Calculate $lim_{nrightarrow infty }left(frac{n^{2}+1}{n+1}right)^{tfrac{n+1}{n^{2}+1}}$












2












$begingroup$



$$lim_{nrightarrow infty }left(frac{n^{2}+1}{n+1}right)^{tfrac{n+1}{n^{2}+1}}$$




I tried to use $f^{g}=e^{g ln f}$ and I got $e^{tfrac{n+1}{n^{2}+1}ln left(tfrac{n^2+1}{n+1} right)}$. How to continue ?






calculus limits







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$endgroup$

















    2












    $begingroup$



    $$lim_{nrightarrow infty }left(frac{n^{2}+1}{n+1}right)^{tfrac{n+1}{n^{2}+1}}$$




    I tried to use $f^{g}=e^{g ln f}$ and I got $e^{tfrac{n+1}{n^{2}+1}ln left(tfrac{n^2+1}{n+1} right)}$. How to continue ?






    calculus limits







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      $$lim_{nrightarrow infty }left(frac{n^{2}+1}{n+1}right)^{tfrac{n+1}{n^{2}+1}}$$




      I tried to use $f^{g}=e^{g ln f}$ and I got $e^{tfrac{n+1}{n^{2}+1}ln left(tfrac{n^2+1}{n+1} right)}$. How to continue ?






      calculus limits







      share|cite|improve this question











      $endgroup$





      $$lim_{nrightarrow infty }left(frac{n^{2}+1}{n+1}right)^{tfrac{n+1}{n^{2}+1}}$$




      I tried to use $f^{g}=e^{g ln f}$ and I got $e^{tfrac{n+1}{n^{2}+1}ln left(tfrac{n^2+1}{n+1} right)}$. How to continue ?






      calculus limits




      calculus limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 26 at 19:28









      TheSimpliFire

      12.5k62460




      12.5k62460










      asked Jan 26 at 18:46









      DaniVajaDaniVaja

      977




      977






















          4 Answers
          4






          active

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          2












          $begingroup$

          Hint. Use substitution: set $x=dfrac{n^2+1}{n+1}$. What is the limit of $x$ when $n$ tends to $infty$?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
            $endgroup$
            – DaniVaja
            Jan 26 at 19:00










          • $begingroup$
            Yes, absolutely.
            $endgroup$
            – Bernard
            Jan 26 at 19:02



















          4












          $begingroup$

          Write $m=frac{n^2+1}{n+1}$, so $mtoinfty$ as $ntoinfty$. Your limit is $$lim_{mtoinfty}expfrac{ln m}{m}=explim_{mtoinfty}frac{ln m}{m}=exp 0=1.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is the same as what I propose, with more details.
            $endgroup$
            – Bernard
            Jan 26 at 19:01



















          0












          $begingroup$

          Making the problem more general, consider
          $$a_n=left(frac{n^{2}+a}{n+b}right)^{tfrac{n+c}{n^{2}+d}}implies log(a_n)={tfrac{n+c}{n^{2}+d}}logleft(frac{n^{2}+a}{n+b}right)$$ Now, use Taylor expansions for large $n$
          $$logleft(frac{n^{2}+a}{n+b}right)=log
          left({n}right)-frac{b}{n}+frac{a+frac{b^2}{2}}{n^2}+Oleft(frac{1}{n^3}right)$$
          $${tfrac{n+c}{n^{2}+d}}=frac{1}{n}+frac{c}{n^2}+Oleft(frac{1}{n^3}right)$$
          $$log(a_n)=frac{log left({n}right)}{n}+frac{-b+c log
          left({n}right)}{n^2}+Oleft(frac{1}{n^3}right)$$           Continuing with Taylor
          $$a_n=e^{log(a_n)}=1+frac{log left({n}right)}{n}+frac{2 left(-b+c log
          left({n}right)right)+log ^2left({n}right)}{2
          n^2}+Oleft(frac{1}{n^3}right)$$
          Then $cdots$ $text{ ???}$ $forall {a,b,c,d}$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            $begin{array}{l}
            displaystylelim_{n to infty}left({n^{2} + 1 over n + 1}right)^{largeleft(n + 1right)/left(n^{2} + 1right)} =
            lim_{n to infty}n^{1/n} =
            expleft(lim_{n to infty}{lnleft(nright) over n}right) \[5mm] =
            displaystyle
            expleft(lim_{n to infty}
            {lnleft(n + 1right) - lnleft(n right) over left[n + 1right] - n}right) =
            expleft(lim_{n to infty}
            lnleft(1 + {1 over n}right)right) = expleft(0right) =
            bbox[10px,#ffd,border:1px groove navy]{1}
            end{array}
            $






            share|cite|improve this answer











            $endgroup$













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              4 Answers
              4






              active

              oldest

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              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              Hint. Use substitution: set $x=dfrac{n^2+1}{n+1}$. What is the limit of $x$ when $n$ tends to $infty$?






