Find the limit of $lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$ without using the identity...












1












$begingroup$


Find the limit of the sequence $$lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}longrightarrowfrac{1}{3}$$



but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    That's a sequence, not a series.
    $endgroup$
    – José Carlos Santos
    Dec 22 '18 at 9:45










  • $begingroup$
    Oops, corrected it.
    $endgroup$
    – Belkan
    Dec 22 '18 at 9:47
















1












$begingroup$


Find the limit of the sequence $$lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}longrightarrowfrac{1}{3}$$



but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    That's a sequence, not a series.
    $endgroup$
    – José Carlos Santos
    Dec 22 '18 at 9:45










  • $begingroup$
    Oops, corrected it.
    $endgroup$
    – Belkan
    Dec 22 '18 at 9:47














1












1








1





$begingroup$


Find the limit of the sequence $$lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}longrightarrowfrac{1}{3}$$



but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.






sequences-and-series







share|cite|improve this question











$endgroup$




Find the limit of the sequence $$lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}longrightarrowfrac{1}{3}$$



but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 11:22









Asaf Karagila♦

305k33435766




305k33435766










asked Dec 22 '18 at 9:41









BelkanBelkan

687




687












  • $begingroup$
    That's a sequence, not a series.
    $endgroup$
    – José Carlos Santos
    Dec 22 '18 at 9:45










  • $begingroup$
    Oops, corrected it.
    $endgroup$
    – Belkan
    Dec 22 '18 at 9:47


















  • $begingroup$
    That's a sequence, not a series.
    $endgroup$
    – José Carlos Santos
    Dec 22 '18 at 9:45










  • $begingroup$
    Oops, corrected it.
    $endgroup$
    – Belkan
    Dec 22 '18 at 9:47
















$begingroup$
That's a sequence, not a series.
$endgroup$
– José Carlos Santos
Dec 22 '18 at 9:45




$begingroup$
That's a sequence, not a series.
$endgroup$
– José Carlos Santos
Dec 22 '18 at 9:45












$begingroup$
Oops, corrected it.
$endgroup$
– Belkan
Dec 22 '18 at 9:47




$begingroup$
Oops, corrected it.
$endgroup$
– Belkan
Dec 22 '18 at 9:47










3 Answers
3






active

oldest

votes


















4












$begingroup$

$$
begin{align}
lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
&=lim_{ntoinfty}nleft(left(1+frac1nright)^{1/3}-left(1+frac1{n^3}right)^{1/3}right)\
&=lim_{ntoinfty}nleft(left[1+frac1{3n}+O!left(frac1{n^2}right)right]-left[1+O!left(frac1{n^3}right)right]right)\
&=lim_{ntoinfty}left(frac13+O!left(frac1nright)right)\
&=frac13
end{align}
$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    My favorite way is to consider
    $$
    f(x)=left(frac{1}{x^3}+frac{1}{x^2}right)^{1/3}-left(frac{1}{x^3}+1right)^{1/3}
    =frac{sqrt[3]{1+x}-sqrt[3]{1+x^3}}{x}
    $$
    so that your sequence is $f(1/n)$ and so you can compute
    $$
    lim_{xto0^+}f(x)=lim_{xto0}frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=frac{1}{3}
    $$
    Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=sqrt[3]{1+x}-sqrt[3]{1+x^3}$, so
    $$
    g'(x)=frac{1}{3sqrt[3]{(1+x)^2}}-frac{3x^2}{3sqrt[3]{(1+x^3)^2}}
    $$
    and $g'(0)=1/3$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      1) $n(1+1/n )^{1/3} = $



      $dfrac{(1+1/n)^{1/3}}{1/n}.$



      2) $n(1+1/n^3 )^{1/3} =$



      $dfrac{(1+1/n^3)^{1/3}}{1/n}.$



      $small{dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$



      $small{=dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$



      First term:



      Let $f(x)=x^{1/3}$:



      $lim_{n rightarrow infty}dfrac{(1+1/n)^{1/3} -1}{1/n}=$



      $f'(x)_{x=1}= (1/3);$



      Second term:



      $small{-lim_{n rightarrow infty}(1/n^2)dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$



      $small{lim_{n rightarrow infty}(1/n^2)×}$



      $small{lim_{n rightarrow infty}dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$



      $small{0 cdot (1/3)=0.}$






      share|cite|improve this answer











      $endgroup$













        Your Answer





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        3 Answers
        3






        active

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        3 Answers
        3






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        active

        oldest

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        active

        oldest

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        4












        $begingroup$

        $$
        begin{align}
        lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
        &=lim_{ntoinfty}nleft(left(1+frac1nright)^{1/3}-left(1+frac1{n^3}right)^{1/3}right)\
        &=lim_{ntoinfty}nleft(left[1+frac1{3n}+O!left(frac1{n^2}right)right]-left[1+O!left(frac1{n^3}right)right]right)\
        &=lim_{ntoinfty}left(frac13+O!left(frac1nright)right)\
        &=frac13
        end{align}
        $$






