How to calculate the two limits?












3












$begingroup$



I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?






limits







share|cite











$endgroup$












  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago
















3












$begingroup$



I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?






limits







share|cite











$endgroup$












  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago














3












3








3





$begingroup$



I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?






limits







share|cite











$endgroup$





I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?






limits




limits






share|cite















share|cite













share|cite




share|cite








edited 1 hour ago







lanse7pty

















asked 2 hours ago









lanse7ptylanse7pty

1,8411823




1,8411823












  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago


















  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago
















$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago




$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago










4 Answers
4






active

oldest

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2












$begingroup$

Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Without L'Hospital
    $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



    Now, by Taylor for large values of $x$
    $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
    $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
    $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
    $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        You can solve the first one using




        • $arctan x + operatorname{arccot}x = frac{pi}{2}$

        • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

        • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


        begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
        & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
        & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
        end{eqnarray*}



        The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





        • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






        share|cite|improve this answer









        $endgroup$














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          4 Answers
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          4 Answers
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          active

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          2












          $begingroup$

          Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






          share|cite|improve this answer











          $endgroup$


















            2












            $begingroup$

            Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






            share|cite|improve this answer











            $endgroup$
















              2












              2








              2





              $begingroup$

              Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






              share|cite|improve this answer











              $endgroup$



              Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 1 hour ago

























              answered 2 hours ago









              Paras KhoslaParas Khosla

              2,736423




              2,736423























                  1












                  $begingroup$

                  Without L'Hospital
                  $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                  Now, by Taylor for large values of $x$
                  $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                  $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                  $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                  $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Without L'Hospital
                    $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                    Now, by Taylor for large values of $x$
                    $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                    $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                    $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                    $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Without L'Hospital
                      $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                      Now, by Taylor for large values of $x$
                      $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                      $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                      $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                      $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






                      share|cite|improve this answer









                      $endgroup$



                      Without L'Hospital
                      $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                      Now, by Taylor for large values of $x$
                      $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                      $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                      $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                      $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 1 hour ago









                      Claude LeiboviciClaude Leibovici

                      125k1158136




                      125k1158136























                          0












                          $begingroup$

                          I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






                              share|cite|improve this answer









                              $endgroup$



                              I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 2 hours ago









                              AdmuthAdmuth

                              285




                              285























                                  0












                                  $begingroup$

                                  You can solve the first one using




                                  • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                  • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                  • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                  begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                  & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                  & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                  end{eqnarray*}



                                  The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                  • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    You can solve the first one using




                                    • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                    • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                    • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                    begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                    & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                    & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                    end{eqnarray*}



                                    The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                    • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      You can solve the first one using




                                      • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                      • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                      • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                      begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                      & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                      & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                      end{eqnarray*}



                                      The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                      • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






                                      share|cite|improve this answer









                                      $endgroup$



                                      You can solve the first one using




                                      • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                      • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                      • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                      begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                      & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                      & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                      end{eqnarray*}



                                      The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                      • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 10 mins ago









                                      trancelocationtrancelocation

                                      13.4k1827




                                      13.4k1827






























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                                          Bundesstraße 106

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                                          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
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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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