Question About Intermediate Step in finding a limit












0














So I have a question in finding the limit of this sequence:



Let $$a_n :=(4^{10}+n2^n)^frac{1}{n}$$



My attempt:



Take out a factor of n and use the result $(n)^frac{1}{n}to1$ as $ntoinfty$



So we get $$a_n :=(frac{2^{20}}{n}+2^n)^frac{1}{n}$$



Now here is my question, can I use the fact that $lim_{ntoinfty}frac{2^{20}}{n} = 0$



And substitute that into the brack to arrive to the answer of 2?$[(2^n)^frac{1}{n} = 2]$






real-analysis







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    0














    So I have a question in finding the limit of this sequence:



    Let $$a_n :=(4^{10}+n2^n)^frac{1}{n}$$



    My attempt:



    Take out a factor of n and use the result $(n)^frac{1}{n}to1$ as $ntoinfty$



    So we get $$a_n :=(frac{2^{20}}{n}+2^n)^frac{1}{n}$$



    Now here is my question, can I use the fact that $lim_{ntoinfty}frac{2^{20}}{n} = 0$



    And substitute that into the brack to arrive to the answer of 2?$[(2^n)^frac{1}{n} = 2]$






    real-analysis







    share|cite|improve this question

























      0












      0








      0


      1





      So I have a question in finding the limit of this sequence:



      Let $$a_n :=(4^{10}+n2^n)^frac{1}{n}$$



      My attempt:



      Take out a factor of n and use the result $(n)^frac{1}{n}to1$ as $ntoinfty$



      So we get $$a_n :=(frac{2^{20}}{n}+2^n)^frac{1}{n}$$



      Now here is my question, can I use the fact that $lim_{ntoinfty}frac{2^{20}}{n} = 0$



      And substitute that into the brack to arrive to the answer of 2?$[(2^n)^frac{1}{n} = 2]$






      real-analysis







      share|cite|improve this question













      So I have a question in finding the limit of this sequence:



      Let $$a_n :=(4^{10}+n2^n)^frac{1}{n}$$



      My attempt:



      Take out a factor of n and use the result $(n)^frac{1}{n}to1$ as $ntoinfty$



      So we get $$a_n :=(frac{2^{20}}{n}+2^n)^frac{1}{n}$$



      Now here is my question, can I use the fact that $lim_{ntoinfty}frac{2^{20}}{n} = 0$



      And substitute that into the brack to arrive to the answer of 2?$[(2^n)^frac{1}{n} = 2]$






      real-analysis




      real-analysis






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      asked 2 days ago









      PolynomialC

      826




      826






















          2 Answers
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          oldest

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          1














          No you are not allowed to take the limit under a power of $n$.



          Here is an approach: For $n geq 20$



          $a_n =(4^{10}+n2^n)^frac{1}{n}=(2^{20}+n2^n)^frac{1}{n} leq (20.2^{20}+n2^n)^frac{1}{n} leq (n.2^{n}+n2^n)^frac{1}{n} leq (n.2^{n+1})^frac{1}{n}$



          and: $(n2^n)^frac{1}{n} leq a_n =(4^{10}+n2^n)^frac{1}{n} $



          Now use Squeeze theorem.






          share|cite|improve this answer





























            1














            No, you can’t treat the limits separately.



            For a very simple and well-known example demonstrating this, take



            $$lim_{n to infty}left(1+frac{x}{n}right)^n = e^x$$



            If you treat $frac{x}{n}$ separately and take the limit as $n to infty$, you will get $(1+0)^n = 1^n = 1$, which is incorrect, as the limit equals $e^x$.






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              active

              oldest

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              active

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              active

              oldest

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              1














              No you are not allowed to take the limit under a power of $n$.



