Function satisfying $limlimits_{xto 0}(f(x)-f(2x))$ but doesn't have $limlimits_{xto 0}f(x)$












0












$begingroup$


I want to find a function that does not have a limit as $limlimits_{xto 0}f(x)$, but the following limit does exist: $limlimits_{xto 0}(f(x)-f(2x))$.



Any direction or hint would be appreciated.






calculus







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$endgroup$








  • 1




    $begingroup$
    I guess $f(x)=sgn(x)$ would do the job. That is $f$ equals $1$, $0$ or $-1$ when $x$ is positive, null or negative, respectively.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 25 '17 at 8:38












  • $begingroup$
    Does $pminfty$ count as existence of the limit?
    $endgroup$
    – Shashi
    Dec 25 '17 at 8:44
















0












$begingroup$


I want to find a function that does not have a limit as $limlimits_{xto 0}f(x)$, but the following limit does exist: $limlimits_{xto 0}(f(x)-f(2x))$.



Any direction or hint would be appreciated.






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I guess $f(x)=sgn(x)$ would do the job. That is $f$ equals $1$, $0$ or $-1$ when $x$ is positive, null or negative, respectively.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 25 '17 at 8:38












  • $begingroup$
    Does $pminfty$ count as existence of the limit?
    $endgroup$
    – Shashi
    Dec 25 '17 at 8:44














0












0








0


0



$begingroup$


I want to find a function that does not have a limit as $limlimits_{xto 0}f(x)$, but the following limit does exist: $limlimits_{xto 0}(f(x)-f(2x))$.



Any direction or hint would be appreciated.






calculus







share|cite|improve this question











$endgroup$




I want to find a function that does not have a limit as $limlimits_{xto 0}f(x)$, but the following limit does exist: $limlimits_{xto 0}(f(x)-f(2x))$.



Any direction or hint would be appreciated.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 8:58









Saad

20.2k92352




20.2k92352










asked Dec 25 '17 at 8:34









Lin TanakaLin Tanaka

265




265








  • 1




    $begingroup$
    I guess $f(x)=sgn(x)$ would do the job. That is $f$ equals $1$, $0$ or $-1$ when $x$ is positive, null or negative, respectively.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 25 '17 at 8:38












  • $begingroup$
    Does $pminfty$ count as existence of the limit?
    $endgroup$
    – Shashi
    Dec 25 '17 at 8:44














  • 1




    $begingroup$
    I guess $f(x)=sgn(x)$ would do the job. That is $f$ equals $1$, $0$ or $-1$ when $x$ is positive, null or negative, respectively.
    $endgroup$
    – Alejandro Nasif Salum
    Dec 25 '17 at 8:38












  • $begingroup$
    Does $pminfty$ count as existence of the limit?
    $endgroup$
    – Shashi
    Dec 25 '17 at 8:44








1




1




$begingroup$
I guess $f(x)=sgn(x)$ would do the job. That is $f$ equals $1$, $0$ or $-1$ when $x$ is positive, null or negative, respectively.
$endgroup$
– Alejandro Nasif Salum
Dec 25 '17 at 8:38






$begingroup$
I guess $f(x)=sgn(x)$ would do the job. That is $f$ equals $1$, $0$ or $-1$ when $x$ is positive, null or negative, respectively.
$endgroup$
– Alejandro Nasif Salum
Dec 25 '17 at 8:38














$begingroup$
Does $pminfty$ count as existence of the limit?
$endgroup$
– Shashi
Dec 25 '17 at 8:44




$begingroup$
Does $pminfty$ count as existence of the limit?
$endgroup$
– Shashi
Dec 25 '17 at 8:44










5 Answers
5






active

oldest

votes


















1












$begingroup$

$f(x)=chi_{{bf{Q}}}(x)$, then $f(2x)=chi_{{bf{Q}}}(2x)=chi_{{bf{Q}}}(x)=f(x)$ but $lim_{xrightarrow 0}f(x)$ does not exist.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What is $chi_{{bf{Q}}}$?
    $endgroup$
    – Holo
    Dec 25 '17 at 11:12










  • $begingroup$
    The function takes the value $1$ for all rational numbers and $0$ otherwise.
    $endgroup$
    – user284331
    Dec 25 '17 at 18:06



















1












$begingroup$

Take



$f(x)=ln(x)$



$f(2x)=ln(2x)=ln(2)+ln(x)$



$g(x)=f(x)-f(2x)=-ln(2)$



Here $lim_{xto0}f(x)$ is undefined but $f(x)-f(2x)$ is a constant






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Btw, if you consider $pminfty$ to be a limit then consider $f(x)=Im(ln(ix))=arg(ix)=arg(i2x)=Im(ln(i2x))=f(2x)implies f(x)-f(2x)=0forall x$ and $lim_{xto0^+}f(x)=pi/2,lim_{xto0^-}f(x)=-pi/2$
    $endgroup$
    – Holo
    Dec 26 '17 at 7:47





















0












$begingroup$

Hint. Take a function such that $$f(x)-f(2x)=fleft(frac{x}{2x} right)=fleft(frac 1 2 right)$$ You already know one of such functions I hope.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Any function such that $f(x)=f(2x)$ will work. Also if they differ in a constant, or if they differ in a function which is continuous at $x=0$ (but of course, that's almost rephrasing your question).






