Verification of $lim_{n rightarrow infty} sqrt[n]{n^3}=1$ [duplicate]











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  • Proof that $lim_{nrightarrow infty} sqrt[n]{n}=1$

    6 answers




I am interested in the limit




$$ lim_{n rightarrow infty} sqrt[n]{n^3}$$




Can we simply conclude that:
$$ lim_{n rightarrow infty} (sqrt[n]{n})^3= 1^3=1.$$
I have proven that $sqrt[n]{n}rightarrow1$ earlier in this textbook. Also since the limit of a power is the power of the limit.










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marked as duplicate by Nosrati, Community Nov 25 at 6:03


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    Yes, if you already know $sqrt[n]nto 1$
    – Hagen von Eitzen
    Nov 21 at 21:08










  • This is a lemma the book had us prove earlier, also the limit of a power is the power of the limit.
    – Wesley Strik
    Nov 21 at 21:08












  • Please include some text in your title besides just mathjax formatting.
    – amWhy
    Nov 21 at 21:13










  • Thanks Hagen, I do now that :)
    – Wesley Strik
    Nov 21 at 21:18










  • "Also since the limit of a power is the power of the limit." Has that been proven? If so... you are good. If not, you must justify it (it is true by the way).
    – fleablood
    Nov 21 at 22:19















up vote
0
down vote

favorite













This question already has an answer here:




  • Proof that $lim_{nrightarrow infty} sqrt[n]{n}=1$

    6 answers




I am interested in the limit




$$ lim_{n rightarrow infty} sqrt[n]{n^3}$$




Can we simply conclude that:
$$ lim_{n rightarrow infty} (sqrt[n]{n})^3= 1^3=1.$$
I have proven that $sqrt[n]{n}rightarrow1$ earlier in this textbook. Also since the limit of a power is the power of the limit.










share|cite|improve this question















marked as duplicate by Nosrati, Community Nov 25 at 6:03


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    Yes, if you already know $sqrt[n]nto 1$
    – Hagen von Eitzen
    Nov 21 at 21:08










  • This is a lemma the book had us prove earlier, also the limit of a power is the power of the limit.
    – Wesley Strik
    Nov 21 at 21:08












  • Please include some text in your title besides just mathjax formatting.
    – amWhy
    Nov 21 at 21:13










  • Thanks Hagen, I do now that :)
    – Wesley Strik
    Nov 21 at 21:18










  • "Also since the limit of a power is the power of the limit." Has that been proven? If so... you are good. If not, you must justify it (it is true by the way).
    – fleablood
    Nov 21 at 22:19













up vote
0
down vote

favorite









up vote
0
down vote

favorite












This question already has an answer here:




  • Proof that $lim_{nrightarrow infty} sqrt[n]{n}=1$

    6 answers




I am interested in the limit




$$ lim_{n rightarrow infty} sqrt[n]{n^3}$$




Can we simply conclude that:
$$ lim_{n rightarrow infty} (sqrt[n]{n})^3= 1^3=1.$$
I have proven that $sqrt[n]{n}rightarrow1$ earlier in this textbook. Also since the limit of a power is the power of the limit.










share|cite|improve this question
















This question already has an answer here:




  • Proof that $lim_{nrightarrow infty} sqrt[n]{n}=1$

    6 answers




I am interested in the limit




$$ lim_{n rightarrow infty} sqrt[n]{n^3}$$




Can we simply conclude that:
$$ lim_{n rightarrow infty} (sqrt[n]{n})^3= 1^3=1.$$
I have proven that $sqrt[n]{n}rightarrow1$ earlier in this textbook. Also since the limit of a power is the power of the limit.





