$lim_xto infty frac1x^2-[x]=0$$



            :)







            share|cite|improve this answer








            New contributor




            Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|cite|improve this answer



            share|cite|improve this answer






            New contributor




            Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            answered 2 days ago









            EurekaEureka

            423112




            423112




            New contributor




            Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.





            New contributor





            Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                1












                $begingroup$

                We always have $lfloor xrfloorle x$.
                For $x ge 2$ we have $x^2 over 2 ge x$.



                Hence, if $x ge 2$, we have $x^2-lfloor xrfloor ge x^2 -x ge x^2 over 2$ and
                so
                $1 over x^2-lfloor xrfloor le 2 over x^2$.






                share|cite|improve this answer











                $endgroup$

















                  1












                  $begingroup$

                  We always have $lfloor xrfloorle x$.
                  For $x ge 2$ we have $x^2 over 2 ge x$.



                  Hence, if $x ge 2$, we have $x^2-lfloor xrfloor ge x^2 -x ge x^2 over 2$ and
                  so
                  $1 over x^2-lfloor xrfloor le 2 over x^2$.






                  share|cite|improve this answer











                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    We always have $lfloor xrfloorle x$.
                    For $x ge 2$ we have $x^2 over 2 ge x$.



                    Hence, if $x ge 2$, we have $x^2-lfloor xrfloor ge x^2 -x ge x^2 over 2$ and
                    so
                    $1 over x^2-lfloor xrfloor le 2 over x^2$.






                    share|cite|improve this answer











                    $endgroup$



                    We always have $lfloor xrfloorle x$.
                    For $x ge 2$ we have $x^2 over 2 ge x$.



                    Hence, if $x ge 2$, we have $x^2-lfloor xrfloor ge x^2 -x ge x^2 over 2$ and
                    so
                    $1 over x^2-lfloor xrfloor le 2 over x^2$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 days ago

























                    answered 2 days ago









                    copper.hatcopper.hat

                    128k559161




                    128k559161





















                        0












                        $begingroup$

                        Is in not obvious that $x^2$ grows proportionally faster than $[x]$ so $x^2 - [x]to infty$?



                        To verify:



                        $x ge [x] > 0$ so $x^2 ge [x]^2$ and $frac 1 x^2 - [x] le frac 1[x]^2 - [x] = frac 1[x]([x] - 1)< frac 1([x]-1)^2to 0$.



                        To formalize:



                        If $epsilon > 0$ and we want $[x]- 1 > sqrtfrac 1epsilon$, which we can get if we choose $x > lceil sqrtfrac 1epsilon rceil + 1=N$.



                        Now if $x > N =lceil sqrtfrac 1epsilon rceil + 1$ then $[x] ge lceil sqrtfrac 1epsilon rceil + 1 ge sqrtfrac 1epsilon + 1$



                        $[x]- 1 ge sqrtfrac 1 epsilon$



                        $x^2 - [x]>[x]^2 - [x]=[x]([x]- 1)>([x]-1)^2 ge frac 1epsilon$



                        $frac 1x^2 -[x] < epsilon$.



                        So $limlimits_xto infty frac 1x^2 -[x] = infty$.






                        share|cite|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          Is in not obvious that $x^2$ grows proportionally faster than $[x]$ so $x^2 - [x]to infty$?



                          To verify:



                          $x ge [x] > 0$ so $x^2 ge [x]^2$ and $frac 1 x^2 - [x] le frac 1[x]^2 - [x] = frac 1[x]([x] - 1)< frac 1([x]-1)^2to 0$.



                          To formalize:



                          If $epsilon > 0$ and we want $[x]- 1 > sqrtfrac 1epsilon$, which we can get if we choose $x > lceil sqrtfrac 1epsilon rceil + 1=N$.



