Why isn't L Hospital Rule applicable here?












2














$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus







share|cite|improve this question






















  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    4 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    4 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    3 hours ago










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    12 mins ago
















2














$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus







share|cite|improve this question






















  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    4 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    4 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    3 hours ago










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    12 mins ago














2












2








2







$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus







share|cite|improve this question













$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









user1752323

324




324












  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    4 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    4 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    3 hours ago










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    12 mins ago


















  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    4 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    4 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    3 hours ago










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    12 mins ago
















Why not? I think it is allright
– José Alejandro Aburto Araneda
4 hours ago




Why not? I think it is allright
– José Alejandro Aburto Araneda
4 hours ago




6




6




The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
4 hours ago






The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
4 hours ago














You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
3 hours ago




You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
3 hours ago












@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
12 mins ago




@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
12 mins ago










3 Answers
3






active

oldest

votes


















2














You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



More generally you need $infty/infty$ or $0/0$ on either side.



If you blindly apply l’Hôpital, you end up with
$$
lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
$$
but this would be incorrect, because
$$
lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
$$
For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
$$
lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
frac{1}{1+0}=1
$$






share|cite|improve this answer































    3














    L'Hopital's rule does not apply because the limit
    $$
    lim_{xto 0}e^{1/x}
    $$
    does not exist, since
    $$
    lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
    $$
    So, consequently
    $$
    lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
    $$






    share|cite|improve this answer



















    • 1




      This is not the reason why l'Hôpital cannot be applied.
      – egreg
      3 hours ago



















    0














    here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
    So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
    $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
    So,
    $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



    so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






    share|cite|improve this answer





















      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%2f3058940%2fwhy-isnt-l-hospital-rule-applicable-here%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



      The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



      More generally you need $infty/infty$ or $0/0$ on either side.



      If you blindly apply l’Hôpital, you end up with
      $$
      lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
      $$
      but this would be incorrect, because
      $$
      lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
      $$
      For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
      $$
      lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
      frac{1}{1+0}=1
      $$






      share|cite|improve this answer




























        2














        You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



        The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



        More generally you need $infty/infty$ or $0/0$ on either side.



        If you blindly apply l’Hôpital, you end up with
        $$
        lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
        $$
        but this would be incorrect, because
        $$
        lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
        $$
        For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
        $$
        lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
        frac{1}{1+0}=1
        $$






        share|cite|improve this answer


























          2












          2








          2






          You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



          The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



          More generally you need $infty/infty$ or $0/0$ on either side.



          If you blindly apply l’Hôpital, you end up with
          $$
          lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
          $$
          but this would be incorrect, because
          $$
          lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
          $$
          For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
          $$
          lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
          frac{1}{1+0}=1
          $$






          share|cite|improve this answer














          You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



          The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



          More generally you need $infty/infty$ or $0/0$ on either side.



          If you blindly apply l’Hôpital, you end up with
          $$
          lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
          $$
          but this would be incorrect, because
          $$
          lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
          $$
          For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
          $$
          lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
          frac{1}{1+0}=1
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 3 hours ago









          egreg

          178k1484201




          178k1484201























              3














              L'Hopital's rule does not apply because the limit
              $$
              lim_{xto 0}e^{1/x}
              $$
              does not exist, since
              $$
              lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
              $$
              So, consequently
              $$
              lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
              $$






              share|cite|improve this answer



















              • 1




                This is not the reason why l'Hôpital cannot be applied.
                – egreg
                3 hours ago
















              3














              L'Hopital's rule does not apply because the limit
              $$
              lim_{xto 0}e^{1/x}
              $$
              does not exist, since
              $$
              lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
              $$
              So, consequently
              $$
              lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
              $$






              share|cite|improve this answer



















              • 1




                This is not the reason why l'Hôpital cannot be applied.
                – egreg
                3 hours ago














              3












              3








              3






              L'Hopital's rule does not apply because the limit
              $$
              lim_{xto 0}e^{1/x}
              $$
              does not exist, since
              $$
              lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
              $$
              So, consequently
              $$
              lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
              $$






              share|cite|improve this answer














              L'Hopital's rule does not apply because the limit
              $$
              lim_{xto 0}e^{1/x}
              $$
              does not exist, since
              $$
              lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
              $$
              So, consequently
              $$
              lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
              $$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 3 hours ago









              Szeto

              6,4362926




              6,4362926










              answered 4 hours ago









              Yiorgos S. Smyrlis

              62.6k1383163




              62.6k1383163








              • 1




                This is not the reason why l'Hôpital cannot be applied.
                – egreg
                3 hours ago














              • 1




                This is not the reason why l'Hôpital cannot be applied.
                – egreg
                3 hours ago








              1




              1




              This is not the reason why l'Hôpital cannot be applied.
              – egreg
              3 hours ago




              This is not the reason why l'Hôpital cannot be applied.
              – egreg
              3 hours ago











              0














              here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
              So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
              $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
              So,
              $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



              so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






              share|cite|improve this answer


























                0














                here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
                So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
                $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
                So,
                $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



                so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






                share|cite|improve this answer
























                  0












                  0








                  0






                  here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
                  So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
                  $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
                  So,
                  $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



                  so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






                  share|cite|improve this answer












                  here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
                  So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
                  $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
                  So,
                  $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



                  so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  Rakibul Islam Prince

                  952211




                  952211






























                      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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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.


                      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%2f3058940%2fwhy-isnt-l-hospital-rule-applicable-here%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

                      Bundesstraße 106

                      Image
                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                      Read more

                      Verónica Boquete

                      Image
                      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
                      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