Compute $lim_{xtoinfty} x lfloor frac{1}{x} rfloor$












1












$begingroup$


I'm working out a limit and I'm not sure if my assumption is considered rigorous
$$lim_{xtoinfty} xleftlfloorfrac1xrightrfloor$$
I supposed that $0leq xleftlfloorfrac1xrightrfloor leq leftlfloorfrac1xrightrfloor$ since $x$ is approaching $infty$ $($thus $x > 1$$)$ and to get the answer $0$.



Any mistakes here?






calculus limits floor-function







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$endgroup$

















    1












    $begingroup$


    I'm working out a limit and I'm not sure if my assumption is considered rigorous
    $$lim_{xtoinfty} xleftlfloorfrac1xrightrfloor$$
    I supposed that $0leq xleftlfloorfrac1xrightrfloor leq leftlfloorfrac1xrightrfloor$ since $x$ is approaching $infty$ $($thus $x > 1$$)$ and to get the answer $0$.



    Any mistakes here?






    calculus limits floor-function







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I'm working out a limit and I'm not sure if my assumption is considered rigorous
      $$lim_{xtoinfty} xleftlfloorfrac1xrightrfloor$$
      I supposed that $0leq xleftlfloorfrac1xrightrfloor leq leftlfloorfrac1xrightrfloor$ since $x$ is approaching $infty$ $($thus $x > 1$$)$ and to get the answer $0$.



      Any mistakes here?






      calculus limits floor-function







      share|cite|improve this question











      $endgroup$




      I'm working out a limit and I'm not sure if my assumption is considered rigorous
      $$lim_{xtoinfty} xleftlfloorfrac1xrightrfloor$$
      I supposed that $0leq xleftlfloorfrac1xrightrfloor leq leftlfloorfrac1xrightrfloor$ since $x$ is approaching $infty$ $($thus $x > 1$$)$ and to get the answer $0$.



      Any mistakes here?






      calculus limits floor-function




      calculus limits floor-function






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 24 '18 at 23:17









      Xander Henderson

      15.2k103556




      15.2k103556










      asked Dec 24 '18 at 12:53







      user531476





























          3 Answers
          3






          active

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          $begingroup$

          No mistakes. But, can you see it in terms of sequences instead? That will be much easier. $limlimits_{x rightarrow infty}$ can be viewed as a sequence $leftlbrace x_n rightrbrace$, where $forall M > 0, exists N in mathbb{N}$ such that $x_n > M$. This tells us something special for the choice of $M geq 1$. Once we choose $M geq 1$, we will get a stage after which $x_n > 1$ and hence $leftlfloor{dfrac{1}{x_n}}rightrfloor = 0$. Hence, the image sequence is evetually zero and $limlimits_{x rightarrow infty} x leftlfloor{dfrac{1}{x}}rightrfloor = 0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
            $endgroup$
            – user531476
            Dec 24 '18 at 13:34





















          6












          $begingroup$

          You make it sound like the reason that $xlfloorfrac{1}{x}rfloorleqlfloorfrac{1}{x}rfloor$ is true for positive $x$ is because $xyleq y$ for positive $x$ and $y$, but this is not so. Instead, the reason it is true is because if $xleq 1$ then $xyleq y$ for positive $y$ and if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ so that $xlfloorfrac{1}{x}rfloor=0$ too.



          It is far simpler just to note that if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ and hence $xlfloorfrac{1}{x}rfloor=0$, giving us $lim_{xtoinfty}xlfloorfrac{1}{x}rfloor=0$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            The function is $0$ for $x>1$. Alternatively, substitute $m=1/x$. This limit is the same as $displaystylelim_{mto0^+}frac{lfloor mrfloor}m$, which easily evaluates to $0$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How does it easily evaluate to 0?
              $endgroup$
              – user531476
              Dec 24 '18 at 13:39












            • $begingroup$
              $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
              $endgroup$
              – Shubham Johri
              Dec 24 '18 at 13:41












            • $begingroup$
              Oh, I thought we take that as $frac{0}{0}$. Thank you!
              $endgroup$
              – user531476
              Dec 24 '18 at 14:28












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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            No mistakes. But, can you see it in terms of sequences instead? That will be much easier. $limlimits_{x rightarrow infty}$ can be viewed as a sequence $leftlbrace x_n rightrbrace$, where $forall M > 0, exists N in mathbb{N}$ such that $x_n > M$. This tells us something special for the choice of $M geq 1$. Once we choose $M geq 1$, we will get a stage after which $x_n > 1$ and hence $leftlfloor{dfrac{1}{x_n}}rightrfloor = 0$. Hence, the image sequence is evetually zero and $limlimits_{x rightarrow infty} x leftlfloor{dfrac{1}{x}}rightrfloor = 0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
              $endgroup$
              – user531476
              Dec 24 '18 at 13:34


