This question already has an answer here:




  • Proof that $lim_{nrightarrow infty} sqrt[n]{n}=1$

    6 answers








real-analysis limits






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share|cite|improve this question













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edited Nov 21 at 21:14









amWhy

191k28223439




191k28223439










asked Nov 21 at 21:06









Wesley Strik

1,373322




1,373322




marked as duplicate by Nosrati, Community Nov 25 at 6:03


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Nosrati, Community Nov 25 at 6:03


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    Yes, if you already know $sqrt[n]nto 1$
    – Hagen von Eitzen
    Nov 21 at 21:08










  • This is a lemma the book had us prove earlier, also the limit of a power is the power of the limit.
    – Wesley Strik
    Nov 21 at 21:08












  • Please include some text in your title besides just mathjax formatting.
    – amWhy
    Nov 21 at 21:13










  • Thanks Hagen, I do now that :)
    – Wesley Strik
    Nov 21 at 21:18










  • "Also since the limit of a power is the power of the limit." Has that been proven? If so... you are good. If not, you must justify it (it is true by the way).
    – fleablood
    Nov 21 at 22:19














  • 1




    Yes, if you already know $sqrt[n]nto 1$
    – Hagen von Eitzen
    Nov 21 at 21:08










  • This is a lemma the book had us prove earlier, also the limit of a power is the power of the limit.
    – Wesley Strik
    Nov 21 at 21:08












  • Please include some text in your title besides just mathjax formatting.
    – amWhy
    Nov 21 at 21:13










  • Thanks Hagen, I do now that :)
    – Wesley Strik
    Nov 21 at 21:18










  • "Also since the limit of a power is the power of the limit." Has that been proven? If so... you are good. If not, you must justify it (it is true by the way).
    – fleablood
    Nov 21 at 22:19








1




1




Yes, if you already know $sqrt[n]nto 1$
– Hagen von Eitzen
Nov 21 at 21:08




Yes, if you already know $sqrt[n]nto 1$
– Hagen von Eitzen
Nov 21 at 21:08












This is a lemma the book had us prove earlier, also the limit of a power is the power of the limit.
– Wesley Strik
Nov 21 at 21:08






This is a lemma the book had us prove earlier, also the limit of a power is the power of the limit.
– Wesley Strik
Nov 21 at 21:08














Please include some text in your title besides just mathjax formatting.
– amWhy
Nov 21 at 21:13




Please include some text in your title besides just mathjax formatting.
– amWhy
Nov 21 at 21:13












Thanks Hagen, I do now that :)
– Wesley Strik
Nov 21 at 21:18




Thanks Hagen, I do now that :)
– Wesley Strik
Nov 21 at 21:18












"Also since the limit of a power is the power of the limit." Has that been proven? If so... you are good. If not, you must justify it (it is true by the way).
– fleablood
Nov 21 at 22:19




"Also since the limit of a power is the power of the limit." Has that been proven? If so... you are good. If not, you must justify it (it is true by the way).
– fleablood
Nov 21 at 22:19










5 Answers
5






active

oldest

votes

















up vote
2
down vote



accepted










Yes, we can simply do that. Since the exponent $^3$ is a constant neutral number (meaning we may interpret it as a fixed number of multiplications) we can move the limit inside of it. So if you already know $lim_{ntoinfty}sqrt[n]n=1$ then that's a full proof.






share|cite|improve this answer























  • This answer is insufficient. It is not enough to state that the exponent is constant. An essential property is missing.
    – Yves Daoust
    Nov 22 at 10:06












  • @YvesDaoust Raising to the power of a constant is continuous around $1$?
    – Arthur
    Nov 22 at 10:46










  • This is precisely the missing property. By the way, even a variable exponent would yield a continuous function.
    – Yves Daoust
    Nov 22 at 14:37












  • @YvesDaoust Sure you can do it with a variable exponent as well, but then you have to be extra careful. You can't put the limit inside of, say, $sqrt[n]{n}^n$ and expect to get a sensible answer. You need a constant exponent for that (or at least an exponent converging to a nice value). And I could argue for this without continuity of exponentiation at all. Just using that $sqrt[n]{n}^3 = sqrt[n]{n}cdotsqrt[n]{n}cdotsqrt[n]{n}$, and you can take the limits of those separately (something which is usually proven quite early and taken as implicit).
    – Arthur
    Nov 22 at 14:40












