solving $limlimits_{xrightarrowinfty} frac{(x^2-1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}}$












4












$begingroup$


To investigate the convergence of a series I have to solve the folliwing limit:



begin{equation}
limlimits_{xrightarrowinfty} frac{(x^2-1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}}
end{equation}



It should be $frac{1}{2}$ but i can't quite seem to get to that solution. I've tried to factor the square root out of the quotient which was:



begin{equation}
limlimits_{xrightarrowinfty} sqrt{frac{((x^2-1) sqrt{x + 2}-x^2sqrt{x+1})^2}{x^2(x + 1)}}
end{equation}



Then i worked out the square in the numerator which was:



begin{equation}
(x^2-1)^2 (x + 2)-2x^2(x^2-1)sqrt{x+1}sqrt{x+2}+x^4(x+1)
end{equation}



I could then factor the terms and take and divide the numerator with the (x+1) from the denominator. Then i could expand the terms in the numerator wich became:



begin{equation}
limlimits_{xrightarrowinfty}sqrt{frac{(x^4+x^3-3x^2-x+2)-sqrt{16x^8+32x^7-12x^6-20x^5+8x^4}+x^4}{x^2}}
end{equation}



Now i can take the $16x^8$ out of the root and then i then looked at the terms with the highest exponent so i had:



begin{equation}
limlimits_{xrightarrowinfty}sqrt{frac{x^4-4x^4+x^4}{x^2}} =limlimits_{xrightarrowinfty}sqrt{frac{-2x^2}{x^2}}
end{equation}



Which could only be solved with complex numbers, so i should be wrong somewhere in my calculations, since i know that i should get $frac{1}{2}$. I also checked my sollution with WolframAlpha which also gave $frac{1}{2}$ so I know that the sollution i have is correct.



Would anyone know where i was wrong or how i could better solve it?






real-analysis calculus convergence







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    To investigate the convergence of a series I have to solve the folliwing limit:



    begin{equation}
    limlimits_{xrightarrowinfty} frac{(x^2-1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}}
    end{equation}



    It should be $frac{1}{2}$ but i can't quite seem to get to that solution. I've tried to factor the square root out of the quotient which was:



    begin{equation}
    limlimits_{xrightarrowinfty} sqrt{frac{((x^2-1) sqrt{x + 2}-x^2sqrt{x+1})^2}{x^2(x + 1)}}
    end{equation}



    Then i worked out the square in the numerator which was:



    begin{equation}
    (x^2-1)^2 (x + 2)-2x^2(x^2-1)sqrt{x+1}sqrt{x+2}+x^4(x+1)
    end{equation}



    I could then factor the terms and take and divide the numerator with the (x+1) from the denominator. Then i could expand the terms in the numerator wich became:



    begin{equation}
    limlimits_{xrightarrowinfty}sqrt{frac{(x^4+x^3-3x^2-x+2)-sqrt{16x^8+32x^7-12x^6-20x^5+8x^4}+x^4}{x^2}}
    end{equation}



    Now i can take the $16x^8$ out of the root and then i then looked at the terms with the highest exponent so i had:



    begin{equation}
    limlimits_{xrightarrowinfty}sqrt{frac{x^4-4x^4+x^4}{x^2}} =limlimits_{xrightarrowinfty}sqrt{frac{-2x^2}{x^2}}
    end{equation}



    Which could only be solved with complex numbers, so i should be wrong somewhere in my calculations, since i know that i should get $frac{1}{2}$. I also checked my sollution with WolframAlpha which also gave $frac{1}{2}$ so I know that the sollution i have is correct.



    Would anyone know where i was wrong or how i could better solve it?






    real-analysis calculus convergence







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      1



      $begingroup$


      To investigate the convergence of a series I have to solve the folliwing limit:



      begin{equation}
      limlimits_{xrightarrowinfty} frac{(x^2-1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}}
      end{equation}



      It should be $frac{1}{2}$ but i can't quite seem to get to that solution. I've tried to factor the square root out of the quotient which was:



      begin{equation}
      limlimits_{xrightarrowinfty} sqrt{frac{((x^2-1) sqrt{x + 2}-x^2sqrt{x+1})^2}{x^2(x + 1)}}
      end{equation}



      Then i worked out the square in the numerator which was:



      begin{equation}
      (x^2-1)^2 (x + 2)-2x^2(x^2-1)sqrt{x+1}sqrt{x+2}+x^4(x+1)
      end{equation}



      I could then factor the terms and take and divide the numerator with the (x+1) from the denominator. Then i could expand the terms in the numerator wich became:



      begin{equation}
      limlimits_{xrightarrowinfty}sqrt{frac{(x^4+x^3-3x^2-x+2)-sqrt{16x^8+32x^7-12x^6-20x^5+8x^4}+x^4}{x^2}}
      end{equation}



      Now i can take the $16x^8$ out of the root and then i then looked at the terms with the highest exponent so i had:



      begin{equation}
      limlimits_{xrightarrowinfty}sqrt{frac{x^4-4x^4+x^4}{x^2}} =limlimits_{xrightarrowinfty}sqrt{frac{-2x^2}{x^2}}
      end{equation}



      Which could only be solved with complex numbers, so i should be wrong somewhere in my calculations, since i know that i should get $frac{1}{2}$. I also checked my sollution with WolframAlpha which also gave $frac{1}{2}$ so I know that the sollution i have is correct.



