Evaluate $limlimits_{nto infty} sqrt[n]{frac{2^n+n}{n^2+1}}$












0












$begingroup$


$$lim_{nto infty} sqrt[n]{frac{2^n+n}{n^2+1}}$$



I know that we can say that:



$$lim_{nto infty} sqrt[n]{frac{2^n+n}{n^2+1}}approx lim_{nto infty} sqrt[n]{frac{2^n}{n^2}}=lim_{nto infty} {frac{2}{sqrt[n]{n^2}}}=2$$



Is there another way?






calculus limits







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  • 1




    $begingroup$
    You can make it more rigorous with squeeze theorem:$$frac{2^n}{2n^2}<frac{2^n+n}{n^2+1}<frac{2cdot2^n}{n^2}$$There is also ratio implies root:$$lim_{ntoinfty}left|frac{a_{n+1}}{a_n}right|=lim_{ntoinfty}sqrt[n]{|a_n|}$$provided the left limit exists.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 at 12:05










  • $begingroup$
    It's fine !!!!!!!
    $endgroup$
    – Felix Marin
    Jan 14 at 19:01
















0












$begingroup$


$$lim_{nto infty} sqrt[n]{frac{2^n+n}{n^2+1}}$$



I know that we can say that:



$$lim_{nto infty} sqrt[n]{frac{2^n+n}{n^2+1}}approx lim_{nto infty} sqrt[n]{frac{2^n}{n^2}}=lim_{nto infty} {frac{2}{sqrt[n]{n^2}}}=2$$



Is there another way?






calculus limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can make it more rigorous with squeeze theorem:$$frac{2^n}{2n^2}<frac{2^n+n}{n^2+1}<frac{2cdot2^n}{n^2}$$There is also ratio implies root:$$lim_{ntoinfty}left|frac{a_{n+1}}{a_n}right|=lim_{ntoinfty}sqrt[n]{|a_n|}$$provided the left limit exists.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 at 12:05










  • $begingroup$
    It's fine !!!!!!!
    $endgroup$
    – Felix Marin
    Jan 14 at 19:01














0












0








0


0



$begingroup$


$$lim_{nto infty} sqrt[n]{frac{2^n+n}{n^2+1}}$$



I know that we can say that:



$$lim_{nto infty} sqrt[n]{frac{2^n+n}{n^2+1}}approx lim_{nto infty} sqrt[n]{frac{2^n}{n^2}}=lim_{nto infty} {frac{2}{sqrt[n]{n^2}}}=2$$



Is there another way?






calculus limits







share|cite|improve this question











$endgroup$




$$lim_{nto infty} sqrt[n]{frac{2^n+n}{n^2+1}}$$



I know that we can say that:



$$lim_{nto infty} sqrt[n]{frac{2^n+n}{n^2+1}}approx lim_{nto infty} sqrt[n]{frac{2^n}{n^2}}=lim_{nto infty} {frac{2}{sqrt[n]{n^2}}}=2$$



Is there another way?






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 22:38









amWhy

1




1










asked Jan 14 at 11:48









newherenewhere

858411




858411








  • 1




    $begingroup$
    You can make it more rigorous with squeeze theorem:$$frac{2^n}{2n^2}<frac{2^n+n}{n^2+1}<frac{2cdot2^n}{n^2}$$There is also ratio implies root:$$lim_{ntoinfty}left|frac{a_{n+1}}{a_n}right|=lim_{ntoinfty}sqrt[n]{|a_n|}$$provided the left limit exists.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 at 12:05










  • $begingroup$
    It's fine !!!!!!!
    $endgroup$
    – Felix Marin
    Jan 14 at 19:01














  • 1




    $begingroup$
    You can make it more rigorous with squeeze theorem:$$frac{2^n}{2n^2}<frac{2^n+n}{n^2+1}<frac{2cdot2^n}{n^2}$$There is also ratio implies root:$$lim_{ntoinfty}left|frac{a_{n+1}}{a_n}right|=lim_{ntoinfty}sqrt[n]{|a_n|}$$provided the left limit exists.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 at 12:05










  • $begingroup$
    It's fine !!!!!!!
    $endgroup$
    – Felix Marin
    Jan 14 at 19:01








1




1




$begingroup$
You can make it more rigorous with squeeze theorem:$$frac{2^n}{2n^2}<frac{2^n+n}{n^2+1}<frac{2cdot2^n}{n^2}$$There is also ratio implies root:$$lim_{ntoinfty}left|frac{a_{n+1}}{a_n}right|=lim_{ntoinfty}sqrt[n]{|a_n|}$$provided the left limit exists.
$endgroup$
– Simply Beautiful Art
Jan 14 at 12:05




