Calculate $limlimits_{ xto + infty}xcdot sin(sqrt{x^{2}+3}-sqrt{x^{2}+2})$












3












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I know that $$limlimits_{ xto + infty}xcdot sin(sqrt{x^{2}+3}-sqrt{x^{2}+2})\=limlimits_{ xto + infty}xcdot sinleft(frac{1}{sqrt{x^{2}+3}+sqrt{x^{2}+2}}right).$$ If $x rightarrow + infty$, then $sinleft(frac{1}{sqrt{x^{2}+3}+sqrt{x^{2}+2}}right)rightarrow sin0 $. However I have also $x$ before $sin x$ and I don't know how to calculate it.






real-analysis limits







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    3












    $begingroup$


    I know that $$limlimits_{ xto + infty}xcdot sin(sqrt{x^{2}+3}-sqrt{x^{2}+2})\=limlimits_{ xto + infty}xcdot sinleft(frac{1}{sqrt{x^{2}+3}+sqrt{x^{2}+2}}right).$$ If $x rightarrow + infty$, then $sinleft(frac{1}{sqrt{x^{2}+3}+sqrt{x^{2}+2}}right)rightarrow sin0 $. However I have also $x$ before $sin x$ and I don't know how to calculate it.






    real-analysis limits







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      0



      $begingroup$


      I know that $$limlimits_{ xto + infty}xcdot sin(sqrt{x^{2}+3}-sqrt{x^{2}+2})\=limlimits_{ xto + infty}xcdot sinleft(frac{1}{sqrt{x^{2}+3}+sqrt{x^{2}+2}}right).$$ If $x rightarrow + infty$, then $sinleft(frac{1}{sqrt{x^{2}+3}+sqrt{x^{2}+2}}right)rightarrow sin0 $. However I have also $x$ before $sin x$ and I don't know how to calculate it.






      real-analysis limits







      share|cite|improve this question











      $endgroup$




      I know that $$limlimits_{ xto + infty}xcdot sin(sqrt{x^{2}+3}-sqrt{x^{2}+2})\=limlimits_{ xto + infty}xcdot sinleft(frac{1}{sqrt{x^{2}+3}+sqrt{x^{2}+2}}right).$$ If $x rightarrow + infty$, then $sinleft(frac{1}{sqrt{x^{2}+3}+sqrt{x^{2}+2}}right)rightarrow sin0 $. However I have also $x$ before $sin x$ and I don't know how to calculate it.






      real-analysis limits




      real-analysis limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 31 '18 at 13:52









      user376343

      3,3582826




      3,3582826










      asked Dec 31 '18 at 13:23









      MP3129MP3129

      1937




      1937






















          3 Answers
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          4












          $begingroup$

          Letting $h=frac1x$:



          $$begin{array}{cl}
          &displaystyle lim_{x to infty} x sin left( sqrt{x^2+3} - sqrt{x^2+2} right) \
          =&displaystyle lim_{x to infty} x sin left( frac 1 {sqrt{x^2+3} + sqrt{x^2+2} } right) \
          =&displaystyle lim_{h to 0^+} frac1h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
          =&displaystyle lim_{h to 0^+} frac 1 {sqrt{1+3h^2} + sqrt{1+2h^2}} frac {sqrt{1+3h^2} + sqrt{1+2h^2}} h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
          =&displaystyle frac12 times lim_{h to 0^+} frac {sqrt{1+3h^2} + sqrt{1+2h^2}} h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
          =&displaystyle frac12 times 1 \
          =&dfrac12
          end{array}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Actually the searched limit is equal to $$frac{1}{2}$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Dec 31 '18 at 13:36










          • $begingroup$
            Fixed.${ }$${ }$
            $endgroup$
            – Kenny Lau
            Dec 31 '18 at 13:37



















          4












          $begingroup$

          begin{align}
          lim_{x to infty} x cdot left( sin left( sqrt{x^2+3}-sqrt{x^2+2}right)right)&=lim_{x to infty} x cdot sin left( frac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)\
          &=lim_{x to infty} frac{x}{sqrt{x^2+3}+sqrt{x^2+2}}\
          &=lim_{x to infty} frac{1}{sqrt{1+frac{3}{x^2}}+sqrt{1+frac{2}{x^2}}}\
          &= frac12
          end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You use the assumption that d.d.d. x $sinx = x$, I understand correctly?
            $endgroup$
            – MP3129
            Dec 31 '18 at 14:02






          • 2




            $begingroup$
            not sure what do those $d$ mean.The main trick is $lim_{h to 0^+} frac{sin h}h=1$, hence that is why I can remove the sine.
            $endgroup$
            – Siong Thye Goh
            Dec 31 '18 at 14:07



















          1












          $begingroup$

          The ordo calculus is very useful for such problems. Sometimes you do not need the full power of Taylor series, only the first couple of coefficients.