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          $$
          begin{align}
          lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
          &=lim_{ntoinfty}nleft(left(1+frac1nright)^{1/3}-left(1+frac1{n^3}right)^{1/3}right)\
          &=lim_{ntoinfty}nleft(left[1+frac1{3n}+O!left(frac1{n^2}right)right]-left[1+O!left(frac1{n^3}right)right]right)\
          &=lim_{ntoinfty}left(frac13+O!left(frac1nright)right)\
          &=frac13
          end{align}
          $$






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            $$
            begin{align}
            lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
            &=lim_{ntoinfty}nleft(left(1+frac1nright)^{1/3}-left(1+frac1{n^3}right)^{1/3}right)\
            &=lim_{ntoinfty}nleft(left[1+frac1{3n}+O!left(frac1{n^2}right)right]-left[1+O!left(frac1{n^3}right)right]right)\
            &=lim_{ntoinfty}left(frac13+O!left(frac1nright)right)\
            &=frac13
            end{align}
            $$






            share|cite|improve this answer









            $endgroup$



            $$
            begin{align}
            lim_{ntoinfty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})
            &=lim_{ntoinfty}nleft(left(1+frac1nright)^{1/3}-left(1+frac1{n^3}right)^{1/3}right)\
            &=lim_{ntoinfty}nleft(left[1+frac1{3n}+O!left(frac1{n^2}right)right]-left[1+O!left(frac1{n^3}right)right]right)\
            &=lim_{ntoinfty}left(frac13+O!left(frac1nright)right)\
            &=frac13
            end{align}
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 22 '18 at 9:50









            robjohn♦robjohn

            268k27309634




            268k27309634























                3












                $begingroup$

                My favorite way is to consider
                $$
                f(x)=left(frac{1}{x^3}+frac{1}{x^2}right)^{1/3}-left(frac{1}{x^3}+1right)^{1/3}
                =frac{sqrt[3]{1+x}-sqrt[3]{1+x^3}}{x}
                $$
                so that your sequence is $f(1/n)$ and so you can compute
                $$
                lim_{xto0^+}f(x)=lim_{xto0}frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=frac{1}{3}
                $$
                Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=sqrt[3]{1+x}-sqrt[3]{1+x^3}$, so
                $$
                g'(x)=frac{1}{3sqrt[3]{(1+x)^2}}-frac{3x^2}{3sqrt[3]{(1+x^3)^2}}
                $$
                and $g'(0)=1/3$.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  My favorite way is to consider
                  $$
                  f(x)=left(frac{1}{x^3}+frac{1}{x^2}right)^{1/3}-left(frac{1}{x^3}+1right)^{1/3}
                  =frac{sqrt[3]{1+x}-sqrt[3]{1+x^3}}{x}
                  $$
                  so that your sequence is $f(1/n)$ and so you can compute
                  $$
                  lim_{xto0^+}f(x)=lim_{xto0}frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=frac{1}{3}
                  $$
                  Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=sqrt[3]{1+x}-sqrt[3]{1+x^3}$, so
                  $$
                  g'(x)=frac{1}{3sqrt[3]{(1+x)^2}}-frac{3x^2}{3sqrt[3]{(1+x^3)^2}}
                  $$
                  and $g'(0)=1/3$.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    My favorite way is to consider
                    $$
                    f(x)=left(frac{1}{x^3}+frac{1}{x^2}right)^{1/3}-left(frac{1}{x^3}+1right)^{1/3}
                    =frac{sqrt[3]{1+x}-sqrt[3]{1+x^3}}{x}
                    $$
                    so that your sequence is $f(1/n)$ and so you can compute
                    $$
                    lim_{xto0^+}f(x)=lim_{xto0}frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=frac{1}{3}
                    $$
                    Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=sqrt[3]{1+x}-sqrt[3]{1+x^3}$, so
                    $$
                    g'(x)=frac{1}{3sqrt[3]{(1+x)^2}}-frac{3x^2}{3sqrt[3]{(1+x^3)^2}}
                    $$
                    and $g'(0)=1/3$.