  • What I mean is that "since the exponent $^3$ is constant" is not the reason.
    – Yves Daoust
    Nov 22 at 15:00


















up vote
2
down vote













Yes we are allowed to do that since for continuity



$$lim_{xto x_0} f(x)=Lin mathbb{R} implies lim_{xto x_0} [f(x)]^k=left[lim_{xto x_0} [f(x)right]^k=L^k$$



Indeed in that particular case since $sqrt[n]{n}to 1$



$$forall epsilon>0 quad exists bar n quad forall n>bar n quad |sqrt[n]{n}-1|<epsilon$$



we have that, assuming $sqrt[n]{n}<2$, $forall bar epsilon=7epsilon>0$



$$|(sqrt[n]{n})^3-1|=|sqrt[n]{n}-1|cdot |sqrt[n]{n^2}+sqrt[n]{n}+1|< 7epsilon=bar epsilon quad forall n>bar n$$



and then $(sqrt[n]{n})^3to 1$.






share|cite|improve this answer



















  • 1




    You would need this to be an "if and only if" for this to apply here.
    – Will Sherwood
    Nov 21 at 21:15










  • @WillSherwood Yes but in that case the asker was interested in that particular implication. Do you think that this is the motivation for the downvoting? Thanks
    – gimusi
    Nov 21 at 21:24


















up vote
2
down vote













You know that
$sqrt[n]{n}
to 1$
.



Also
$sqrt[n]{n^k}
=(sqrt[n]{n})^k
$

so



$begin{array}\
sqrt[n]{n^k}-1
&=(sqrt[n]{n})^k-1\
&=(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j\
text{so}\
|sqrt[n]{n^k}-1|
&=|(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j|\
&=|(sqrt[n]{n}-1)||sum_{j=0}^{k-1} sqrt[n]{n}^j|\
&le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^j|\
&le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^n|\
&=|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |n|\
&=|(sqrt[n]{n}-1)|kn\
end{array}
$



Therefore,
to make
$|sqrt[n]{n^k}-1|
le epsilon$
,
choose $n$ large enough
so that
$|(sqrt[n]{n}-1)|
le frac{epsilon}{kn}
$
.






share|cite|improve this answer




























    up vote
    2
    down vote













    You can if you know four things.



    1) If $f$ is continuous and $a_n to M$ and for all $a_n$ that $f(a_n)$ is defined and $f(M)$ is defined then $a_nto M$ means $f(a_n) to f(M)$.



    But you DO have to prove this sometime. I'm sure your text has proven that somewhere.



    And



    2) you need to know that $sqrt[n]{n} to 1$.



    But you claim you have already shown that.



    And 3) that $sqrt[n]{n^3} = (sqrt[n]{n})^3$.



    Which is basic. It follows that as for all $M > 0$ and $nin mathbb N$ there is a unique $k = sqrt[n]{M}$ so that $k^n = M$. And as $(k^j)^n= (k^n)^j = M^j; for j in mathbb N$ it follows that $(sqrt[n]{M})^j= sqrt[n]{M^3}$.



    And finally you need to know 4) that $()^3: mathbb R to mathbb R$ is continuous.



    Which is .... basic. But you should have proven that sometime.



    So $lim sqrt[n]{n^3} = lim (sqrt[n]{n})^3 = (lim sqrt[n]{n})^3 = 1^3 =1$.






    share|cite|improve this answer




























      up vote
      2
      down vote













      Yes, because the cubic root function is continuous, so that you can swap the limit and the root.