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Here's an example that is rather continuous.



      Take $f(x) = sinleft(2pifrac{ln |x|}{ln 2}right)$ defined for $x neq 0$ (you can set $f(0)=0$ if you want $f$ to be definied for all $x in mathbb R$). Then
      $$f(2x) = sinleft(2pifrac{ln |2x|}{ln 2}right) = sinleft(2pifrac{ln2 + ln |x|}{ln 2}right) \= sinleft(2pi+2pifrac{ln |x|}{ln 2}right) = sinleft(2pifrac{ln |x|}{ln 2}right) \= f(x)$$
      so
      $$lim_{xto 0}(f(x)-f(2x)) = 0.$$
      But $lim_{xto 0} f(x)$ is not defined since $f$ oscillates quickly as $x to 0$.






      share|cite|improve this answer









      $endgroup$













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        5 Answers
        5






        active

        oldest

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        5 Answers
        5






        active

        oldest

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        active

        oldest

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        active

        oldest

        votes









        1












        $begingroup$

        $f(x)=chi_{{bf{Q}}}(x)$, then $f(2x)=chi_{{bf{Q}}}(2x)=chi_{{bf{Q}}}(x)=f(x)$ but $lim_{xrightarrow 0}f(x)$ does not exist.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          What is $chi_{{bf{Q}}}$?
          $endgroup$
          – Holo
          Dec 25 '17 at 11:12










        • $begingroup$
          The function takes the value $1$ for all rational numbers and $0$ otherwise.
          $endgroup$
          – user284331
          Dec 25 '17 at 18:06
















        1












        $begingroup$

        $f(x)=chi_{{bf{Q}}}(x)$, then $f(2x)=chi_{{bf{Q}}}(2x)=chi_{{bf{Q}}}(x)=f(x)$ but $lim_{xrightarrow 0}f(x)$ does not exist.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          What is $chi_{{bf{Q}}}$?
          $endgroup$
          – Holo
          Dec 25 '17 at 11:12










        • $begingroup$
          The function takes the value $1$ for all rational numbers and $0$ otherwise.
          $endgroup$
          – user284331
          Dec 25 '17 at 18:06














        1












        1








        1





        $begingroup$

        $f(x)=chi_{{bf{Q}}}(x)$, then $f(2x)=chi_{{bf{Q}}}(2x)=chi_{{bf{Q}}}(x)=f(x)$ but $lim_{xrightarrow 0}f(x)$ does not exist.






        share|cite|improve this answer









        $endgroup$



        $f(x)=chi_{{bf{Q}}}(x)$, then $f(2x)=chi_{{bf{Q}}}(2x)=chi_{{bf{Q}}}(x)=f(x)$ but $lim_{xrightarrow 0}f(x)$ does not exist.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '17 at 8:39









        user284331user284331

        35.4k31646




        35.4k31646












        • $begingroup$
          What is $chi_{{bf{Q}}}$?
          $endgroup$
          – Holo
          Dec 25 '17 at 11:12










        • $begingroup$
          The function takes the value $1$ for all rational numbers and $0$ otherwise.
          $endgroup$
          – user284331
          Dec 25 '17 at 18:06


















        • $begingroup$
          What is $chi_{{bf{Q}}}$?
          $endgroup$
          – Holo
          Dec 25 '17 at 11:12










        • $begingroup$
          The function takes the value $1$ for all rational numbers and $0$ otherwise.
          $endgroup$
          – user284331
          Dec 25 '17 at 18:06
















        $begingroup$
        What is $chi_{{bf{Q}}}$?
        $endgroup$
        – Holo
        Dec 25 '17 at 11:12




        $begingroup$
        What is $chi_{{bf{Q}}}$?
        $endgroup$
        – Holo
        Dec 25 '17 at 11:12












        $begingroup$
        The function takes the value $1$ for all rational numbers and $0$ otherwise.
        $endgroup$
        – user284331
        Dec 25 '17 at 18:06