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
                $endgroup$
                – DaniVaja
                Jan 26 at 19:00










              • $begingroup$
                Yes, absolutely.
                $endgroup$
                – Bernard
                Jan 26 at 19:02
















              2












              $begingroup$

              Hint. Use substitution: set $x=dfrac{n^2+1}{n+1}$. What is the limit of $x$ when $n$ tends to $infty$?






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
                $endgroup$
                – DaniVaja
                Jan 26 at 19:00










              • $begingroup$
                Yes, absolutely.
                $endgroup$
                – Bernard
                Jan 26 at 19:02














              2












              2








              2





              $begingroup$

              Hint. Use substitution: set $x=dfrac{n^2+1}{n+1}$. What is the limit of $x$ when $n$ tends to $infty$?






              share|cite|improve this answer











              $endgroup$



              Hint. Use substitution: set $x=dfrac{n^2+1}{n+1}$. What is the limit of $x$ when $n$ tends to $infty$?







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 26 at 18:57

























              answered Jan 26 at 18:53









              BernardBernard

              121k740116




              121k740116








              • 1




                $begingroup$
                I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
                $endgroup$
                – DaniVaja
                Jan 26 at 19:00










              • $begingroup$
                Yes, absolutely.
                $endgroup$
                – Bernard
                Jan 26 at 19:02














              • 1




                $begingroup$
                I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
                $endgroup$
                – DaniVaja
                Jan 26 at 19:00










              • $begingroup$
                Yes, absolutely.
                $endgroup$
                – Bernard
                Jan 26 at 19:02








              1




              1




              $begingroup$
              I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
              $endgroup$
              – DaniVaja
              Jan 26 at 19:00




              $begingroup$
              I got $e^{lim_{xrightarrow 0}frac{lnx}{x}}=e^{0}=1$ is that right ?
              $endgroup$
              – DaniVaja
              Jan 26 at 19:00












              $begingroup$
              Yes, absolutely.
              $endgroup$
              – Bernard
              Jan 26 at 19:02




              $begingroup$
              Yes, absolutely.
              $endgroup$
              – Bernard
              Jan 26 at 19:02











              4












              $begingroup$

              Write $m=frac{n^2+1}{n+1}$, so $mtoinfty$ as $ntoinfty$. Your limit is $$lim_{mtoinfty}expfrac{ln m}{m}=explim_{mtoinfty}frac{ln m}{m}=exp 0=1.$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This is the same as what I propose, with more details.
                $endgroup$
                – Bernard
                Jan 26 at 19:01
















              4












              $begingroup$

              Write $m=frac{n^2+1}{n+1}$, so $mtoinfty$ as $ntoinfty$. Your limit is $$lim_{mtoinfty}expfrac{ln m}{m}=explim_{mtoinfty}frac{ln m}{m}=exp 0=1.$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This is the same as what I propose, with more details.
                $endgroup$
                – Bernard
                Jan 26 at 19:01














              4












              4








              4





              $begingroup$

              Write $m=frac{n^2+1}{n+1}$, so $mtoinfty$ as $ntoinfty$. Your limit is $$lim_{mtoinfty}expfrac{ln m}{m}=explim_{mtoinfty}frac{ln m}{m}=exp 0=1.$$






              share|cite|improve this answer









              $endgroup$



              Write $m=frac{n^2+1}{n+1}$, so $mtoinfty$ as $ntoinfty$. Your limit is $$lim_{mtoinfty}expfrac{ln m}{m}=explim_{mtoinfty}frac{ln m}{m}=exp 0=1.$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 26 at 19:00









              J.G.J.G.

              27.1k22843




              27.1k22843












              • $begingroup$
                This is the same as what I propose, with more details.
                $endgroup$
                – Bernard
                Jan 26 at 19:01


















              • $begingroup$
                This is the same as what I propose, with more details.
                $endgroup$
                – Bernard
                Jan 26 at 19:01
















              $begingroup$
              This is the same as what I propose, with more details.
              $endgroup$
              – Bernard
              Jan 26 at 19:01




              $begingroup$
              This is the same as what I propose, with more details.
              $endgroup$
              – Bernard
              Jan 26 at 19:01