              Here is an approach: For $n geq 20$



              $a_n =(4^{10}+n2^n)^frac{1}{n}=(2^{20}+n2^n)^frac{1}{n} leq (20.2^{20}+n2^n)^frac{1}{n} leq (n.2^{n}+n2^n)^frac{1}{n} leq (n.2^{n+1})^frac{1}{n}$



              and: $(n2^n)^frac{1}{n} leq a_n =(4^{10}+n2^n)^frac{1}{n} $



              Now use Squeeze theorem.






              share|cite|improve this answer


























                1














                No you are not allowed to take the limit under a power of $n$.



                Here is an approach: For $n geq 20$



                $a_n =(4^{10}+n2^n)^frac{1}{n}=(2^{20}+n2^n)^frac{1}{n} leq (20.2^{20}+n2^n)^frac{1}{n} leq (n.2^{n}+n2^n)^frac{1}{n} leq (n.2^{n+1})^frac{1}{n}$



                and: $(n2^n)^frac{1}{n} leq a_n =(4^{10}+n2^n)^frac{1}{n} $



                Now use Squeeze theorem.






                share|cite|improve this answer
























                  1












                  1








                  1






                  No you are not allowed to take the limit under a power of $n$.



                  Here is an approach: For $n geq 20$



                  $a_n =(4^{10}+n2^n)^frac{1}{n}=(2^{20}+n2^n)^frac{1}{n} leq (20.2^{20}+n2^n)^frac{1}{n} leq (n.2^{n}+n2^n)^frac{1}{n} leq (n.2^{n+1})^frac{1}{n}$



                  and: $(n2^n)^frac{1}{n} leq a_n =(4^{10}+n2^n)^frac{1}{n} $



                  Now use Squeeze theorem.






                  share|cite|improve this answer












                  No you are not allowed to take the limit under a power of $n$.



                  Here is an approach: For $n geq 20$



                  $a_n =(4^{10}+n2^n)^frac{1}{n}=(2^{20}+n2^n)^frac{1}{n} leq (20.2^{20}+n2^n)^frac{1}{n} leq (n.2^{n}+n2^n)^frac{1}{n} leq (n.2^{n+1})^frac{1}{n}$



                  and: $(n2^n)^frac{1}{n} leq a_n =(4^{10}+n2^n)^frac{1}{n} $



                  Now use Squeeze theorem.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  John11

                  1,0321821




                  1,0321821























                      1














                      No, you can’t treat the limits separately.



                      For a very simple and well-known example demonstrating this, take



                      $$lim_{n to infty}left(1+frac{x}{n}right)^n = e^x$$



                      If you treat $frac{x}{n}$ separately and take the limit as $n to infty$, you will get $(1+0)^n = 1^n = 1$, which is incorrect, as the limit equals $e^x$.






                      share|cite|improve this answer


























                        1














                        No, you can’t treat the limits separately.



                        For a very simple and well-known example demonstrating this, take



                        $$lim_{n to infty}left(1+frac{x}{n}right)^n = e^x$$



                        If you treat $frac{x}{n}$ separately and take the limit as $n to infty$, you will get $(1+0)^n = 1^n = 1$, which is incorrect, as the limit equals $e^x$.






                        share|cite|improve this answer
























                          1












                          1








                          1






                          No, you can’t treat the limits separately.



                          For a very simple and well-known example demonstrating this, take



                          $$lim_{n to infty}left(1+frac{x}{n}right)^n = e^x$$



                          If you treat $frac{x}{n}$ separately and take the limit as $n to infty$, you will get $(1+0)^n = 1^n = 1$, which is incorrect, as the limit equals $e^x$.






                          share|cite|improve this answer












                          No, you can’t treat the limits separately.



                          For a very simple and well-known example demonstrating this, take



                          $$lim_{n to infty}left(1+frac{x}{n}right)^n = e^x$$



                          If you treat $frac{x}{n}$ separately and take the limit as $n to infty$, you will get $(1+0)^n = 1^n = 1$, which is incorrect, as the limit equals $e^x$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 days ago









                          KM101

                          5,4711423




                          5,4711423






























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