                          Now if $x > N =lceil sqrtfrac 1epsilon rceil + 1$ then $[x] ge lceil sqrtfrac 1epsilon rceil + 1 ge sqrtfrac 1epsilon + 1$



                          $[x]- 1 ge sqrtfrac 1 epsilon$



                          $x^2 - [x]>[x]^2 - [x]=[x]([x]- 1)>([x]-1)^2 ge frac 1epsilon$



                          $frac 1x^2 -[x] < epsilon$.



                          So $limlimits_xto infty frac 1x^2 -[x] = infty$.






                          share|cite|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            Is in not obvious that $x^2$ grows proportionally faster than $[x]$ so $x^2 - [x]to infty$?



                            To verify:



                            $x ge [x] > 0$ so $x^2 ge [x]^2$ and $frac 1 x^2 - [x] le frac 1[x]^2 - [x] = frac 1[x]([x] - 1)< frac 1([x]-1)^2to 0$.



                            To formalize:



                            If $epsilon > 0$ and we want $[x]- 1 > sqrtfrac 1epsilon$, which we can get if we choose $x > lceil sqrtfrac 1epsilon rceil + 1=N$.



                            Now if $x > N =lceil sqrtfrac 1epsilon rceil + 1$ then $[x] ge lceil sqrtfrac 1epsilon rceil + 1 ge sqrtfrac 1epsilon + 1$



                            $[x]- 1 ge sqrtfrac 1 epsilon$



                            $x^2 - [x]>[x]^2 - [x]=[x]([x]- 1)>([x]-1)^2 ge frac 1epsilon$



                            $frac 1x^2 -[x] < epsilon$.



                            So $limlimits_xto infty frac 1x^2 -[x] = infty$.






                            share|cite|improve this answer









                            $endgroup$



                            Is in not obvious that $x^2$ grows proportionally faster than $[x]$ so $x^2 - [x]to infty$?



                            To verify:



                            $x ge [x] > 0$ so $x^2 ge [x]^2$ and $frac 1 x^2 - [x] le frac 1[x]^2 - [x] = frac 1[x]([x] - 1)< frac 1([x]-1)^2to 0$.



                            To formalize:



                            If $epsilon > 0$ and we want $[x]- 1 > sqrtfrac 1epsilon$, which we can get if we choose $x > lceil sqrtfrac 1epsilon rceil + 1=N$.



                            Now if $x > N =lceil sqrtfrac 1epsilon rceil + 1$ then $[x] ge lceil sqrtfrac 1epsilon rceil + 1 ge sqrtfrac 1epsilon + 1$



                            $[x]- 1 ge sqrtfrac 1 epsilon$



                            $x^2 - [x]>[x]^2 - [x]=[x]([x]- 1)>([x]-1)^2 ge frac 1epsilon$



                            $frac 1x^2 -[x] < epsilon$.



                            So $limlimits_xto infty frac 1x^2 -[x] = infty$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 days ago









                            fleabloodfleablood

                            73.6k22891




                            73.6k22891



























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                                HânzestêdPlak yn Lincolnshire bestjoerlike gebietBostonIngelskegreefskipLincolnshireWithamThe WashThe Fens1545Boston Boston (Lincolnshire) Ut Wikipedy Jump to navigation Jump to search Boston De Boston Stump Sifers Ynwennertal 35.124 (2001) Befolkingstichtens 1244 km² Polityk Lân Feriene Keninkryk Regio East Midlands Greefskip Lincolnshire Oar Tiidsône GMT Simmertiid (UTC+1) Webside Boston Borough Council Lizzing fan Boston yn Ingelân Boston is in lytse âlde havenstêd ( town ) yn it bestjoerlike gebiet Boston, yn it Ingelske greefskip Lincolnshire. It plak hie 35.124 ynwenners yn 2001. De stêd leit oan de rivier de Witham, sa'n seis kilometer fan it plak dêr't de Witham yn it estuarium The Wash streamt. It plak leit yn it flakke, sompige en agrarysk gebiet The Fens. Boston syn meast opfallende gebou is de St Botolph's-tsjerke ( The Stump ), boud fan 1309 oant 1460; it is de grutste parish -tsjerke yn Ingelân mei ien fan de heechste t
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