Real Analysis Limits Question












5












$begingroup$


Currently preparing for a first course in analysis exam and my answer disagrees with the model solutions but I can't see where my logic fails



The question:



Find the limit of:



$$a_n :=frac{2^{3n}-n3^n}{n^{1729}+8^n}$$



My attempt:



First rewrite $8^n$ as $2^{3n}$
and divide through by $2^{3n}$



This gives us



$$= frac{1-nleft(frac{3}{8}right)^n}{large frac{n^{1729}}{2^{3n}}+1}$$



and then divide through by $n$ to give



$$= frac{frac{1}{n}-left(frac{3}{8}right)^n}{large frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$



My answer is zero using the following limits:



$lim_{ntoinfty}frac{1}{n} = 0$



and since $frac{3}{8} <1$ $implieslim_{ntoinfty}frac{3}{8}^n = 0$



The model solutions however suggest the limit is $1$, any help would be amazing.






sequences-and-series analysis







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    Currently preparing for a first course in analysis exam and my answer disagrees with the model solutions but I can't see where my logic fails



    The question:



    Find the limit of:



    $$a_n :=frac{2^{3n}-n3^n}{n^{1729}+8^n}$$



    My attempt:



    First rewrite $8^n$ as $2^{3n}$
    and divide through by $2^{3n}$



    This gives us



    $$= frac{1-nleft(frac{3}{8}right)^n}{large frac{n^{1729}}{2^{3n}}+1}$$



    and then divide through by $n$ to give



    $$= frac{frac{1}{n}-left(frac{3}{8}right)^n}{large frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$



    My answer is zero using the following limits:



    $lim_{ntoinfty}frac{1}{n} = 0$



    and since $frac{3}{8} <1$ $implieslim_{ntoinfty}frac{3}{8}^n = 0$



    The model solutions however suggest the limit is $1$, any help would be amazing.






    sequences-and-series analysis







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      0



      $begingroup$


      Currently preparing for a first course in analysis exam and my answer disagrees with the model solutions but I can't see where my logic fails



      The question:



      Find the limit of:



      $$a_n :=frac{2^{3n}-n3^n}{n^{1729}+8^n}$$



      My attempt:



      First rewrite $8^n$ as $2^{3n}$
      and divide through by $2^{3n}$



      This gives us



      $$= frac{1-nleft(frac{3}{8}right)^n}{large frac{n^{1729}}{2^{3n}}+1}$$



      and then divide through by $n$ to give



      $$= frac{frac{1}{n}-left(frac{3}{8}right)^n}{large frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$



      My answer is zero using the following limits:



      $lim_{ntoinfty}frac{1}{n} = 0$



      and since $frac{3}{8} <1$ $implieslim_{ntoinfty}frac{3}{8}^n = 0$



      The model solutions however suggest the limit is $1$, any help would be amazing.






      sequences-and-series analysis







      share|cite|improve this question











      $endgroup$




      Currently preparing for a first course in analysis exam and my answer disagrees with the model solutions but I can't see where my logic fails



      The question:



      Find the limit of:



      $$a_n :=frac{2^{3n}-n3^n}{n^{1729}+8^n}$$



      My attempt:



      First rewrite $8^n$ as $2^{3n}$
      and divide through by $2^{3n}$



      This gives us



      $$= frac{1-nleft(frac{3}{8}right)^n}{large frac{n^{1729}}{2^{3n}}+1}$$



      and then divide through by $n$ to give



      $$= frac{frac{1}{n}-left(frac{3}{8}right)^n}{large frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$



      My answer is zero using the following limits:



      $lim_{ntoinfty}frac{1}{n} = 0$



      and since $frac{3}{8} <1$ $implieslim_{ntoinfty}frac{3}{8}^n = 0$



      The model solutions however suggest the limit is $1$, any help would be amazing.






      sequences-and-series analysis




      sequences-and-series analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 at 12:35









      Chinnapparaj R

      5,7032928




      5,7032928










      asked Jan 3 at 12:24









      PolynomialCPolynomialC

      876




      876






















          2 Answers
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          active

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          11












          $begingroup$


          and then divide through by $n$ to give



          $$= frac{frac{1}{n}-(frac{3}{8})^n}{frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$



          My answer is zero using the following limits:



          $lim_{ntoinfty}frac{1}{n} = 0$



          and since $frac{3}{8} <1$ $implieslim_{ntoinfty}frac{3}{8}^n = 0$




          But not only the numerator tends to zero, the denominator does too...! That would leave you with an indeterminate form, so you cannot conclude (from this) that the limit is $0$.



          Instead, go one step back to the form:




          $$= frac{1-color{blue}{n(frac{3}{8})^n}}{color{red}{frac{n^{1729}}{2^{3n}}}+1}$$




          and try to reason why the blue and red expressions both tend to zero, so?






          share|cite|improve this answer









          $endgroup$





















            5












            $begingroup$

            The expression $$frac{frac{1}{n}-(frac{3}{8})^n}{frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$ cannot immediatelly be used to find the limit. Sure, the numerator tends to $0$, but the denominator does too.