        $begingroup$
        The function takes the value $1$ for all rational numbers and $0$ otherwise.
        $endgroup$
        – user284331
        Dec 25 '17 at 18:06











        1












        $begingroup$

        Take



        $f(x)=ln(x)$



        $f(2x)=ln(2x)=ln(2)+ln(x)$



        $g(x)=f(x)-f(2x)=-ln(2)$



        Here $lim_{xto0}f(x)$ is undefined but $f(x)-f(2x)$ is a constant






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Btw, if you consider $pminfty$ to be a limit then consider $f(x)=Im(ln(ix))=arg(ix)=arg(i2x)=Im(ln(i2x))=f(2x)implies f(x)-f(2x)=0forall x$ and $lim_{xto0^+}f(x)=pi/2,lim_{xto0^-}f(x)=-pi/2$
          $endgroup$
          – Holo
          Dec 26 '17 at 7:47


















        1












        $begingroup$

        Take



        $f(x)=ln(x)$



        $f(2x)=ln(2x)=ln(2)+ln(x)$



        $g(x)=f(x)-f(2x)=-ln(2)$



        Here $lim_{xto0}f(x)$ is undefined but $f(x)-f(2x)$ is a constant






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Btw, if you consider $pminfty$ to be a limit then consider $f(x)=Im(ln(ix))=arg(ix)=arg(i2x)=Im(ln(i2x))=f(2x)implies f(x)-f(2x)=0forall x$ and $lim_{xto0^+}f(x)=pi/2,lim_{xto0^-}f(x)=-pi/2$
          $endgroup$
          – Holo
          Dec 26 '17 at 7:47
















        1












        1








        1





        $begingroup$

        Take



        $f(x)=ln(x)$



        $f(2x)=ln(2x)=ln(2)+ln(x)$



        $g(x)=f(x)-f(2x)=-ln(2)$



        Here $lim_{xto0}f(x)$ is undefined but $f(x)-f(2x)$ is a constant






        share|cite|improve this answer











        $endgroup$



        Take



        $f(x)=ln(x)$



        $f(2x)=ln(2x)=ln(2)+ln(x)$



        $g(x)=f(x)-f(2x)=-ln(2)$



        Here $lim_{xto0}f(x)$ is undefined but $f(x)-f(2x)$ is a constant







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 25 '17 at 9:44

























        answered Dec 25 '17 at 9:36









        HoloHolo

        6,08421131




        6,08421131












        • $begingroup$
          Btw, if you consider $pminfty$ to be a limit then consider $f(x)=Im(ln(ix))=arg(ix)=arg(i2x)=Im(ln(i2x))=f(2x)implies f(x)-f(2x)=0forall x$ and $lim_{xto0^+}f(x)=pi/2,lim_{xto0^-}f(x)=-pi/2$
          $endgroup$
          – Holo
          Dec 26 '17 at 7:47




















        • $begingroup$
          Btw, if you consider $pminfty$ to be a limit then consider $f(x)=Im(ln(ix))=arg(ix)=arg(i2x)=Im(ln(i2x))=f(2x)implies f(x)-f(2x)=0forall x$ and $lim_{xto0^+}f(x)=pi/2,lim_{xto0^-}f(x)=-pi/2$
          $endgroup$
          – Holo
          Dec 26 '17 at 7:47


















        $begingroup$
        Btw, if you consider $pminfty$ to be a limit then consider $f(x)=Im(ln(ix))=arg(ix)=arg(i2x)=Im(ln(i2x))=f(2x)implies f(x)-f(2x)=0forall x$ and $lim_{xto0^+}f(x)=pi/2,lim_{xto0^-}f(x)=-pi/2$
        $endgroup$
        – Holo
        Dec 26 '17 at 7:47






        $begingroup$
        Btw, if you consider $pminfty$ to be a limit then consider $f(x)=Im(ln(ix))=arg(ix)=arg(i2x)=Im(ln(i2x))=f(2x)implies f(x)-f(2x)=0forall x$ and $lim_{xto0^+}f(x)=pi/2,lim_{xto0^-}f(x)=-pi/2$
        $endgroup$
        – Holo
        Dec 26 '17 at 7:47













        0












        $begingroup$

        Hint. Take a function such that $$f(x)-f(2x)=fleft(frac{x}{2x} right)=fleft(frac 1 2 right)$$ You already know one of such functions I hope.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          Hint. Take a function such that $$f(x)-f(2x)=fleft(frac{x}{2x} right)=fleft(frac 1 2 right)$$ You already know one of such functions I hope.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            Hint. Take a function such that $$f(x)-f(2x)=fleft(frac{x}{2x} right)=fleft(frac 1 2 right)$$ You already know one of such functions I hope.






            share|cite|improve this answer









            $endgroup$



            Hint. Take a function such that $$f(x)-f(2x)=fleft(frac{x}{2x} right)=fleft(frac 1 2 right)$$ You already know one of such functions I hope.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 25 '17 at 8:39









            ShashiShashi

            7,3151628




            7,3151628























                0












                $begingroup$

                Any function such that $f(x)=f(2x)$ will work. Also if they differ in a constant, or if they differ in a function which is continuous at $x=0$ (but of course, that's almost rephrasing your question).