              0












              $begingroup$

              Making the problem more general, consider
              $$a_n=left(frac{n^{2}+a}{n+b}right)^{tfrac{n+c}{n^{2}+d}}implies log(a_n)={tfrac{n+c}{n^{2}+d}}logleft(frac{n^{2}+a}{n+b}right)$$ Now, use Taylor expansions for large $n$
              $$logleft(frac{n^{2}+a}{n+b}right)=log
              left({n}right)-frac{b}{n}+frac{a+frac{b^2}{2}}{n^2}+Oleft(frac{1}{n^3}right)$$
              $${tfrac{n+c}{n^{2}+d}}=frac{1}{n}+frac{c}{n^2}+Oleft(frac{1}{n^3}right)$$
              $$log(a_n)=frac{log left({n}right)}{n}+frac{-b+c log
              left({n}right)}{n^2}+Oleft(frac{1}{n^3}right)$$               Continuing with Taylor
              $$a_n=e^{log(a_n)}=1+frac{log left({n}right)}{n}+frac{2 left(-b+c log
              left({n}right)right)+log ^2left({n}right)}{2
              n^2}+Oleft(frac{1}{n^3}right)$$
              Then $cdots$ $text{ ???}$ $forall {a,b,c,d}$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Making the problem more general, consider
                $$a_n=left(frac{n^{2}+a}{n+b}right)^{tfrac{n+c}{n^{2}+d}}implies log(a_n)={tfrac{n+c}{n^{2}+d}}logleft(frac{n^{2}+a}{n+b}right)$$ Now, use Taylor expansions for large $n$
                $$logleft(frac{n^{2}+a}{n+b}right)=log
                left({n}right)-frac{b}{n}+frac{a+frac{b^2}{2}}{n^2}+Oleft(frac{1}{n^3}right)$$
                $${tfrac{n+c}{n^{2}+d}}=frac{1}{n}+frac{c}{n^2}+Oleft(frac{1}{n^3}right)$$
                $$log(a_n)=frac{log left({n}right)}{n}+frac{-b+c log
                left({n}right)}{n^2}+Oleft(frac{1}{n^3}right)$$                 Continuing with Taylor
                $$a_n=e^{log(a_n)}=1+frac{log left({n}right)}{n}+frac{2 left(-b+c log
                left({n}right)right)+log ^2left({n}right)}{2
                n^2}+Oleft(frac{1}{n^3}right)$$
                Then $cdots$ $text{ ???}$ $forall {a,b,c,d}$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Making the problem more general, consider
                  $$a_n=left(frac{n^{2}+a}{n+b}right)^{tfrac{n+c}{n^{2}+d}}implies log(a_n)={tfrac{n+c}{n^{2}+d}}logleft(frac{n^{2}+a}{n+b}right)$$ Now, use Taylor expansions for large $n$
                  $$logleft(frac{n^{2}+a}{n+b}right)=log
                  left({n}right)-frac{b}{n}+frac{a+frac{b^2}{2}}{n^2}+Oleft(frac{1}{n^3}right)$$
                  $${tfrac{n+c}{n^{2}+d}}=frac{1}{n}+frac{c}{n^2}+Oleft(frac{1}{n^3}right)$$
                  $$log(a_n)=frac{log left({n}right)}{n}+frac{-b+c log
                  left({n}right)}{n^2}+Oleft(frac{1}{n^3}right)$$                   Continuing with Taylor
                  $$a_n=e^{log(a_n)}=1+frac{log left({n}right)}{n}+frac{2 left(-b+c log
                  left({n}right)right)+log ^2left({n}right)}{2
                  n^2}+Oleft(frac{1}{n^3}right)$$
                  Then $cdots$ $text{ ???}$ $forall {a,b,c,d}$






                  share|cite|improve this answer









                  $endgroup$



                  Making the problem more general, consider
                  $$a_n=left(frac{n^{2}+a}{n+b}right)^{tfrac{n+c}{n^{2}+d}}implies log(a_n)={tfrac{n+c}{n^{2}+d}}logleft(frac{n^{2}+a}{n+b}right)$$ Now, use Taylor expansions for large $n$
                  $$logleft(frac{n^{2}+a}{n+b}right)=log
                  left({n}right)-frac{b}{n}+frac{a+frac{b^2}{2}}{n^2}+Oleft(frac{1}{n^3}right)$$
                  $${tfrac{n+c}{n^{2}+d}}=frac{1}{n}+frac{c}{n^2}+Oleft(frac{1}{n^3}right)$$
                  $$log(a_n)=frac{log left({n}right)}{n}+frac{-b+c log
                  left({n}right)}{n^2}+Oleft(frac{1}{n^3}right)$$                   Continuing with Taylor
                  $$a_n=e^{log(a_n)}=1+frac{log left({n}right)}{n}+frac{2 left(-b+c log
                  left({n}right)right)+log ^2left({n}right)}{2
                  n^2}+Oleft(frac{1}{n^3}right)$$
                  Then $cdots$ $text{ ???}$ $forall {a,b,c,d}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 27 at 3:06









