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

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      3,9584829










      asked Dec 31 '18 at 13:23









      MP3129MP3129

      607110




      607110






















          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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            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

            20k2160




            20k2160












            • $begingroup$
              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

            103k1468119




            103k1468119












            • $begingroup$
              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

                5,9981929




                5,9981929






























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                    Wiesbaden

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                    Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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                    Marschland

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                    Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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                    Dieringhausen

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                    Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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