      Would anyone know where i was wrong or how i could better solve it?






      real-analysis calculus convergence







      share|cite|improve this question









      $endgroup$




      To investigate the convergence of a series I have to solve the folliwing limit:



      begin{equation}
      limlimits_{xrightarrowinfty} frac{(x^2-1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}}
      end{equation}



      It should be $frac{1}{2}$ but i can't quite seem to get to that solution. I've tried to factor the square root out of the quotient which was:



      begin{equation}
      limlimits_{xrightarrowinfty} sqrt{frac{((x^2-1) sqrt{x + 2}-x^2sqrt{x+1})^2}{x^2(x + 1)}}
      end{equation}



      Then i worked out the square in the numerator which was:



      begin{equation}
      (x^2-1)^2 (x + 2)-2x^2(x^2-1)sqrt{x+1}sqrt{x+2}+x^4(x+1)
      end{equation}



      I could then factor the terms and take and divide the numerator with the (x+1) from the denominator. Then i could expand the terms in the numerator wich became:



      begin{equation}
      limlimits_{xrightarrowinfty}sqrt{frac{(x^4+x^3-3x^2-x+2)-sqrt{16x^8+32x^7-12x^6-20x^5+8x^4}+x^4}{x^2}}
      end{equation}



      Now i can take the $16x^8$ out of the root and then i then looked at the terms with the highest exponent so i had:



      begin{equation}
      limlimits_{xrightarrowinfty}sqrt{frac{x^4-4x^4+x^4}{x^2}} =limlimits_{xrightarrowinfty}sqrt{frac{-2x^2}{x^2}}
      end{equation}



      Which could only be solved with complex numbers, so i should be wrong somewhere in my calculations, since i know that i should get $frac{1}{2}$. I also checked my sollution with WolframAlpha which also gave $frac{1}{2}$ so I know that the sollution i have is correct.



      Would anyone know where i was wrong or how i could better solve it?






      real-analysis calculus convergence




      real-analysis calculus convergence






      share|cite|improve this question













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      asked Jan 7 at 15:49









      ViktorViktor

      1389




      1389






















          2 Answers
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          $begingroup$

          $$lim_{xtoinfty}frac{(x^2-1)sqrt{x+2}-x^2sqrt{x+1}}{xsqrt{x+1}}=lim_{xtoinfty}frac{x^2big[sqrt{x+2}-sqrt{x+1}big]}{xsqrt{x+1}}-frac{sqrt{x+2}}{xsqrt{x+1}}$$



          The latter goes to $0$ since $$lim_{xtoinfty}frac{sqrt{x+2}}{xsqrt{x+1}}=lim_{xtoinfty}frac1xcdotsqrt{frac{x+2}{x+1}}=lim_{xtoinfty}frac1xcdotsqrt{1+frac1{x+1}}=0$$



          You are left with $$lim_{xtoinfty}frac{xbig[sqrt{x+2}-sqrt{x+1}big]}{sqrt{x+1}}=lim_{xtoinfty}frac{xbig[sqrt{x+2}-sqrt{x+1}big]}{sqrt{x+1}}cdotfrac{big[sqrt{x+2}+sqrt{x+1}big]}{big[sqrt{x+2}+sqrt{x+1}big]}\=lim_{xtoinfty}frac x{sqrt{x+1}big[sqrt{x+2}+sqrt{x+1}big]}=lim_{xtoinfty}frac 1{sqrt{1+frac1x}Big[sqrt{1+frac2x}+sqrt{1+frac1x}Big]}=1/2$$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Another way:
            begin{eqnarray*} frac{(x^2-1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}}
            & = & frac{(x-1)(x+1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}} \
            & = & frac{(x-1)sqrt{(x+1)(x + 2)}-x^2}{x} \
            & stackrel{x^2 = x(x-1)+x}{=} & underbrace{frac{x-1}{x}}_{stackrel{x to +infty}{longrightarrow}1}underbrace{left(sqrt{(x+1)(x + 2)} - xright)}_{= frac{3x+2}{sqrt{(x+1)(x + 2)} + x}stackrel{x to +infty}{longrightarrow}frac{3}{2}} - 1 \
            & stackrel{x to +infty}{longrightarrow} & 1 cdot frac{3}{2} -1 = frac{1}{2}
            end{eqnarray*}






            share|cite|improve this answer









            $endgroup$













              Your Answer





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              2 Answers
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              2 Answers
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              5