            In fact, you should have stopped when you got to $$frac{1-n(frac{3}{8})^n}{frac{n^{1729}}{2^{3n}}+1}$$



            where you should notice that both the denominator and the numerator tend to $1$.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

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              active

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              active

              oldest

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              11












              $begingroup$


              and then divide through by $n$ to give



              $$= frac{frac{1}{n}-(frac{3}{8})^n}{frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$



              My answer is zero using the following limits:



              $lim_{ntoinfty}frac{1}{n} = 0$



              and since $frac{3}{8} <1$ $implieslim_{ntoinfty}frac{3}{8}^n = 0$




              But not only the numerator tends to zero, the denominator does too...! That would leave you with an indeterminate form, so you cannot conclude (from this) that the limit is $0$.



              Instead, go one step back to the form:




              $$= frac{1-color{blue}{n(frac{3}{8})^n}}{color{red}{frac{n^{1729}}{2^{3n}}}+1}$$




              and try to reason why the blue and red expressions both tend to zero, so?






              share|cite|improve this answer









              $endgroup$


















                11












                $begingroup$


                and then divide through by $n$ to give



                $$= frac{frac{1}{n}-(frac{3}{8})^n}{frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$



                My answer is zero using the following limits:



                $lim_{ntoinfty}frac{1}{n} = 0$



                and since $frac{3}{8} <1$ $implieslim_{ntoinfty}frac{3}{8}^n = 0$




                But not only the numerator tends to zero, the denominator does too...! That would leave you with an indeterminate form, so you cannot conclude (from this) that the limit is $0$.



                Instead, go one step back to the form:




                $$= frac{1-color{blue}{n(frac{3}{8})^n}}{color{red}{frac{n^{1729}}{2^{3n}}}+1}$$




                and try to reason why the blue and red expressions both tend to zero, so?






                share|cite|improve this answer









                $endgroup$
















                  11












                  11








                  11





                  $begingroup$


                  and then divide through by $n$ to give



                  $$= frac{frac{1}{n}-(frac{3}{8})^n}{frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$



                  My answer is zero using the following limits:



                  $lim_{ntoinfty}frac{1}{n} = 0$



                  and since $frac{3}{8} <1$ $implieslim_{ntoinfty}frac{3}{8}^n = 0$




                  But not only the numerator tends to zero, the denominator does too...! That would leave you with an indeterminate form, so you cannot conclude (from this) that the limit is $0$.



                  Instead, go one step back to the form:




                  $$= frac{1-color{blue}{n(frac{3}{8})^n}}{color{red}{frac{n^{1729}}{2^{3n}}}+1}$$




                  and try to reason why the blue and red expressions both tend to zero, so?






                  share|cite|improve this answer









                  $endgroup$




                  and then divide through by $n$ to give



                  $$= frac{frac{1}{n}-(frac{3}{8})^n}{frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$



                  My answer is zero using the following limits:



                  $lim_{ntoinfty}frac{1}{n} = 0$



                  and since $frac{3}{8} <1$ $implieslim_{ntoinfty}frac{3}{8}^n = 0$




                  But not only the numerator tends to zero, the denominator does too...! That would leave you with an indeterminate form, so you cannot conclude (from this) that the limit is $0$.



                  Instead, go one step back to the form:




                  $$= frac{1-color{blue}{n(frac{3}{8})^n}}{color{red}{frac{n^{1729}}{2^{3n}}}+1}$$




                  and try to reason why the blue and red expressions both tend to zero, so?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 3 at 12:28









                  StackTDStackTD

                  23.2k2153




                  23.2k2153























                      5












                      $begingroup$

                      The expression $$frac{frac{1}{n}-(frac{3}{8})^n}{frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$ cannot immediatelly be used to find the limit. Sure, the numerator tends to $0$, but the denominator does too.



                      In fact, you should have stopped when you got to $$frac{1-n(frac{3}{8})^n}{frac{n^{1729}}{2^{3n}}+1}$$



                      where you should notice that both the denominator and the numerator tend to $1$.






                      share|cite|improve this answer









                      $endgroup$


















                        5












                        $begingroup$

                        The expression $$frac{frac{1}{n}-(frac{3}{8})^n}{frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$ cannot immediatelly be used to find the limit. Sure, the numerator tends to $0$, but the denominator does too.



                        In fact, you should have stopped when you got to $$frac{1-n(frac{3}{8})^n}{frac{n^{1729}}{2^{3n}}+1}$$



                        where you should notice that both the denominator and the numerator tend to $1$.






                        share|cite|improve this answer









                        $endgroup$
















                          5












                          5








                          5





                          $begingroup$

                          The expression $$frac{frac{1}{n}-(frac{3}{8})^n}{frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$ cannot immediatelly be used to find the limit. Sure, the numerator tends to $0$, but the denominator does too.



                          In fact, you should have stopped when you got to $$frac{1-n(frac{3}{8})^n}{frac{n^{1729}}{2^{3n}}+1}$$



                          where you should notice that both the denominator and the numerator tend to $1$.






                          share|cite|improve this answer









                          $endgroup$



                          The expression $$frac{frac{1}{n}-(frac{3}{8})^n}{frac{n^{1728}}{2^{3n}}+frac{1}{n}}$$ cannot immediatelly be used to find the limit. Sure, the numerator tends to $0$, but the denominator does too.



                          In fact, you should have stopped when you got to $$frac{1-n(frac{3}{8})^n}{frac{n^{1729}}{2^{3n}}+1}$$



                          where you should notice that both the denominator and the numerator tend to $1$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 3 at 12:30









                          5xum5xum

                          91.3k394161




                          91.3k394161






























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