          $sqrt{x^2+3}- sqrt{x^2+3} = x(sqrt{1+3/x^2}- sqrt{1+2/x^2})=$



          $= x(1+frac{1}{2}cdot(3/x^2)+O(1/x^4)- 1-frac{1}{2}cdot(2/x^2)+O(1/x^4))= frac{1}{2x}+O(1/x^3)$.



          Now $sin(frac{1}{2x}+O(1/x^3))= frac{1}{2x}+O(1/x^2)$ as $xrightarrow infty$, thus the expression is $frac{1}{2}+ O(frac{1}{x})$.






          share|cite|improve this answer











          $endgroup$













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            3 Answers
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            active

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            3 Answers
            3






            active

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            active

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            votes






            active

            oldest

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            4












            $begingroup$

            Letting $h=frac1x$:



            $$begin{array}{cl}
            &displaystyle lim_{x to infty} x sin left( sqrt{x^2+3} - sqrt{x^2+2} right) \
            =&displaystyle lim_{x to infty} x sin left( frac 1 {sqrt{x^2+3} + sqrt{x^2+2} } right) \
            =&displaystyle lim_{h to 0^+} frac1h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle lim_{h to 0^+} frac 1 {sqrt{1+3h^2} + sqrt{1+2h^2}} frac {sqrt{1+3h^2} + sqrt{1+2h^2}} h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle frac12 times lim_{h to 0^+} frac {sqrt{1+3h^2} + sqrt{1+2h^2}} h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle frac12 times 1 \
            =&dfrac12
            end{array}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Actually the searched limit is equal to $$frac{1}{2}$$
              $endgroup$
              – Dr. Sonnhard Graubner
              Dec 31 '18 at 13:36










            • $begingroup$
              Fixed.${ }$${ }$
              $endgroup$
              – Kenny Lau
              Dec 31 '18 at 13:37
















            4












            $begingroup$

            Letting $h=frac1x$:



            $$begin{array}{cl}
            &displaystyle lim_{x to infty} x sin left( sqrt{x^2+3} - sqrt{x^2+2} right) \
            =&displaystyle lim_{x to infty} x sin left( frac 1 {sqrt{x^2+3} + sqrt{x^2+2} } right) \
            =&displaystyle lim_{h to 0^+} frac1h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle lim_{h to 0^+} frac 1 {sqrt{1+3h^2} + sqrt{1+2h^2}} frac {sqrt{1+3h^2} + sqrt{1+2h^2}} h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle frac12 times lim_{h to 0^+} frac {sqrt{1+3h^2} + sqrt{1+2h^2}} h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle frac12 times 1 \
            =&dfrac12
            end{array}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Actually the searched limit is equal to $$frac{1}{2}$$
              $endgroup$
              – Dr. Sonnhard Graubner
              Dec 31 '18 at 13:36










            • $begingroup$
              Fixed.${ }$${ }$
              $endgroup$
              – Kenny Lau
              Dec 31 '18 at 13:37














            4












            4








            4





            $begingroup$

            Letting $h=frac1x$:



            $$begin{array}{cl}
            &displaystyle lim_{x to infty} x sin left( sqrt{x^2+3} - sqrt{x^2+2} right) \
            =&displaystyle lim_{x to infty} x sin left( frac 1 {sqrt{x^2+3} + sqrt{x^2+2} } right) \
            =&displaystyle lim_{h to 0^+} frac1h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle lim_{h to 0^+} frac 1 {sqrt{1+3h^2} + sqrt{1+2h^2}} frac {sqrt{1+3h^2} + sqrt{1+2h^2}} h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle frac12 times lim_{h to 0^+} frac {sqrt{1+3h^2} + sqrt{1+2h^2}} h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle frac12 times 1 \
            =&dfrac12
            end{array}$$






            share|cite|improve this answer











            $endgroup$



            Letting $h=frac1x$:



            $$begin{array}{cl}
            &displaystyle lim_{x to infty} x sin left( sqrt{x^2+3} - sqrt{x^2+2} right) \
            =&displaystyle lim_{x to infty} x sin left( frac 1 {sqrt{x^2+3} + sqrt{x^2+2} } right) \
            =&displaystyle lim_{h to 0^+} frac1h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle lim_{h to 0^+} frac 1 {sqrt{1+3h^2} + sqrt{1+2h^2}} frac {sqrt{1+3h^2} + sqrt{1+2h^2}} h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle frac12 times lim_{h to 0^+} frac {sqrt{1+3h^2} + sqrt{1+2h^2}} h sin left( frac h {sqrt{1+3h^2} + sqrt{1+2h^2} } right) \
            =&displaystyle frac12 times 1 \
            =&dfrac12
            end{array}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 31 '18 at 13:37

























            answered Dec 31 '18 at 13:30









            Kenny LauKenny Lau

            19.9k2159




            19.9k2159












            • $begingroup$
              Actually the searched limit is equal to $$frac{1}{2}$$
              $endgroup$
              – Dr. Sonnhard Graubner
              Dec 31 '18 at 13:36