            2












            $begingroup$

            No mistakes. But, can you see it in terms of sequences instead? That will be much easier. $limlimits_{x rightarrow infty}$ can be viewed as a sequence $leftlbrace x_n rightrbrace$, where $forall M > 0, exists N in mathbb{N}$ such that $x_n > M$. This tells us something special for the choice of $M geq 1$. Once we choose $M geq 1$, we will get a stage after which $x_n > 1$ and hence $leftlfloor{dfrac{1}{x_n}}rightrfloor = 0$. Hence, the image sequence is evetually zero and $limlimits_{x rightarrow infty} x leftlfloor{dfrac{1}{x}}rightrfloor = 0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
              $endgroup$
              – user531476
              Dec 24 '18 at 13:34
















            2












            2








            2





            $begingroup$

            No mistakes. But, can you see it in terms of sequences instead? That will be much easier. $limlimits_{x rightarrow infty}$ can be viewed as a sequence $leftlbrace x_n rightrbrace$, where $forall M > 0, exists N in mathbb{N}$ such that $x_n > M$. This tells us something special for the choice of $M geq 1$. Once we choose $M geq 1$, we will get a stage after which $x_n > 1$ and hence $leftlfloor{dfrac{1}{x_n}}rightrfloor = 0$. Hence, the image sequence is evetually zero and $limlimits_{x rightarrow infty} x leftlfloor{dfrac{1}{x}}rightrfloor = 0$.






            share|cite|improve this answer











            $endgroup$



            No mistakes. But, can you see it in terms of sequences instead? That will be much easier. $limlimits_{x rightarrow infty}$ can be viewed as a sequence $leftlbrace x_n rightrbrace$, where $forall M > 0, exists N in mathbb{N}$ such that $x_n > M$. This tells us something special for the choice of $M geq 1$. Once we choose $M geq 1$, we will get a stage after which $x_n > 1$ and hence $leftlfloor{dfrac{1}{x_n}}rightrfloor = 0$. Hence, the image sequence is evetually zero and $limlimits_{x rightarrow infty} x leftlfloor{dfrac{1}{x}}rightrfloor = 0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 24 '18 at 15:41









            BPP

            2,204927




            2,204927










            answered Dec 24 '18 at 13:00









            Aniruddha DeshmukhAniruddha Deshmukh

            1,261519




            1,261519












            • $begingroup$
              Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
              $endgroup$
              – user531476
              Dec 24 '18 at 13:34




















            • $begingroup$
              Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
              $endgroup$
              – user531476
              Dec 24 '18 at 13:34


















            $begingroup$
            Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
            $endgroup$
            – user531476
            Dec 24 '18 at 13:34






            $begingroup$
            Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
            $endgroup$
            – user531476
            Dec 24 '18 at 13:34













            6












            $begingroup$

            You make it sound like the reason that $xlfloorfrac{1}{x}rfloorleqlfloorfrac{1}{x}rfloor$ is true for positive $x$ is because $xyleq y$ for positive $x$ and $y$, but this is not so. Instead, the reason it is true is because if $xleq 1$ then $xyleq y$ for positive $y$ and if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ so that $xlfloorfrac{1}{x}rfloor=0$ too.



            It is far simpler just to note that if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ and hence $xlfloorfrac{1}{x}rfloor=0$, giving us $lim_{xtoinfty}xlfloorfrac{1}{x}rfloor=0$.






            share|cite|improve this answer









            $endgroup$


















              6












              $begingroup$

              You make it sound like the reason that $xlfloorfrac{1}{x}rfloorleqlfloorfrac{1}{x}rfloor$ is true for positive $x$ is because $xyleq y$ for positive $x$ and $y$, but this is not so. Instead, the reason it is true is because if $xleq 1$ then $xyleq y$ for positive $y$ and if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ so that $xlfloorfrac{1}{x}rfloor=0$ too.



              It is far simpler just to note that if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ and hence $xlfloorfrac{1}{x}rfloor=0$, giving us $lim_{xtoinfty}xlfloorfrac{1}{x}rfloor=0$.






              share|cite|improve this answer









              $endgroup$
















                6












                6








                6





                $begingroup$

                You make it sound like the reason that $xlfloorfrac{1}{x}rfloorleqlfloorfrac{1}{x}rfloor$ is true for positive $x$ is because $xyleq y$ for positive $x$ and $y$, but this is not so. Instead, the reason it is true is because if $xleq 1$ then $xyleq y$ for positive $y$ and if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ so that $xlfloorfrac{1}{x}rfloor=0$ too.



