Does $ (sqrt[n]n)^p to 1 textas n to infty textif p < 1 $?

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We know that $ sqrt[n]n to 1 textas n to infty $.



We can show that $ (sqrt[n]n)^p to 1 textas n to infty textif p geq 1 $



My question is-



Does $ (sqrt[n]n)^p to 1 textas n to infty textif p < 1 $ ?




sequences-and-series limits




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  • Why do you doubt the result when $p<1$?
    – Paramanand Singh
    Aug 29 at 10:51














up vote
2
down vote

favorite
1












We know that $ sqrt[n]n to 1 textas n to infty $.



We can show that $ (sqrt[n]n)^p to 1 textas n to infty textif p geq 1 $



My question is-



Does $ (sqrt[n]n)^p to 1 textas n to infty textif p < 1 $ ?




sequences-and-series limits




share|cite|improve this question






















  • Why do you doubt the result when $p<1$?
    – Paramanand Singh
    Aug 29 at 10:51












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





We know that $ sqrt[n]n to 1 textas n to infty $.



We can show that $ (sqrt[n]n)^p to 1 textas n to infty textif p geq 1 $



My question is-



Does $ (sqrt[n]n)^p to 1 textas n to infty textif p < 1 $ ?




sequences-and-series limits




share|cite|improve this question














We know that $ sqrt[n]n to 1 textas n to infty $.



We can show that $ (sqrt[n]n)^p to 1 textas n to infty textif p geq 1 $



My question is-



Does $ (sqrt[n]n)^p to 1 textas n to infty textif p < 1 $ ?





sequences-and-series limits





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 29 at 8:47









gimusi

71.2k73786




71.2k73786










asked Aug 29 at 8:23









yourmath

1,8111617




1,8111617











  • Why do you doubt the result when $p<1$?
    – Paramanand Singh
    Aug 29 at 10:51
















  • Why do you doubt the result when $p<1$?
    – Paramanand Singh
    Aug 29 at 10:51















Why do you doubt the result when $p<1$?
– Paramanand Singh
Aug 29 at 10:51




Why do you doubt the result when $p<1$?
– Paramanand Singh
Aug 29 at 10:51










3 Answers
3






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up vote
2
down vote



accepted










For $p=0$



$$(sqrt[n]n)^p=(sqrt[n]n)^0=1$$



For $pneq 0$



$$(sqrt[n]n)^p=e^frac plog nnto e^0=1$$






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    up vote
    4
    down vote













    The power function $x^p$ is continuous, so that



    $$lim_ntoinftyleft(sqrt[n]nright)^p=left(lim_ntoinftysqrt[n]nright)^p$$



    and $1^p=1$ for all $p$.






    share|cite|improve this answer
















    • 1




      for $ p<0 $ also
      – yourmath
      Aug 29 at 8:30






    • 2




      @yourmath: "all $p$" indeed includes the negatives.
      – Yves Daoust
      Aug 29 at 8:31

















    up vote
    4
    down vote













    Yes, for all $p in mathbb R$. Let $a_n = sqrt[n]n$, then $a_n to 1$ as you said, and by definition of continuity of $x mapsto x^p$, $a_n^p to 1^p=1$.






    share|cite|improve this answer




















      Your Answer




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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes








      up vote
      2
      down vote



      accepted










      For $p=0$



      $$(sqrt[n]n)^p=(sqrt[n]n)^0=1$$



      For $pneq 0$



      $$(sqrt[n]n)^p=e^frac plog nnto e^0=1$$






      share|cite|improve this answer


























        up vote
        2
        down vote



        accepted










        For $p=0$



        $$(sqrt[n]n)^p=(sqrt[n]n)^0=1$$



        For $pneq 0$



        $$(sqrt[n]n)^p=e^frac plog nnto e^0=1$$






        share|cite|improve this answer
























          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          For $p=0$



          $$(sqrt[n]n)^p=(sqrt[n]n)^0=1$$



          For $pneq 0$



          $$(sqrt[n]n)^p=e^frac plog nnto e^0=1$$






          share|cite|improve this answer














          For $p=0$



          $$(sqrt[n]n)^p=(sqrt[n]n)^0=1$$



          For $pneq 0$



          $$(sqrt[n]n)^p=e^frac plog nnto e^0=1$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 29 at 8:50

