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Any function such that $f(x)=f(2x)$ will work. Also if they differ in a constant, or if they differ in a function which is continuous at $x=0$ (but of course, that's almost rephrasing your question).






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Any function such that $f(x)=f(2x)$ will work. Also if they differ in a constant, or if they differ in a function which is continuous at $x=0$ (but of course, that's almost rephrasing your question).






                    share|cite|improve this answer









                    $endgroup$



                    Any function such that $f(x)=f(2x)$ will work. Also if they differ in a constant, or if they differ in a function which is continuous at $x=0$ (but of course, that's almost rephrasing your question).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 25 '17 at 8:42









                    Alejandro Nasif SalumAlejandro Nasif Salum

                    4,765118




                    4,765118























                        0












                        $begingroup$

                        Here's an example that is rather continuous.



                        Take $f(x) = sinleft(2pifrac{ln |x|}{ln 2}right)$ defined for $x neq 0$ (you can set $f(0)=0$ if you want $f$ to be definied for all $x in mathbb R$). Then
                        $$f(2x) = sinleft(2pifrac{ln |2x|}{ln 2}right) = sinleft(2pifrac{ln2 + ln |x|}{ln 2}right) \= sinleft(2pi+2pifrac{ln |x|}{ln 2}right) = sinleft(2pifrac{ln |x|}{ln 2}right) \= f(x)$$
                        so
                        $$lim_{xto 0}(f(x)-f(2x)) = 0.$$
                        But $lim_{xto 0} f(x)$ is not defined since $f$ oscillates quickly as $x to 0$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Here's an example that is rather continuous.



                          Take $f(x) = sinleft(2pifrac{ln |x|}{ln 2}right)$ defined for $x neq 0$ (you can set $f(0)=0$ if you want $f$ to be definied for all $x in mathbb R$). Then
                          $$f(2x) = sinleft(2pifrac{ln |2x|}{ln 2}right) = sinleft(2pifrac{ln2 + ln |x|}{ln 2}right) \= sinleft(2pi+2pifrac{ln |x|}{ln 2}right) = sinleft(2pifrac{ln |x|}{ln 2}right) \= f(x)$$
                          so
                          $$lim_{xto 0}(f(x)-f(2x)) = 0.$$
                          But $lim_{xto 0} f(x)$ is not defined since $f$ oscillates quickly as $x to 0$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Here's an example that is rather continuous.



                            Take $f(x) = sinleft(2pifrac{ln |x|}{ln 2}right)$ defined for $x neq 0$ (you can set $f(0)=0$ if you want $f$ to be definied for all $x in mathbb R$). Then
                            $$f(2x) = sinleft(2pifrac{ln |2x|}{ln 2}right) = sinleft(2pifrac{ln2 + ln |x|}{ln 2}right) \= sinleft(2pi+2pifrac{ln |x|}{ln 2}right) = sinleft(2pifrac{ln |x|}{ln 2}right) \= f(x)$$
                            so
                            $$lim_{xto 0}(f(x)-f(2x)) = 0.$$
                            But $lim_{xto 0} f(x)$ is not defined since $f$ oscillates quickly as $x to 0$.






                            share|cite|improve this answer









                            $endgroup$



                            Here's an example that is rather continuous.



                            Take $f(x) = sinleft(2pifrac{ln |x|}{ln 2}right)$ defined for $x neq 0$ (you can set $f(0)=0$ if you want $f$ to be definied for all $x in mathbb R$). Then
                            $$f(2x) = sinleft(2pifrac{ln |2x|}{ln 2}right) = sinleft(2pifrac{ln2 + ln |x|}{ln 2}right) \= sinleft(2pi+2pifrac{ln |x|}{ln 2}right) = sinleft(2pifrac{ln |x|}{ln 2}right) \= f(x)$$
                            so
                            $$lim_{xto 0}(f(x)-f(2x)) = 0.$$
                            But $lim_{xto 0} f(x)$ is not defined since $f$ oscillates quickly as $x to 0$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 25 '17 at 9:25









                            md2perpemd2perpe

                            8,30911028




                            8,30911028






























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                                Verónica Boquete

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                                Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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                                Ida-Boy-Ed-Garten

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                                Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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