$begingroup$
You can make it more rigorous with squeeze theorem:$$frac{2^n}{2n^2}<frac{2^n+n}{n^2+1}<frac{2cdot2^n}{n^2}$$There is also ratio implies root:$$lim_{ntoinfty}left|frac{a_{n+1}}{a_n}right|=lim_{ntoinfty}sqrt[n]{|a_n|}$$provided the left limit exists.
$endgroup$
– Simply Beautiful Art
Jan 14 at 12:05












$begingroup$
It's fine !!!!!!!
$endgroup$
– Felix Marin
Jan 14 at 19:01




$begingroup$
It's fine !!!!!!!
$endgroup$
– Felix Marin
Jan 14 at 19:01










2 Answers
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$begingroup$

That approach is good, although $approx$ is a bit vague.



You can also use the fact that$$lim_{ntoinfty}sqrt[n]{frac{2^n+n}{n^2+1}}=lim_{ntoinfty}left(frac{2^n+n}{n^2+1}right)^frac1n=lim_{ntoinfty}expleft(frac{logleft(frac{2^n+n}{n^2+1}right)}nright).$$






share|cite|improve this answer









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    0












    $begingroup$

    By using the squeeze theorem:
    It is $2^n+nleq n^22^n+2^n$ so you get that your limit is $leq2$. On the other hand, $2^n+ngeq2^n$ and it is $(n^2+1)^{1/n}to1$, hence your limits is $geq2$.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      0












      $begingroup$

      That approach is good, although $approx$ is a bit vague.



      You can also use the fact that$$lim_{ntoinfty}sqrt[n]{frac{2^n+n}{n^2+1}}=lim_{ntoinfty}left(frac{2^n+n}{n^2+1}right)^frac1n=lim_{ntoinfty}expleft(frac{logleft(frac{2^n+n}{n^2+1}right)}nright).$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        That approach is good, although $approx$ is a bit vague.



        You can also use the fact that$$lim_{ntoinfty}sqrt[n]{frac{2^n+n}{n^2+1}}=lim_{ntoinfty}left(frac{2^n+n}{n^2+1}right)^frac1n=lim_{ntoinfty}expleft(frac{logleft(frac{2^n+n}{n^2+1}right)}nright).$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          That approach is good, although $approx$ is a bit vague.



          You can also use the fact that$$lim_{ntoinfty}sqrt[n]{frac{2^n+n}{n^2+1}}=lim_{ntoinfty}left(frac{2^n+n}{n^2+1}right)^frac1n=lim_{ntoinfty}expleft(frac{logleft(frac{2^n+n}{n^2+1}right)}nright).$$






          share|cite|improve this answer









          $endgroup$



          That approach is good, although $approx$ is a bit vague.



          You can also use the fact that$$lim_{ntoinfty}sqrt[n]{frac{2^n+n}{n^2+1}}=lim_{ntoinfty}left(frac{2^n+n}{n^2+1}right)^frac1n=lim_{ntoinfty}expleft(frac{logleft(frac{2^n+n}{n^2+1}right)}nright).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 14 at 12:01









          José Carlos SantosJosé Carlos Santos

          161k22128232




          161k22128232























              0












              $begingroup$

              By using the squeeze theorem:
              It is $2^n+nleq n^22^n+2^n$ so you get that your limit is $leq2$. On the other hand, $2^n+ngeq2^n$ and it is $(n^2+1)^{1/n}to1$, hence your limits is $geq2$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                By using the squeeze theorem:
                It is $2^n+nleq n^22^n+2^n$ so you get that your limit is $leq2$. On the other hand, $2^n+ngeq2^n$ and it is $(n^2+1)^{1/n}to1$, hence your limits is $geq2$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  By using the squeeze theorem:
                  It is $2^n+nleq n^22^n+2^n$ so you get that your limit is $leq2$. On the other hand, $2^n+ngeq2^n$ and it is $(n^2+1)^{1/n}to1$, hence your limits is $geq2$.






                  share|cite|improve this answer









                  $endgroup$



                  By using the squeeze theorem:
                  It is $2^n+nleq n^22^n+2^n$ so you get that your limit is $leq2$. On the other hand, $2^n+ngeq2^n$ and it is $(n^2+1)^{1/n}to1$, hence your limits is $geq2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 14 at 12:06









                  JustDroppedInJustDroppedIn

                  2,044420




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