          answered Aug 29 at 8:43









          gimusi

          71.2k73786




          71.2k73786




















              up vote
              4
              down vote













              The power function $x^p$ is continuous, so that



              $$lim_ntoinftyleft(sqrt[n]nright)^p=left(lim_ntoinftysqrt[n]nright)^p$$



              and $1^p=1$ for all $p$.






              share|cite|improve this answer
















              • 1




                for $ p<0 $ also
                – yourmath
                Aug 29 at 8:30






              • 2




                @yourmath: "all $p$" indeed includes the negatives.
                – Yves Daoust
                Aug 29 at 8:31














              up vote
              4
              down vote













              The power function $x^p$ is continuous, so that



              $$lim_ntoinftyleft(sqrt[n]nright)^p=left(lim_ntoinftysqrt[n]nright)^p$$



              and $1^p=1$ for all $p$.






              share|cite|improve this answer
















              • 1




                for $ p<0 $ also
                – yourmath
                Aug 29 at 8:30






              • 2




                @yourmath: "all $p$" indeed includes the negatives.
                – Yves Daoust
                Aug 29 at 8:31












              up vote
              4
              down vote










              up vote
              4
              down vote









              The power function $x^p$ is continuous, so that



              $$lim_ntoinftyleft(sqrt[n]nright)^p=left(lim_ntoinftysqrt[n]nright)^p$$



              and $1^p=1$ for all $p$.






              share|cite|improve this answer












              The power function $x^p$ is continuous, so that



              $$lim_ntoinftyleft(sqrt[n]nright)^p=left(lim_ntoinftysqrt[n]nright)^p$$



              and $1^p=1$ for all $p$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Aug 29 at 8:29









              Yves Daoust

              114k665208




              114k665208







              • 1




                for $ p<0 $ also
                – yourmath
                Aug 29 at 8:30






              • 2




                @yourmath: "all $p$" indeed includes the negatives.
                – Yves Daoust
                Aug 29 at 8:31












              • 1




                for $ p<0 $ also
                – yourmath
                Aug 29 at 8:30






              • 2




                @yourmath: "all $p$" indeed includes the negatives.
                – Yves Daoust
                Aug 29 at 8:31







              1




              1




              for $ p<0 $ also
              – yourmath
              Aug 29 at 8:30




              for $ p<0 $ also
              – yourmath
              Aug 29 at 8:30




              2




              2




              @yourmath: "all $p$" indeed includes the negatives.
              – Yves Daoust
              Aug 29 at 8:31




              @yourmath: "all $p$" indeed includes the negatives.
              – Yves Daoust
              Aug 29 at 8:31










              up vote
              4
              down vote













              Yes, for all $p in mathbb R$. Let $a_n = sqrt[n]n$, then $a_n to 1$ as you said, and by definition of continuity of $x mapsto x^p$, $a_n^p to 1^p=1$.






              share|cite|improve this answer
























                up vote
                4
                down vote













                Yes, for all $p in mathbb R$. Let $a_n = sqrt[n]n$, then $a_n to 1$ as you said, and by definition of continuity of $x mapsto x^p$, $a_n^p to 1^p=1$.






                share|cite|improve this answer






















                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  Yes, for all $p in mathbb R$. Let $a_n = sqrt[n]n$, then $a_n to 1$ as you said, and by definition of continuity of $x mapsto x^p$, $a_n^p to 1^p=1$.






                  share|cite|improve this answer












                  Yes, for all $p in mathbb R$. Let $a_n = sqrt[n]n$, then $a_n to 1$ as you said, and by definition of continuity of $x mapsto x^p$, $a_n^p to 1^p=1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 29 at 8:30









                  barto

                  13.4k32581




                  13.4k32581



























                       

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