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












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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

      3,9584829




      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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                    Redirect URL with Chrome Remote Debugging Android Devices

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                    1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
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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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