                    share|cite|improve this answer









                    $endgroup$



                    My favorite way is to consider
                    $$
                    f(x)=left(frac{1}{x^3}+frac{1}{x^2}right)^{1/3}-left(frac{1}{x^3}+1right)^{1/3}
                    =frac{sqrt[3]{1+x}-sqrt[3]{1+x^3}}{x}
                    $$
                    so that your sequence is $f(1/n)$ and so you can compute
                    $$
                    lim_{xto0^+}f(x)=lim_{xto0}frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=frac{1}{3}
                    $$
                    Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=sqrt[3]{1+x}-sqrt[3]{1+x^3}$, so
                    $$
                    g'(x)=frac{1}{3sqrt[3]{(1+x)^2}}-frac{3x^2}{3sqrt[3]{(1+x^3)^2}}
                    $$
                    and $g'(0)=1/3$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 22 '18 at 10:01









                    egregegreg

                    183k1486205




                    183k1486205























                        1












                        $begingroup$

                        1) $n(1+1/n )^{1/3} = $



                        $dfrac{(1+1/n)^{1/3}}{1/n}.$



                        2) $n(1+1/n^3 )^{1/3} =$



                        $dfrac{(1+1/n^3)^{1/3}}{1/n}.$



                        $small{dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$



                        $small{=dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$



                        First term:



                        Let $f(x)=x^{1/3}$:



                        $lim_{n rightarrow infty}dfrac{(1+1/n)^{1/3} -1}{1/n}=$



                        $f'(x)_{x=1}= (1/3);$



                        Second term:



                        $small{-lim_{n rightarrow infty}(1/n^2)dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$



                        $small{lim_{n rightarrow infty}(1/n^2)×}$



                        $small{lim_{n rightarrow infty}dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$



                        $small{0 cdot (1/3)=0.}$






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          1) $n(1+1/n )^{1/3} = $



                          $dfrac{(1+1/n)^{1/3}}{1/n}.$



                          2) $n(1+1/n^3 )^{1/3} =$



                          $dfrac{(1+1/n^3)^{1/3}}{1/n}.$



                          $small{dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$



                          $small{=dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$



                          First term:



                          Let $f(x)=x^{1/3}$:



                          $lim_{n rightarrow infty}dfrac{(1+1/n)^{1/3} -1}{1/n}=$



                          $f'(x)_{x=1}= (1/3);$



                          Second term:



                          $small{-lim_{n rightarrow infty}(1/n^2)dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$



                          $small{lim_{n rightarrow infty}(1/n^2)×}$



                          $small{lim_{n rightarrow infty}dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$



                          $small{0 cdot (1/3)=0.}$






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            1) $n(1+1/n )^{1/3} = $



                            $dfrac{(1+1/n)^{1/3}}{1/n}.$



                            2) $n(1+1/n^3 )^{1/3} =$



                            $dfrac{(1+1/n^3)^{1/3}}{1/n}.$



                            $small{dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$



                            $small{=dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$



                            First term:



                            Let $f(x)=x^{1/3}$:



                            $lim_{n rightarrow infty}dfrac{(1+1/n)^{1/3} -1}{1/n}=$



                            $f'(x)_{x=1}= (1/3);$



                            Second term:



                            $small{-lim_{n rightarrow infty}(1/n^2)dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$



                            $small{lim_{n rightarrow infty}(1/n^2)×}$



                            $small{lim_{n rightarrow infty}dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$



                            $small{0 cdot (1/3)=0.}$






                            share|cite|improve this answer











                            $endgroup$



                            1) $n(1+1/n )^{1/3} = $



                            $dfrac{(1+1/n)^{1/3}}{1/n}.$



                            2) $n(1+1/n^3 )^{1/3} =$



                            $dfrac{(1+1/n^3)^{1/3}}{1/n}.$



                            $small{dfrac{((1+1/n)^{1/3} -1) -((1+1/n^3)^{1/3}-1)}{1/n}}$



                            $small{=dfrac{(1+1/n)^{1/3} -1}{1/n} - (1/n^2)dfrac{(1+1/n^3)^{1/3} -1}{1/n^3}.}$



                            First term:



                            Let $f(x)=x^{1/3}$:



                            $lim_{n rightarrow infty}dfrac{(1+1/n)^{1/3} -1}{1/n}=$



                            $f'(x)_{x=1}= (1/3);$



                            Second term:



                            $small{-lim_{n rightarrow infty}(1/n^2)dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}= }$



                            $small{lim_{n rightarrow infty}(1/n^2)×}$



                            $small{lim_{n rightarrow infty}dfrac{(1+1/n^3)^{1/3}-1}{1/n^3}=}$



                            $small{0 cdot (1/3)=0.}$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 22 '18 at 10:56

























                            answered Dec 22 '18 at 10:30









                            Peter SzilasPeter Szilas

                            11.4k2822




                            11.4k2822






























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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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