                  Claude LeiboviciClaude Leibovici

                  121k1157134




                  121k1157134























                      0












                      $begingroup$

                      $begin{array}{l}
                      displaystylelim_{n to infty}left({n^{2} + 1 over n + 1}right)^{largeleft(n + 1right)/left(n^{2} + 1right)} =
                      lim_{n to infty}n^{1/n} =
                      expleft(lim_{n to infty}{lnleft(nright) over n}right) \[5mm] =
                      displaystyle
                      expleft(lim_{n to infty}
                      {lnleft(n + 1right) - lnleft(n right) over left[n + 1right] - n}right) =
                      expleft(lim_{n to infty}
                      lnleft(1 + {1 over n}right)right) = expleft(0right) =
                      bbox[10px,#ffd,border:1px groove navy]{1}
                      end{array}
                      $






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        $begin{array}{l}
                        displaystylelim_{n to infty}left({n^{2} + 1 over n + 1}right)^{largeleft(n + 1right)/left(n^{2} + 1right)} =
                        lim_{n to infty}n^{1/n} =
                        expleft(lim_{n to infty}{lnleft(nright) over n}right) \[5mm] =
                        displaystyle
                        expleft(lim_{n to infty}
                        {lnleft(n + 1right) - lnleft(n right) over left[n + 1right] - n}right) =
                        expleft(lim_{n to infty}
                        lnleft(1 + {1 over n}right)right) = expleft(0right) =
                        bbox[10px,#ffd,border:1px groove navy]{1}
                        end{array}
                        $






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          $begin{array}{l}
                          displaystylelim_{n to infty}left({n^{2} + 1 over n + 1}right)^{largeleft(n + 1right)/left(n^{2} + 1right)} =
                          lim_{n to infty}n^{1/n} =
                          expleft(lim_{n to infty}{lnleft(nright) over n}right) \[5mm] =
                          displaystyle
                          expleft(lim_{n to infty}
                          {lnleft(n + 1right) - lnleft(n right) over left[n + 1right] - n}right) =
                          expleft(lim_{n to infty}
                          lnleft(1 + {1 over n}right)right) = expleft(0right) =
                          bbox[10px,#ffd,border:1px groove navy]{1}
                          end{array}
                          $






                          share|cite|improve this answer











                          $endgroup$



                          $begin{array}{l}
                          displaystylelim_{n to infty}left({n^{2} + 1 over n + 1}right)^{largeleft(n + 1right)/left(n^{2} + 1right)} =
                          lim_{n to infty}n^{1/n} =
                          expleft(lim_{n to infty}{lnleft(nright) over n}right) \[5mm] =
                          displaystyle
                          expleft(lim_{n to infty}
                          {lnleft(n + 1right) - lnleft(n right) over left[n + 1right] - n}right) =
                          expleft(lim_{n to infty}
                          lnleft(1 + {1 over n}right)right) = expleft(0right) =
                          bbox[10px,#ffd,border:1px groove navy]{1}
                          end{array}
                          $







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 28 at 6:42

























                          answered Jan 28 at 2:27









                          Felix MarinFelix Marin

                          68k7108142




                          68k7108142






























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                              Eduardo VII do Reino Unido

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                              Eduardo VII Rei do Reino Unido da Grã-Bretanha e Irlanda e dos Domínios Britânicos de Além-mar Imperador da Índia Rei do Reino Unido e Imperador da Índia Reinado 22 de janeiro de 1901 a 6 de maio de 1910 Coroação 9 de agosto de 1902 Delhi Durbar 1 de janeiro de 1903 Predecessora Vitória Sucessor Jorge V   Esposa Alexandra da Dinamarca Descendência Alberto Vitor, Duque de Clarence e Avondale Jorge V do Reino Unido Luísa, Princesa Real Vitória de Gales Maud de Gales Alexandre João de Gales Casa Saxe-Coburgo-Gota Nome completo Alberto Eduardo Nascimento 9 de novembro de 1841   Palácio de Buckingham, Londres, Reino Unido Morte 6 de maio de 1910 (68 anos)   Palácio de Buckingham, Londres, Reino Unido Enterro Capela de São Jorge, Windsor, Berkshire, Reino Unido 20 de maio de 1910 Pai Alberto de Saxe-Coburgo-Gota Mãe Vitória do Reino Unido Assinatura ...
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