      Continuity at $1$ is ensured by the fact that



      $$|sqrt[3]{1+delta}-1|=left|frac{delta}{(sqrt[3]{1+delta})^2+sqrt[3]{1+delta}+1}right|<fracdelta3<epsilon$$ holds with $delta<3epsilon$.






      share|cite|improve this answer






























        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote



        accepted










        Yes, we can simply do that. Since the exponent $^3$ is a constant neutral number (meaning we may interpret it as a fixed number of multiplications) we can move the limit inside of it. So if you already know $lim_{ntoinfty}sqrt[n]n=1$ then that's a full proof.






        share|cite|improve this answer























        • This answer is insufficient. It is not enough to state that the exponent is constant. An essential property is missing.
          – Yves Daoust
          Nov 22 at 10:06












        • @YvesDaoust Raising to the power of a constant is continuous around $1$?
          – Arthur
          Nov 22 at 10:46










        • This is precisely the missing property. By the way, even a variable exponent would yield a continuous function.
          – Yves Daoust
          Nov 22 at 14:37












        • @YvesDaoust Sure you can do it with a variable exponent as well, but then you have to be extra careful. You can't put the limit inside of, say, $sqrt[n]{n}^n$ and expect to get a sensible answer. You need a constant exponent for that (or at least an exponent converging to a nice value). And I could argue for this without continuity of exponentiation at all. Just using that $sqrt[n]{n}^3 = sqrt[n]{n}cdotsqrt[n]{n}cdotsqrt[n]{n}$, and you can take the limits of those separately (something which is usually proven quite early and taken as implicit).
          – Arthur
          Nov 22 at 14:40












        • What I mean is that "since the exponent $^3$ is constant" is not the reason.
          – Yves Daoust
          Nov 22 at 15:00















        up vote
        2
        down vote



        accepted










        Yes, we can simply do that. Since the exponent $^3$ is a constant neutral number (meaning we may interpret it as a fixed number of multiplications) we can move the limit inside of it. So if you already know $lim_{ntoinfty}sqrt[n]n=1$ then that's a full proof.






        share|cite|improve this answer























        • This answer is insufficient. It is not enough to state that the exponent is constant. An essential property is missing.
          – Yves Daoust
          Nov 22 at 10:06












        • @YvesDaoust Raising to the power of a constant is continuous around $1$?
          – Arthur
          Nov 22 at 10:46










        • This is precisely the missing property. By the way, even a variable exponent would yield a continuous function.
          – Yves Daoust
          Nov 22 at 14:37












        • @YvesDaoust Sure you can do it with a variable exponent as well, but then you have to be extra careful. You can't put the limit inside of, say, $sqrt[n]{n}^n$ and expect to get a sensible answer. You need a constant exponent for that (or at least an exponent converging to a nice value). And I could argue for this without continuity of exponentiation at all. Just using that $sqrt[n]{n}^3 = sqrt[n]{n}cdotsqrt[n]{n}cdotsqrt[n]{n}$, and you can take the limits of those separately (something which is usually proven quite early and taken as implicit).
          – Arthur
          Nov 22 at 14:40












        • What I mean is that "since the exponent $^3$ is constant" is not the reason.
          – Yves Daoust
          Nov 22 at 15:00













        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Yes, we can simply do that. Since the exponent $^3$ is a constant neutral number (meaning we may interpret it as a fixed number of multiplications) we can move the limit inside of it. So if you already know $lim_{ntoinfty}sqrt[n]n=1$ then that's a full proof.






        share|cite|improve this answer














        Yes, we can simply do that. Since the exponent $^3$ is a constant neutral number (meaning we may interpret it as a fixed number of multiplications) we can move the limit inside of it. So if you already know $lim_{ntoinfty}sqrt[n]n=1$ then that's a full proof.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 22 at 15:13

























        answered Nov 21 at 21:08









        Arthur

        110k7104186




        110k7104186












        • This answer is insufficient. It is not enough to state that the exponent is constant. An essential property is missing.
          – Yves Daoust
          Nov 22 at 10:06












        • @YvesDaoust Raising to the power of a constant is continuous around $1$?
          – Arthur
          Nov 22 at 10:46










        • This is precisely the missing property. By the way, even a variable exponent would yield a continuous function.
          – Yves Daoust
          Nov 22 at 14:37












