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







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












$begingroup$


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









      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2579482%2ffunction-satisfying-lim-limits-x-to-0fx-f2x-but-doesnt-have-lim-li%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        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






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2579482%2ffunction-satisfying-lim-limits-x-to-0fx-f2x-but-doesnt-have-lim-li%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Le Mesnil-Réaume

                                Image
                                Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
                                Read more

                                Ida-Boy-Ed-Garten

                                Image
                                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
                                Read more

                                web3.py web3.isConnected() returns false always

                                Image
                                1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
                                Read more