              $begingroup$

              $$lim_{xtoinfty}frac{(x^2-1)sqrt{x+2}-x^2sqrt{x+1}}{xsqrt{x+1}}=lim_{xtoinfty}frac{x^2big[sqrt{x+2}-sqrt{x+1}big]}{xsqrt{x+1}}-frac{sqrt{x+2}}{xsqrt{x+1}}$$



              The latter goes to $0$ since $$lim_{xtoinfty}frac{sqrt{x+2}}{xsqrt{x+1}}=lim_{xtoinfty}frac1xcdotsqrt{frac{x+2}{x+1}}=lim_{xtoinfty}frac1xcdotsqrt{1+frac1{x+1}}=0$$



              You are left with $$lim_{xtoinfty}frac{xbig[sqrt{x+2}-sqrt{x+1}big]}{sqrt{x+1}}=lim_{xtoinfty}frac{xbig[sqrt{x+2}-sqrt{x+1}big]}{sqrt{x+1}}cdotfrac{big[sqrt{x+2}+sqrt{x+1}big]}{big[sqrt{x+2}+sqrt{x+1}big]}\=lim_{xtoinfty}frac x{sqrt{x+1}big[sqrt{x+2}+sqrt{x+1}big]}=lim_{xtoinfty}frac 1{sqrt{1+frac1x}Big[sqrt{1+frac2x}+sqrt{1+frac1x}Big]}=1/2$$






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                $$lim_{xtoinfty}frac{(x^2-1)sqrt{x+2}-x^2sqrt{x+1}}{xsqrt{x+1}}=lim_{xtoinfty}frac{x^2big[sqrt{x+2}-sqrt{x+1}big]}{xsqrt{x+1}}-frac{sqrt{x+2}}{xsqrt{x+1}}$$



                The latter goes to $0$ since $$lim_{xtoinfty}frac{sqrt{x+2}}{xsqrt{x+1}}=lim_{xtoinfty}frac1xcdotsqrt{frac{x+2}{x+1}}=lim_{xtoinfty}frac1xcdotsqrt{1+frac1{x+1}}=0$$



                You are left with $$lim_{xtoinfty}frac{xbig[sqrt{x+2}-sqrt{x+1}big]}{sqrt{x+1}}=lim_{xtoinfty}frac{xbig[sqrt{x+2}-sqrt{x+1}big]}{sqrt{x+1}}cdotfrac{big[sqrt{x+2}+sqrt{x+1}big]}{big[sqrt{x+2}+sqrt{x+1}big]}\=lim_{xtoinfty}frac x{sqrt{x+1}big[sqrt{x+2}+sqrt{x+1}big]}=lim_{xtoinfty}frac 1{sqrt{1+frac1x}Big[sqrt{1+frac2x}+sqrt{1+frac1x}Big]}=1/2$$






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  $$lim_{xtoinfty}frac{(x^2-1)sqrt{x+2}-x^2sqrt{x+1}}{xsqrt{x+1}}=lim_{xtoinfty}frac{x^2big[sqrt{x+2}-sqrt{x+1}big]}{xsqrt{x+1}}-frac{sqrt{x+2}}{xsqrt{x+1}}$$



                  The latter goes to $0$ since $$lim_{xtoinfty}frac{sqrt{x+2}}{xsqrt{x+1}}=lim_{xtoinfty}frac1xcdotsqrt{frac{x+2}{x+1}}=lim_{xtoinfty}frac1xcdotsqrt{1+frac1{x+1}}=0$$



                  You are left with $$lim_{xtoinfty}frac{xbig[sqrt{x+2}-sqrt{x+1}big]}{sqrt{x+1}}=lim_{xtoinfty}frac{xbig[sqrt{x+2}-sqrt{x+1}big]}{sqrt{x+1}}cdotfrac{big[sqrt{x+2}+sqrt{x+1}big]}{big[sqrt{x+2}+sqrt{x+1}big]}\=lim_{xtoinfty}frac x{sqrt{x+1}big[sqrt{x+2}+sqrt{x+1}big]}=lim_{xtoinfty}frac 1{sqrt{1+frac1x}Big[sqrt{1+frac2x}+sqrt{1+frac1x}Big]}=1/2$$






                  share|cite|improve this answer









                  $endgroup$



                  $$lim_{xtoinfty}frac{(x^2-1)sqrt{x+2}-x^2sqrt{x+1}}{xsqrt{x+1}}=lim_{xtoinfty}frac{x^2big[sqrt{x+2}-sqrt{x+1}big]}{xsqrt{x+1}}-frac{sqrt{x+2}}{xsqrt{x+1}}$$