                It is far simpler just to note that if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ and hence $xlfloorfrac{1}{x}rfloor=0$, giving us $lim_{xtoinfty}xlfloorfrac{1}{x}rfloor=0$.






                share|cite|improve this answer









                $endgroup$



                You make it sound like the reason that $xlfloorfrac{1}{x}rfloorleqlfloorfrac{1}{x}rfloor$ is true for positive $x$ is because $xyleq y$ for positive $x$ and $y$, but this is not so. Instead, the reason it is true is because if $xleq 1$ then $xyleq y$ for positive $y$ and if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ so that $xlfloorfrac{1}{x}rfloor=0$ too.



                It is far simpler just to note that if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ and hence $xlfloorfrac{1}{x}rfloor=0$, giving us $lim_{xtoinfty}xlfloorfrac{1}{x}rfloor=0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 24 '18 at 13:03









                Ben WBen W

                2,734918




                2,734918























                    2












                    $begingroup$

                    The function is $0$ for $x>1$. Alternatively, substitute $m=1/x$. This limit is the same as $displaystylelim_{mto0^+}frac{lfloor mrfloor}m$, which easily evaluates to $0$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      How does it easily evaluate to 0?
                      $endgroup$
                      – user531476
                      Dec 24 '18 at 13:39












                    • $begingroup$
                      $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                      $endgroup$
                      – Shubham Johri
                      Dec 24 '18 at 13:41












                    • $begingroup$
                      Oh, I thought we take that as $frac{0}{0}$. Thank you!
                      $endgroup$
                      – user531476
                      Dec 24 '18 at 14:28
















                    2












                    $begingroup$

                    The function is $0$ for $x>1$. Alternatively, substitute $m=1/x$. This limit is the same as $displaystylelim_{mto0^+}frac{lfloor mrfloor}m$, which easily evaluates to $0$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      How does it easily evaluate to 0?
                      $endgroup$
                      – user531476
                      Dec 24 '18 at 13:39












                    • $begingroup$
                      $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                      $endgroup$
                      – Shubham Johri
                      Dec 24 '18 at 13:41












                    • $begingroup$
                      Oh, I thought we take that as $frac{0}{0}$. Thank you!
                      $endgroup$
                      – user531476
                      Dec 24 '18 at 14:28














                    2












                    2








                    2





                    $begingroup$

                    The function is $0$ for $x>1$. Alternatively, substitute $m=1/x$. This limit is the same as $displaystylelim_{mto0^+}frac{lfloor mrfloor}m$, which easily evaluates to $0$.






                    share|cite|improve this answer









                    $endgroup$



                    The function is $0$ for $x>1$. Alternatively, substitute $m=1/x$. This limit is the same as $displaystylelim_{mto0^+}frac{lfloor mrfloor}m$, which easily evaluates to $0$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 24 '18 at 13:33









                    Shubham JohriShubham Johri

                    5,658918




                    5,658918












                    • $begingroup$
                      How does it easily evaluate to 0?
                      $endgroup$
                      – user531476
                      Dec 24 '18 at 13:39












                    • $begingroup$
                      $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                      $endgroup$
                      – Shubham Johri
                      Dec 24 '18 at 13:41












                    • $begingroup$
                      Oh, I thought we take that as $frac{0}{0}$. Thank you!
                      $endgroup$
                      – user531476
                      Dec 24 '18 at 14:28


















                    • $begingroup$
                      How does it easily evaluate to 0?
                      $endgroup$
                      – user531476
                      Dec 24 '18 at 13:39












                    • $begingroup$
                      $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                      $endgroup$
                      – Shubham Johri
                      Dec 24 '18 at 13:41












                    • $begingroup$
                      Oh, I thought we take that as $frac{0}{0}$. Thank you!
                      $endgroup$
                      – user531476
                      Dec 24 '18 at 14:28
















                    $begingroup$
                    How does it easily evaluate to 0?
                    $endgroup$
                    – user531476
                    Dec 24 '18 at 13:39






                    $begingroup$
                    How does it easily evaluate to 0?
                    $endgroup$
                    – user531476
                    Dec 24 '18 at 13:39














                    $begingroup$
                    $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                    $endgroup$
                    – Shubham Johri
                    Dec 24 '18 at 13:41






                    $begingroup$
                    $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                    $endgroup$
                    – Shubham Johri
                    Dec 24 '18 at 13:41














                    $begingroup$
                    Oh, I thought we take that as $frac{0}{0}$. Thank you!
                    $endgroup$
                    – user531476
                    Dec 24 '18 at 14:28




                    $begingroup$
                    Oh, I thought we take that as $frac{0}{0}$. Thank you!
                    $endgroup$
                    – user531476
                    Dec 24 '18 at 14:28


















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