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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        oldest

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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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                                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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