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











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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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    up vote
    3
    down vote

    favorite













    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


























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite












      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
















      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 at 13:26









      Arjang

      5,56662363




      5,56662363










      asked Dec 5 at 10:47









      Moshe

      376




      376






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          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























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










          • @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.
            – gimusi
            Dec 5 at 11:16












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










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










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


















          up vote
          6
          down vote













          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





















          • very nice alternative approach (+1).
            – gimusi
            Dec 5 at 12:58










          • Thank you gimusi.
            – farruhota
            Dec 5 at 13:01


















          up vote
          0
          down vote













          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





















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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote



            accepted










            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























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










            • @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.
              – gimusi
              Dec 5 at 11:16












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










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










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















            up vote
            4
            down vote



            accepted










            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























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










            • @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.
              – gimusi
              Dec 5 at 11:16












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










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










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













            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            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














            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 at 15:13

























            answered Dec 5 at 10:49









            gimusi

            92.9k94495




            92.9k94495












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










            • @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.
              – gimusi
              Dec 5 at 11:16












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










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










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


















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










            • @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.
              – gimusi
              Dec 5 at 11:16












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










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










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
















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




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












            @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.
            – gimusi
            Dec 5 at 11:16






            @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.
            – gimusi
            Dec 5 at 11:16














            Nice and simple (as usual !). Cheers.
            – Claude Leibovici
            Dec 5 at 11:25




            Nice and simple (as usual !). Cheers.
            – Claude Leibovici
            Dec 5 at 11:25












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




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












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




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










            up vote
            6
            down vote













            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





















            • very nice alternative approach (+1).
              – gimusi
              Dec 5 at 12:58










            • Thank you gimusi.
              – farruhota
              Dec 5 at 13:01















            up vote
            6
            down vote













            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





















            • very nice alternative approach (+1).
              – gimusi
              Dec 5 at 12:58










            • Thank you gimusi.
              – farruhota
              Dec 5 at 13:01













            up vote
            6
            down vote










            up vote
            6
            down vote









            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












            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 at 12:15









            farruhota

            18.7k2736




            18.7k2736












            • very nice alternative approach (+1).
              – gimusi
              Dec 5 at 12:58










            • Thank you gimusi.
              – farruhota
              Dec 5 at 13:01


















            • very nice alternative approach (+1).
              – gimusi
              Dec 5 at 12:58










            • Thank you gimusi.
              – farruhota
              Dec 5 at 13:01
















            very nice alternative approach (+1).
            – gimusi
            Dec 5 at 12:58




            very nice alternative approach (+1).
            – gimusi
            Dec 5 at 12:58












            Thank you gimusi.
            – farruhota
            Dec 5 at 13:01




            Thank you gimusi.
            – farruhota
            Dec 5 at 13:01










            up vote
            0
            down vote













            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

























              up vote
              0
              down vote













              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























                up vote
                0
                down vote










                up vote
                0
                down vote









                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












                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 at 14:14









                AmbretteOrrisey

                54110




                54110






























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