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.
            Check out our Code of Conduct.






            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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                                Σερβία Πίνακας περιεχομένων Γεωγραφία | Ιστορία | Πολιτική | Δημογραφία | Οικονομία | Τουρισμός | Εκπαίδευση και επιστήμη | Πολιτισμός | Δείτε επίσης | Παραπομπές | Εξωτερικοί σύνδεσμοι | Μενού πλοήγησης43°49′00″N 21°08′00″E / 43.8167°N 21.1333°E / 43.8167; 21.133344°49′14″N 20°27′44″E / 44.8206°N 20.4622°E / 44.8206; 20.4622 (Βελιγράδι)Επίσημη εκτίμηση«Σερβία»«Human Development Report 2018»Παγκόσμιος Οργανισμός Υγείας, Προσδόκιμο ζωής και υγιές προσδόκιμο ζωής, Δεδομένα ανά χώρα2003 statistics2004 statistics2005 statistics2006 statistics2007 statistics2008 statistics2009-2013 statistics2014 statisticsStatistical Yearbook of the Republic of Serbia – Tourism, 20152016 statisticsStatistical Yearbook of the Republic of Serbia – Tourism, 2015Πληροφορίες σχετικά με τη Σερβία και τον πολιτισμό τηςΣερβική ΠροεδρίαΕθνικός Οργανισμός Τουρισμού της ΣερβίαςΣερβική ΕθνοσυνέλευσηΣερβίαεε

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                                ΣερβίαΠερίκλειστα κράτηΕυρωπαϊκά κράτηΓιουγκοσλαβικές Δημοκρατίες σερβικάσερβικάκυρίαρχο κράτοςΚεντρικήςΝοτιοανατολικής ΕυρώπηςΠεδιάδας της ΠαννονίαςΒαλκάνιαμεσόγειαΟυγγαρία...
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