        • @YvesDaoust Sure you can do it with a variable exponent as well, but then you have to be extra careful. You can't put the limit inside of, say, $sqrt[n]{n}^n$ and expect to get a sensible answer. You need a constant exponent for that (or at least an exponent converging to a nice value). And I could argue for this without continuity of exponentiation at all. Just using that $sqrt[n]{n}^3 = sqrt[n]{n}cdotsqrt[n]{n}cdotsqrt[n]{n}$, and you can take the limits of those separately (something which is usually proven quite early and taken as implicit).
          – Arthur
          Nov 22 at 14:40












        • What I mean is that "since the exponent $^3$ is constant" is not the reason.
          – Yves Daoust
          Nov 22 at 15:00


















        • This answer is insufficient. It is not enough to state that the exponent is constant. An essential property is missing.
          – Yves Daoust
          Nov 22 at 10:06












        • @YvesDaoust Raising to the power of a constant is continuous around $1$?
          – Arthur
          Nov 22 at 10:46










        • This is precisely the missing property. By the way, even a variable exponent would yield a continuous function.
          – Yves Daoust
          Nov 22 at 14:37












        • @YvesDaoust Sure you can do it with a variable exponent as well, but then you have to be extra careful. You can't put the limit inside of, say, $sqrt[n]{n}^n$ and expect to get a sensible answer. You need a constant exponent for that (or at least an exponent converging to a nice value). And I could argue for this without continuity of exponentiation at all. Just using that $sqrt[n]{n}^3 = sqrt[n]{n}cdotsqrt[n]{n}cdotsqrt[n]{n}$, and you can take the limits of those separately (something which is usually proven quite early and taken as implicit).
          – Arthur
          Nov 22 at 14:40












        • What I mean is that "since the exponent $^3$ is constant" is not the reason.
          – Yves Daoust
          Nov 22 at 15:00
















        This answer is insufficient. It is not enough to state that the exponent is constant. An essential property is missing.
        – Yves Daoust
        Nov 22 at 10:06






        This answer is insufficient. It is not enough to state that the exponent is constant. An essential property is missing.
        – Yves Daoust
        Nov 22 at 10:06














        @YvesDaoust Raising to the power of a constant is continuous around $1$?
        – Arthur
        Nov 22 at 10:46




        @YvesDaoust Raising to the power of a constant is continuous around $1$?
        – Arthur
        Nov 22 at 10:46












        This is precisely the missing property. By the way, even a variable exponent would yield a continuous function.
        – Yves Daoust
        Nov 22 at 14:37






        This is precisely the missing property. By the way, even a variable exponent would yield a continuous function.
        – Yves Daoust
        Nov 22 at 14:37














        @YvesDaoust Sure you can do it with a variable exponent as well, but then you have to be extra careful. You can't put the limit inside of, say, $sqrt[n]{n}^n$ and expect to get a sensible answer. You need a constant exponent for that (or at least an exponent converging to a nice value). And I could argue for this without continuity of exponentiation at all. Just using that $sqrt[n]{n}^3 = sqrt[n]{n}cdotsqrt[n]{n}cdotsqrt[n]{n}$, and you can take the limits of those separately (something which is usually proven quite early and taken as implicit).
        – Arthur
        Nov 22 at 14:40






        @YvesDaoust Sure you can do it with a variable exponent as well, but then you have to be extra careful. You can't put the limit inside of, say, $sqrt[n]{n}^n$ and expect to get a sensible answer. You need a constant exponent for that (or at least an exponent converging to a nice value). And I could argue for this without continuity of exponentiation at all. Just using that $sqrt[n]{n}^3 = sqrt[n]{n}cdotsqrt[n]{n}cdotsqrt[n]{n}$, and you can take the limits of those separately (something which is usually proven quite early and taken as implicit).
        – Arthur
        Nov 22 at 14:40














        What I mean is that "since the exponent $^3$ is constant" is not the reason.
        – Yves Daoust
        Nov 22 at 15:00




        What I mean is that "since the exponent $^3$ is constant" is not the reason.
        – Yves Daoust
        Nov 22 at 15:00










        up vote
        2
        down vote













        Yes we are allowed to do that since for continuity



        $$lim_{xto x_0} f(x)=Lin mathbb{R} implies lim_{xto x_0} [f(x)]^k=left[lim_{xto x_0} [f(x)right]^k=L^k$$