                  The latter goes to $0$ since $$lim_{xtoinfty}frac{sqrt{x+2}}{xsqrt{x+1}}=lim_{xtoinfty}frac1xcdotsqrt{frac{x+2}{x+1}}=lim_{xtoinfty}frac1xcdotsqrt{1+frac1{x+1}}=0$$



                  You are left with $$lim_{xtoinfty}frac{xbig[sqrt{x+2}-sqrt{x+1}big]}{sqrt{x+1}}=lim_{xtoinfty}frac{xbig[sqrt{x+2}-sqrt{x+1}big]}{sqrt{x+1}}cdotfrac{big[sqrt{x+2}+sqrt{x+1}big]}{big[sqrt{x+2}+sqrt{x+1}big]}\=lim_{xtoinfty}frac x{sqrt{x+1}big[sqrt{x+2}+sqrt{x+1}big]}=lim_{xtoinfty}frac 1{sqrt{1+frac1x}Big[sqrt{1+frac2x}+sqrt{1+frac1x}Big]}=1/2$$







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Jan 7 at 16:13









                  Shubham JohriShubham Johri

                  5,017717




                  5,017717























                      2












                      $begingroup$

                      Another way:
                      begin{eqnarray*} frac{(x^2-1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}}
                      & = & frac{(x-1)(x+1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}} \
                      & = & frac{(x-1)sqrt{(x+1)(x + 2)}-x^2}{x} \
                      & stackrel{x^2 = x(x-1)+x}{=} & underbrace{frac{x-1}{x}}_{stackrel{x to +infty}{longrightarrow}1}underbrace{left(sqrt{(x+1)(x + 2)} - xright)}_{= frac{3x+2}{sqrt{(x+1)(x + 2)} + x}stackrel{x to +infty}{longrightarrow}frac{3}{2}} - 1 \
                      & stackrel{x to +infty}{longrightarrow} & 1 cdot frac{3}{2} -1 = frac{1}{2}
                      end{eqnarray*}






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Another way:
                        begin{eqnarray*} frac{(x^2-1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}}
                        & = & frac{(x-1)(x+1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}} \
                        & = & frac{(x-1)sqrt{(x+1)(x + 2)}-x^2}{x} \
                        & stackrel{x^2 = x(x-1)+x}{=} & underbrace{frac{x-1}{x}}_{stackrel{x to +infty}{longrightarrow}1}underbrace{left(sqrt{(x+1)(x + 2)} - xright)}_{= frac{3x+2}{sqrt{(x+1)(x + 2)} + x}stackrel{x to +infty}{longrightarrow}frac{3}{2}} - 1 \
                        & stackrel{x to +infty}{longrightarrow} & 1 cdot frac{3}{2} -1 = frac{1}{2}
                        end{eqnarray*}






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Another way:
                          begin{eqnarray*} frac{(x^2-1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}}
                          & = & frac{(x-1)(x+1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}} \
                          & = & frac{(x-1)sqrt{(x+1)(x + 2)}-x^2}{x} \
                          & stackrel{x^2 = x(x-1)+x}{=} & underbrace{frac{x-1}{x}}_{stackrel{x to +infty}{longrightarrow}1}underbrace{left(sqrt{(x+1)(x + 2)} - xright)}_{= frac{3x+2}{sqrt{(x+1)(x + 2)} + x}stackrel{x to +infty}{longrightarrow}frac{3}{2}} - 1 \
                          & stackrel{x to +infty}{longrightarrow} & 1 cdot frac{3}{2} -1 = frac{1}{2}
                          end{eqnarray*}






                          share|cite|improve this answer









                          $endgroup$



                          Another way:
                          begin{eqnarray*} frac{(x^2-1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}}
                          & = & frac{(x-1)(x+1) sqrt{x + 2}-x^2sqrt{x+1}}{xsqrt{x + 1}} \
                          & = & frac{(x-1)sqrt{(x+1)(x + 2)}-x^2}{x} \
                          & stackrel{x^2 = x(x-1)+x}{=} & underbrace{frac{x-1}{x}}_{stackrel{x to +infty}{longrightarrow}1}underbrace{left(sqrt{(x+1)(x + 2)} - xright)}_{= frac{3x+2}{sqrt{(x+1)(x + 2)} + x}stackrel{x to +infty}{longrightarrow}frac{3}{2}} - 1 \
                          & stackrel{x to +infty}{longrightarrow} & 1 cdot frac{3}{2} -1 = frac{1}{2}
                          end{eqnarray*}







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 7 at 17:07









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