            • $begingroup$
              Fixed.${ }$${ }$
              $endgroup$
              – Kenny Lau
              Dec 31 '18 at 13:37


















            • $begingroup$
              Actually the searched limit is equal to $$frac{1}{2}$$
              $endgroup$
              – Dr. Sonnhard Graubner
              Dec 31 '18 at 13:36










            • $begingroup$
              Fixed.${ }$${ }$
              $endgroup$
              – Kenny Lau
              Dec 31 '18 at 13:37
















            $begingroup$
            Actually the searched limit is equal to $$frac{1}{2}$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Dec 31 '18 at 13:36




            $begingroup$
            Actually the searched limit is equal to $$frac{1}{2}$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Dec 31 '18 at 13:36












            $begingroup$
            Fixed.${ }$${ }$
            $endgroup$
            – Kenny Lau
            Dec 31 '18 at 13:37




            $begingroup$
            Fixed.${ }$${ }$
            $endgroup$
            – Kenny Lau
            Dec 31 '18 at 13:37











            4












            $begingroup$

            begin{align}
            lim_{x to infty} x cdot left( sin left( sqrt{x^2+3}-sqrt{x^2+2}right)right)&=lim_{x to infty} x cdot sin left( frac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)\
            &=lim_{x to infty} frac{x}{sqrt{x^2+3}+sqrt{x^2+2}}\
            &=lim_{x to infty} frac{1}{sqrt{1+frac{3}{x^2}}+sqrt{1+frac{2}{x^2}}}\
            &= frac12
            end{align}






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You use the assumption that d.d.d. x $sinx = x$, I understand correctly?
              $endgroup$
              – MP3129
              Dec 31 '18 at 14:02






            • 2




              $begingroup$
              not sure what do those $d$ mean.The main trick is $lim_{h to 0^+} frac{sin h}h=1$, hence that is why I can remove the sine.
              $endgroup$
              – Siong Thye Goh
              Dec 31 '18 at 14:07
















            4












            $begingroup$

            begin{align}
            lim_{x to infty} x cdot left( sin left( sqrt{x^2+3}-sqrt{x^2+2}right)right)&=lim_{x to infty} x cdot sin left( frac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)\
            &=lim_{x to infty} frac{x}{sqrt{x^2+3}+sqrt{x^2+2}}\
            &=lim_{x to infty} frac{1}{sqrt{1+frac{3}{x^2}}+sqrt{1+frac{2}{x^2}}}\
            &= frac12
            end{align}






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You use the assumption that d.d.d. x $sinx = x$, I understand correctly?
              $endgroup$
              – MP3129
              Dec 31 '18 at 14:02






            • 2




              $begingroup$
              not sure what do those $d$ mean.The main trick is $lim_{h to 0^+} frac{sin h}h=1$, hence that is why I can remove the sine.
              $endgroup$
              – Siong Thye Goh
              Dec 31 '18 at 14:07














            4












            4








            4





            $begingroup$

            begin{align}
            lim_{x to infty} x cdot left( sin left( sqrt{x^2+3}-sqrt{x^2+2}right)right)&=lim_{x to infty} x cdot sin left( frac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)\
            &=lim_{x to infty} frac{x}{sqrt{x^2+3}+sqrt{x^2+2}}\
            &=lim_{x to infty} frac{1}{sqrt{1+frac{3}{x^2}}+sqrt{1+frac{2}{x^2}}}\
            &= frac12
            end{align}






            share|cite|improve this answer









            $endgroup$



            begin{align}
            lim_{x to infty} x cdot left( sin left( sqrt{x^2+3}-sqrt{x^2+2}right)right)&=lim_{x to infty} x cdot sin left( frac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)\
            &=lim_{x to infty} frac{x}{sqrt{x^2+3}+sqrt{x^2+2}}\
            &=lim_{x to infty} frac{1}{sqrt{1+frac{3}{x^2}}+sqrt{1+frac{2}{x^2}}}\
            &= frac12
            end{align}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 31 '18 at 13:37









            Siong Thye GohSiong Thye Goh

            100k1465117




            100k1465117












            • $begingroup$
              You use the assumption that d.d.d. x $sinx = x$, I understand correctly?
              $endgroup$
              – MP3129
              Dec 31 '18 at 14:02






