Real Analysis Limits Question












2














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





























    2














    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



























      2












      2








      2


      1





      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















      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




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      edited 1 hour ago









      Chinnapparaj R

      5,2601826




      5,2601826










      asked 1 hour ago









      PolynomialC

      656




      656






















          2 Answers
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          4















          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





























            2














            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





















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              2 Answers
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              2 Answers
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              4















              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


























                4















                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
























                  4












                  4








                  4







                  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













                  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 1 hour ago









                  StackTD

                  22.1k1947




                  22.1k1947























                      2














                      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


























                        2














                        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
























                          2












                          2








                          2






                          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












                          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 1 hour ago









                          5xum

                          89.6k393161




                          89.6k393161






























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