Limit with floor function












2














I'm working out a limit and I'm not sure if my assumption is considered rigorous
$$lim_{xtoinfty} xlfloor frac{1}{x} rfloor$$
I supposed that $0leq xlfloor frac{1}{x} rfloor leq lfloor frac{1}{x} rfloor$ 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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    2














    I'm working out a limit and I'm not sure if my assumption is considered rigorous
    $$lim_{xtoinfty} xlfloor frac{1}{x} rfloor$$
    I supposed that $0leq xlfloor frac{1}{x} rfloor leq lfloor frac{1}{x} rfloor$ 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



























      2












      2








      2


      1





      I'm working out a limit and I'm not sure if my assumption is considered rigorous
      $$lim_{xtoinfty} xlfloor frac{1}{x} rfloor$$
      I supposed that $0leq xlfloor frac{1}{x} rfloor leq lfloor frac{1}{x} rfloor$ 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















      I'm working out a limit and I'm not sure if my assumption is considered rigorous
      $$lim_{xtoinfty} xlfloor frac{1}{x} rfloor$$
      I supposed that $0leq xlfloor frac{1}{x} rfloor leq lfloor frac{1}{x} rfloor$ 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






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      share|cite|improve this question




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      edited 2 hours ago

























      asked 2 hours ago









      archaic

      816




      816






















          3 Answers
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          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























          • Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
            – archaic
            2 hours ago





















          4














          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





























            2














            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





















            • How does it easily evaluate to 0?
              – archaic
              2 hours ago












            • $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
              – Shubham Johri
              2 hours ago












            • Oh, I thought we take that as $frac{0}{0}$. Thank you!
              – archaic
              1 hour ago











            Your Answer





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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            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























            • Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
              – archaic
              2 hours ago


















            2














            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























            • Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
              – archaic
              2 hours ago
















            2












            2








            2






            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














            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 4 mins ago









            BPP

            2,190927




            2,190927










            answered 2 hours ago









            Aniruddha Deshmukh

            889418




            889418












            • Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
              – archaic
              2 hours ago




















            • Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
              – archaic
              2 hours ago


















            Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
            – archaic
            2 hours ago






            Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
            – archaic
            2 hours ago













            4














            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


























              4














              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
























                4












                4








                4






                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












                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 2 hours ago









                Ben W

                1,368513




                1,368513























                    2














                    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





















                    • How does it easily evaluate to 0?
                      – archaic
                      2 hours ago












                    • $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                      – Shubham Johri
                      2 hours ago












                    • Oh, I thought we take that as $frac{0}{0}$. Thank you!
                      – archaic
                      1 hour ago
















                    2














                    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





















                    • How does it easily evaluate to 0?
                      – archaic
                      2 hours ago












                    • $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                      – Shubham Johri
                      2 hours ago












                    • Oh, I thought we take that as $frac{0}{0}$. Thank you!
                      – archaic
                      1 hour ago














                    2












                    2








                    2






                    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












                    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 2 hours ago









                    Shubham Johri

                    3,621716




                    3,621716












                    • How does it easily evaluate to 0?
                      – archaic
                      2 hours ago












                    • $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                      – Shubham Johri
                      2 hours ago












                    • Oh, I thought we take that as $frac{0}{0}$. Thank you!
                      – archaic
                      1 hour ago


















                    • How does it easily evaluate to 0?
                      – archaic
                      2 hours ago












                    • $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                      – Shubham Johri
                      2 hours ago












                    • Oh, I thought we take that as $frac{0}{0}$. Thank you!
                      – archaic
                      1 hour ago
















                    How does it easily evaluate to 0?
                    – archaic
                    2 hours ago






                    How does it easily evaluate to 0?
                    – archaic
                    2 hours ago














                    $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                    – Shubham Johri
                    2 hours ago






                    $displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
                    – Shubham Johri
                    2 hours ago














                    Oh, I thought we take that as $frac{0}{0}$. Thank you!
                    – archaic
                    1 hour ago




                    Oh, I thought we take that as $frac{0}{0}$. Thank you!
                    – archaic
                    1 hour ago


















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