        Indeed in that particular case since $sqrt[n]{n}to 1$



        $$forall epsilon>0 quad exists bar n quad forall n>bar n quad |sqrt[n]{n}-1|<epsilon$$



        we have that, assuming $sqrt[n]{n}<2$, $forall bar epsilon=7epsilon>0$



        $$|(sqrt[n]{n})^3-1|=|sqrt[n]{n}-1|cdot |sqrt[n]{n^2}+sqrt[n]{n}+1|< 7epsilon=bar epsilon quad forall n>bar n$$



        and then $(sqrt[n]{n})^3to 1$.






        share|cite|improve this answer



















        • 1




          You would need this to be an "if and only if" for this to apply here.
          – Will Sherwood
          Nov 21 at 21:15










        • @WillSherwood Yes but in that case the asker was interested in that particular implication. Do you think that this is the motivation for the downvoting? Thanks
          – gimusi
          Nov 21 at 21:24















        up vote
        2
        down vote













        Yes we are allowed to do that since for continuity



        $$lim_{xto x_0} f(x)=Lin mathbb{R} implies lim_{xto x_0} [f(x)]^k=left[lim_{xto x_0} [f(x)right]^k=L^k$$



        Indeed in that particular case since $sqrt[n]{n}to 1$



        $$forall epsilon>0 quad exists bar n quad forall n>bar n quad |sqrt[n]{n}-1|<epsilon$$



        we have that, assuming $sqrt[n]{n}<2$, $forall bar epsilon=7epsilon>0$



        $$|(sqrt[n]{n})^3-1|=|sqrt[n]{n}-1|cdot |sqrt[n]{n^2}+sqrt[n]{n}+1|< 7epsilon=bar epsilon quad forall n>bar n$$



        and then $(sqrt[n]{n})^3to 1$.






        share|cite|improve this answer



















        • 1




          You would need this to be an "if and only if" for this to apply here.
          – Will Sherwood
          Nov 21 at 21:15










        • @WillSherwood Yes but in that case the asker was interested in that particular implication. Do you think that this is the motivation for the downvoting? Thanks
          – gimusi
          Nov 21 at 21:24













        up vote
        2
        down vote










        up vote
        2
        down vote









        Yes we are allowed to do that since for continuity



        $$lim_{xto x_0} f(x)=Lin mathbb{R} implies lim_{xto x_0} [f(x)]^k=left[lim_{xto x_0} [f(x)right]^k=L^k$$



        Indeed in that particular case since $sqrt[n]{n}to 1$



        $$forall epsilon>0 quad exists bar n quad forall n>bar n quad |sqrt[n]{n}-1|<epsilon$$



        we have that, assuming $sqrt[n]{n}<2$, $forall bar epsilon=7epsilon>0$



        $$|(sqrt[n]{n})^3-1|=|sqrt[n]{n}-1|cdot |sqrt[n]{n^2}+sqrt[n]{n}+1|< 7epsilon=bar epsilon quad forall n>bar n$$



        and then $(sqrt[n]{n})^3to 1$.






        share|cite|improve this answer














        Yes we are allowed to do that since for continuity



        $$lim_{xto x_0} f(x)=Lin mathbb{R} implies lim_{xto x_0} [f(x)]^k=left[lim_{xto x_0} [f(x)right]^k=L^k$$



        Indeed in that particular case since $sqrt[n]{n}to 1$



        $$forall epsilon>0 quad exists bar n quad forall n>bar n quad |sqrt[n]{n}-1|<epsilon$$



        we have that, assuming $sqrt[n]{n}<2$, $forall bar epsilon=7epsilon>0$



        $$|(sqrt[n]{n})^3-1|=|sqrt[n]{n}-1|cdot |sqrt[n]{n^2}+sqrt[n]{n}+1|< 7epsilon=bar epsilon quad forall n>bar n$$



        and then $(sqrt[n]{n})^3to 1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 21 at 21:36

























        answered Nov 21 at 21:10









        gimusi

        91.7k84495




        91.7k84495








        • 1




          You would need this to be an "if and only if" for this to apply here.
          – Will Sherwood
          Nov 21 at 21:15