            • 2




              $begingroup$
              not sure what do those $d$ mean.The main trick is $lim_{h to 0^+} frac{sin h}h=1$, hence that is why I can remove the sine.
              $endgroup$
              – Siong Thye Goh
              Dec 31 '18 at 14:07


















            • $begingroup$
              You use the assumption that d.d.d. x $sinx = x$, I understand correctly?
              $endgroup$
              – MP3129
              Dec 31 '18 at 14:02






            • 2




              $begingroup$
              not sure what do those $d$ mean.The main trick is $lim_{h to 0^+} frac{sin h}h=1$, hence that is why I can remove the sine.
              $endgroup$
              – Siong Thye Goh
              Dec 31 '18 at 14:07
















            $begingroup$
            You use the assumption that d.d.d. x $sinx = x$, I understand correctly?
            $endgroup$
            – MP3129
            Dec 31 '18 at 14:02




            $begingroup$
            You use the assumption that d.d.d. x $sinx = x$, I understand correctly?
            $endgroup$
            – MP3129
            Dec 31 '18 at 14:02




            2




            2




            $begingroup$
            not sure what do those $d$ mean.The main trick is $lim_{h to 0^+} frac{sin h}h=1$, hence that is why I can remove the sine.
            $endgroup$
            – Siong Thye Goh
            Dec 31 '18 at 14:07




            $begingroup$
            not sure what do those $d$ mean.The main trick is $lim_{h to 0^+} frac{sin h}h=1$, hence that is why I can remove the sine.
            $endgroup$
            – Siong Thye Goh
            Dec 31 '18 at 14:07











            1












            $begingroup$

            The ordo calculus is very useful for such problems. Sometimes you do not need the full power of Taylor series, only the first couple of coefficients.



            $sqrt{x^2+3}- sqrt{x^2+3} = x(sqrt{1+3/x^2}- sqrt{1+2/x^2})=$



            $= x(1+frac{1}{2}cdot(3/x^2)+O(1/x^4)- 1-frac{1}{2}cdot(2/x^2)+O(1/x^4))= frac{1}{2x}+O(1/x^3)$.



            Now $sin(frac{1}{2x}+O(1/x^3))= frac{1}{2x}+O(1/x^2)$ as $xrightarrow infty$, thus the expression is $frac{1}{2}+ O(frac{1}{x})$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              The ordo calculus is very useful for such problems. Sometimes you do not need the full power of Taylor series, only the first couple of coefficients.



              $sqrt{x^2+3}- sqrt{x^2+3} = x(sqrt{1+3/x^2}- sqrt{1+2/x^2})=$



              $= x(1+frac{1}{2}cdot(3/x^2)+O(1/x^4)- 1-frac{1}{2}cdot(2/x^2)+O(1/x^4))= frac{1}{2x}+O(1/x^3)$.



              Now $sin(frac{1}{2x}+O(1/x^3))= frac{1}{2x}+O(1/x^2)$ as $xrightarrow infty$, thus the expression is $frac{1}{2}+ O(frac{1}{x})$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                The ordo calculus is very useful for such problems. Sometimes you do not need the full power of Taylor series, only the first couple of coefficients.



                $sqrt{x^2+3}- sqrt{x^2+3} = x(sqrt{1+3/x^2}- sqrt{1+2/x^2})=$



                $= x(1+frac{1}{2}cdot(3/x^2)+O(1/x^4)- 1-frac{1}{2}cdot(2/x^2)+O(1/x^4))= frac{1}{2x}+O(1/x^3)$.



                Now $sin(frac{1}{2x}+O(1/x^3))= frac{1}{2x}+O(1/x^2)$ as $xrightarrow infty$, thus the expression is $frac{1}{2}+ O(frac{1}{x})$.






                share|cite|improve this answer











                $endgroup$



                The ordo calculus is very useful for such problems. Sometimes you do not need the full power of Taylor series, only the first couple of coefficients.



                $sqrt{x^2+3}- sqrt{x^2+3} = x(sqrt{1+3/x^2}- sqrt{1+2/x^2})=$



                $= x(1+frac{1}{2}cdot(3/x^2)+O(1/x^4)- 1-frac{1}{2}cdot(2/x^2)+O(1/x^4))= frac{1}{2x}+O(1/x^3)$.



                Now $sin(frac{1}{2x}+O(1/x^3))= frac{1}{2x}+O(1/x^2)$ as $xrightarrow infty$, thus the expression is $frac{1}{2}+ O(frac{1}{x})$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 31 '18 at 13:52

























                answered Dec 31 '18 at 13:40









                A. PongráczA. Pongrácz

                5,9831929




                5,9831929






























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                    Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
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                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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                    0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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                    張江高科駅

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                    張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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