Limit $ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$












3












$begingroup$



I'm required to compute the limit of the following sequence:
$$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$




I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.



Let $a_n$ denote the base.



$$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$



From there though I don't know how to compute:
$$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$



I tried using the ln function though I couldn't get any further. I'd be glad for help :)






real-analysis calculus limits







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$endgroup$

















    3












    $begingroup$



    I'm required to compute the limit of the following sequence:
    $$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$




    I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.



    Let $a_n$ denote the base.



    $$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$



    From there though I don't know how to compute:
    $$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$



    I tried using the ln function though I couldn't get any further. I'd be glad for help :)






    real-analysis calculus limits







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      I'm required to compute the limit of the following sequence:
      $$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$




      I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.



      Let $a_n$ denote the base.



      $$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$



      From there though I don't know how to compute:
      $$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$



      I tried using the ln function though I couldn't get any further. I'd be glad for help :)






      real-analysis calculus limits







      share|cite|improve this question











      $endgroup$





      I'm required to compute the limit of the following sequence:
      $$ lim_{ntoinfty} left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}$$




      I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle.



      Let $a_n$ denote the base.



      $$ lim_{ntoinfty}(a_n)^{frac{n^2}{1-n}}= left(lim_{ntoinfty}(a_n)^{large frac{1}{1-n}}right)^{n^2}=lim_{ntoinfty} left(sqrt[1-n]{a_n}right)^{n^2}=1^{infty} $$



      From there though I don't know how to compute:
      $$e^{lim_{ntoinfty} left(sqrt[n-1]{a_n}n^2right)}$$



      I tried using the ln function though I couldn't get any further. I'd be glad for help :)






      real-analysis calculus limits




      real-analysis calculus limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 '18 at 13:26









      Arjang

      5,59162363




      5,59162363










      asked Dec 5 '18 at 10:47









      MosheMoshe

      396




      396






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          HINT



          $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$



          indeed



          $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$



          and



          $$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
            $endgroup$
            – Moshe
            Dec 5 '18 at 11:05










          • $begingroup$
            @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:16












          • $begingroup$
            Nice and simple (as usual !). Cheers.
            $endgroup$
            – Claude Leibovici
            Dec 5 '18 at 11:25










          • $begingroup$
            @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:28










          • $begingroup$
            "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
            $endgroup$
            – Did
            Dec 5 '18 at 15:01



















          6












          $begingroup$

          You can estimate:
          $$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$



          $$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            very nice alternative approach (+1).
            $endgroup$
            – gimusi
            Dec 5 '18 at 12:58










          • $begingroup$
            Thank you gimusi.
            $endgroup$
            – farruhota
            Dec 5 '18 at 13:01



















          0












          $begingroup$

          You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.






          share|cite|improve this answer









          $endgroup$













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            HINT



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$



            indeed



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$



            and



            $$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
              $endgroup$
              – Moshe
              Dec 5 '18 at 11:05










            • $begingroup$
              @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:16












            • $begingroup$
              Nice and simple (as usual !). Cheers.
              $endgroup$
              – Claude Leibovici
              Dec 5 '18 at 11:25










            • $begingroup$
              @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:28










            • $begingroup$
              "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
              $endgroup$
              – Did
              Dec 5 '18 at 15:01
















            4












            $begingroup$

            HINT



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$



            indeed



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$



            and



            $$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
              $endgroup$
              – Moshe
              Dec 5 '18 at 11:05










            • $begingroup$
              @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:16












            • $begingroup$
              Nice and simple (as usual !). Cheers.
              $endgroup$
              – Claude Leibovici
              Dec 5 '18 at 11:25










            • $begingroup$
              @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:28










            • $begingroup$
              "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
              $endgroup$
              – Did
              Dec 5 '18 at 15:01














            4












            4








            4





            $begingroup$

            HINT



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$



            indeed



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$



            and



            $$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$






            share|cite|improve this answer











            $endgroup$



            HINT



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}sim ccdot left(frac32right)^{-n}$$



            indeed



            $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac32right)^{-n}left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}$$



            and



            $$left(frac{2n^2-frac23n+frac23}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left[left(1-frac{frac53n+frac13}{2n^2+n+1}right)^{frac{2n^2+n+1}{frac53n+frac13}}right]^{frac{frac53n^3+frac13n^2}{(2n^2+n+1)(1-n)}}toleft(frac1eright)^{-frac56}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 5 '18 at 15:13

























            answered Dec 5 '18 at 10:49









            gimusigimusi

            92.8k84494




            92.8k84494












            • $begingroup$
              I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
              $endgroup$
              – Moshe
              Dec 5 '18 at 11:05










            • $begingroup$
              @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:16












            • $begingroup$
              Nice and simple (as usual !). Cheers.
              $endgroup$
              – Claude Leibovici
              Dec 5 '18 at 11:25










            • $begingroup$
              @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:28










