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

      636110




      636110






















          4 Answers
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          $begingroup$

          Letting $h=frac1x$:



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






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          • $begingroup$
            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



















          5












          $begingroup$

          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











          $endgroup$





















            0












            $begingroup$

            begin{align}
            x sin left( sqrt{x^2+3}-sqrt{x^2+2}right)
            &=x sin left( frac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)\
            &=left(dfrac{x}{sqrt{x^2+3}+sqrt{x^2+2}}right)
            dfrac{sin left( dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}
            {left(dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}\
            &=left(dfrac{1}{sqrt{1+frac{3}{x^2}}+sqrt{1+frac{2}{x^2}}}right)
            dfrac{sin left( dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}
            {left(dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}\
            &to dfrac 12 cdot 1 text{as $x to infty$}\
            &to dfrac 12 text{as $x to infty$}
            end{align}






            share|cite|improve this answer









            $endgroup$













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              4 Answers
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              4 Answers
              4






              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











              5












              $begingroup$

              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
















              5












              $begingroup$

              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














              5












              5








              5





              $begingroup$

              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

                  6,0231929




                  6,0231929























                      0












                      $begingroup$

                      begin{align}
                      x sin left( sqrt{x^2+3}-sqrt{x^2+2}right)
                      &=x sin left( frac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)\
                      &=left(dfrac{x}{sqrt{x^2+3}+sqrt{x^2+2}}right)
                      dfrac{sin left( dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}
                      {left(dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}\
                      &=left(dfrac{1}{sqrt{1+frac{3}{x^2}}+sqrt{1+frac{2}{x^2}}}right)
                      dfrac{sin left( dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}
                      {left(dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}\
                      &to dfrac 12 cdot 1 text{as $x to infty$}\
                      &to dfrac 12 text{as $x to infty$}
                      end{align}






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        begin{align}
                        x sin left( sqrt{x^2+3}-sqrt{x^2+2}right)
                        &=x sin left( frac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)\
                        &=left(dfrac{x}{sqrt{x^2+3}+sqrt{x^2+2}}right)
                        dfrac{sin left( dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}
                        {left(dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}\
                        &=left(dfrac{1}{sqrt{1+frac{3}{x^2}}+sqrt{1+frac{2}{x^2}}}right)
                        dfrac{sin left( dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}
                        {left(dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}\
                        &to dfrac 12 cdot 1 text{as $x to infty$}\
                        &to dfrac 12 text{as $x to infty$}
                        end{align}






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          begin{align}
                          x sin left( sqrt{x^2+3}-sqrt{x^2+2}right)
                          &=x sin left( frac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)\
                          &=left(dfrac{x}{sqrt{x^2+3}+sqrt{x^2+2}}right)
                          dfrac{sin left( dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}
                          {left(dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}\
                          &=left(dfrac{1}{sqrt{1+frac{3}{x^2}}+sqrt{1+frac{2}{x^2}}}right)
                          dfrac{sin left( dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}
                          {left(dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}\
                          &to dfrac 12 cdot 1 text{as $x to infty$}\
                          &to dfrac 12 text{as $x to infty$}
                          end{align}






                          share|cite|improve this answer









                          $endgroup$



                          begin{align}
                          x sin left( sqrt{x^2+3}-sqrt{x^2+2}right)
                          &=x sin left( frac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)\
                          &=left(dfrac{x}{sqrt{x^2+3}+sqrt{x^2+2}}right)
                          dfrac{sin left( dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}
                          {left(dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}\
                          &=left(dfrac{1}{sqrt{1+frac{3}{x^2}}+sqrt{1+frac{2}{x^2}}}right)
                          dfrac{sin left( dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}
                          {left(dfrac{1}{sqrt{x^2+3}+sqrt{x^2+2}}right)}\
                          &to dfrac 12 cdot 1 text{as $x to infty$}\
                          &to dfrac 12 text{as $x to infty$}
                          end{align}







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 12 hours ago









                          steven gregorysteven gregory

                          18.3k32358




                          18.3k32358






























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