            • $begingroup$
              "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
              $endgroup$
              – Did
              Dec 5 '18 at 15:01


















            • $begingroup$
              I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
              $endgroup$
              – Moshe
              Dec 5 '18 at 11:05










            • $begingroup$
              @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:16












            • $begingroup$
              Nice and simple (as usual !). Cheers.
              $endgroup$
              – Claude Leibovici
              Dec 5 '18 at 11:25










            • $begingroup$
              @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
              $endgroup$
              – gimusi
              Dec 5 '18 at 11:28










            • $begingroup$
              "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
              $endgroup$
              – Did
              Dec 5 '18 at 15:01
















            $begingroup$
            I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
            $endgroup$
            – Moshe
            Dec 5 '18 at 11:05




            $begingroup$
            I observed that, it leads to 0 but subtituing larger values in the original form gives me other value
            $endgroup$
            – Moshe
            Dec 5 '18 at 11:05












            $begingroup$
            @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:16






            $begingroup$
            @Moshe To prove that we can proceed as follows $$left(frac{3n^2-n+1}{2n^2+n+1}right)^{frac{n^2}{1-n}}=left(frac{3n^2}{2n^2}right)^{-frac{n^2}{n-1}}cdot left(frac{1-1/3n+1/3n^2}{1+1/2n+1/2n^2}right)^{-frac{n^2}{n-1}}$$ and the second part on the RHS tends to 1.
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:16














            $begingroup$
            Nice and simple (as usual !). Cheers.
            $endgroup$
            – Claude Leibovici
            Dec 5 '18 at 11:25




            $begingroup$
            Nice and simple (as usual !). Cheers.
            $endgroup$
            – Claude Leibovici
            Dec 5 '18 at 11:25












            $begingroup$
            @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:28




            $begingroup$
            @ClaudeLeibovici Thanks Claude! Much appreciative from you! Many cheers :)
            $endgroup$
            – gimusi
            Dec 5 '18 at 11:28












            $begingroup$
            "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
            $endgroup$
            – Did
            Dec 5 '18 at 15:01




            $begingroup$
            "[T]he second part on the RHS" does not converge to 1 (hence the equivalence sign in the answer is wrong).
            $endgroup$
            – Did
            Dec 5 '18 at 15:01











            6












            $begingroup$

            You can estimate:
            $$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$



            $$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              very nice alternative approach (+1).
              $endgroup$
              – gimusi
              Dec 5 '18 at 12:58










            • $begingroup$
              Thank you gimusi.
              $endgroup$
              – farruhota
              Dec 5 '18 at 13:01
















            6












            $begingroup$

            You can estimate:
            $$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$



            $$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              very nice alternative approach (+1).
              $endgroup$
              – gimusi
              Dec 5 '18 at 12:58










            • $begingroup$
              Thank you gimusi.
              $endgroup$
              – farruhota
              Dec 5 '18 at 13:01














            6












            6








            6





            $begingroup$

            You can estimate:
            $$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$



            $$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$






            share|cite|improve this answer









            $endgroup$



            You can estimate:
            $$frac{2n^2+n+1}{3n^2-n+1}<frac{3}{4}, n>7;$$



            $$0<left(frac{3n^2-n+1}{2n^2+n+1}right)^{large frac{n^2}{1-n}}=left(frac{2n^2+n+1}{3n^2-n+1}right)^{large frac{n^2}{n-1}}<left(frac34right)^{large frac{n^2}{n-1}}=left(frac34right)^{n+1+large frac{1}{n-1}}to 0.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 5 '18 at 12:15









            farruhotafarruhota

            20.2k2738




            20.2k2738












            • $begingroup$
              very nice alternative approach (+1).
              $endgroup$
              – gimusi
              Dec 5 '18 at 12:58










            • $begingroup$
              Thank you gimusi.
              $endgroup$
              – farruhota
              Dec 5 '18 at 13:01


















            • $begingroup$
              very nice alternative approach (+1).
              $endgroup$
              – gimusi
              Dec 5 '18 at 12:58










            • $begingroup$
              Thank you gimusi.
              $endgroup$
              – farruhota
              Dec 5 '18 at 13:01
















            $begingroup$
            very nice alternative approach (+1).
            $endgroup$
            – gimusi
            Dec 5 '18 at 12:58




            $begingroup$
            very nice alternative approach (+1).
            $endgroup$
            – gimusi
            Dec 5 '18 at 12:58












            $begingroup$
            Thank you gimusi.
            $endgroup$
            – farruhota
            Dec 5 '18 at 13:01




            $begingroup$
            Thank you gimusi.
            $endgroup$
            – farruhota
            Dec 5 '18 at 13:01











            0












            $begingroup$

            You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.






                share|cite|improve this answer









                $endgroup$



                You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $ntoinfty$; so you have the the reciprocal of a positive-exponentially increasing number.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 5 '18 at 14:14









                AmbretteOrriseyAmbretteOrrisey

                54210




                54210






























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