        • @WillSherwood Yes but in that case the asker was interested in that particular implication. Do you think that this is the motivation for the downvoting? Thanks
          – gimusi
          Nov 21 at 21:24














        • 1




          You would need this to be an "if and only if" for this to apply here.
          – Will Sherwood
          Nov 21 at 21:15










        • @WillSherwood Yes but in that case the asker was interested in that particular implication. Do you think that this is the motivation for the downvoting? Thanks
          – gimusi
          Nov 21 at 21:24








        1




        1




        You would need this to be an "if and only if" for this to apply here.
        – Will Sherwood
        Nov 21 at 21:15




        You would need this to be an "if and only if" for this to apply here.
        – Will Sherwood
        Nov 21 at 21:15












        @WillSherwood Yes but in that case the asker was interested in that particular implication. Do you think that this is the motivation for the downvoting? Thanks
        – gimusi
        Nov 21 at 21:24




        @WillSherwood Yes but in that case the asker was interested in that particular implication. Do you think that this is the motivation for the downvoting? Thanks
        – gimusi
        Nov 21 at 21:24










        up vote
        2
        down vote













        You know that
        $sqrt[n]{n}
        to 1$
        .



        Also
        $sqrt[n]{n^k}
        =(sqrt[n]{n})^k
        $

        so



        $begin{array}\
        sqrt[n]{n^k}-1
        &=(sqrt[n]{n})^k-1\
        &=(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j\
        text{so}\
        |sqrt[n]{n^k}-1|
        &=|(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j|\
        &=|(sqrt[n]{n}-1)||sum_{j=0}^{k-1} sqrt[n]{n}^j|\
        &le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^j|\
        &le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^n|\
        &=|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |n|\
        &=|(sqrt[n]{n}-1)|kn\
        end{array}
        $



        Therefore,
        to make
        $|sqrt[n]{n^k}-1|
        le epsilon$
        ,
        choose $n$ large enough
        so that
        $|(sqrt[n]{n}-1)|
        le frac{epsilon}{kn}
        $
        .






        share|cite|improve this answer

























          up vote
          2
          down vote













          You know that
          $sqrt[n]{n}
          to 1$
          .



          Also
          $sqrt[n]{n^k}
          =(sqrt[n]{n})^k
          $

          so



          $begin{array}\
          sqrt[n]{n^k}-1
          &=(sqrt[n]{n})^k-1\
          &=(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j\
          text{so}\
          |sqrt[n]{n^k}-1|
          &=|(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j|\
          &=|(sqrt[n]{n}-1)||sum_{j=0}^{k-1} sqrt[n]{n}^j|\
          &le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^j|\
          &le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^n|\
          &=|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |n|\
          &=|(sqrt[n]{n}-1)|kn\
          end{array}
          $



          Therefore,
          to make
          $|sqrt[n]{n^k}-1|
          le epsilon$
          ,
          choose $n$ large enough
          so that
          $|(sqrt[n]{n}-1)|
          le frac{epsilon}{kn}
          $
          .






          share|cite|improve this answer























            up vote
            2
            down vote










            up vote
            2
            down vote









            You know that
            $sqrt[n]{n}
            to 1$
            .



            Also
            $sqrt[n]{n^k}
            =(sqrt[n]{n})^k
            $

            so



            $begin{array}\
            sqrt[n]{n^k}-1
            &=(sqrt[n]{n})^k-1\
            &=(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j\
            text{so}\
            |sqrt[n]{n^k}-1|
            &=|(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j|\
            &=|(sqrt[n]{n}-1)||sum_{j=0}^{k-1} sqrt[n]{n}^j|\
            &le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^j|\
            &le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^n|\
            &=|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |n|\
            &=|(sqrt[n]{n}-1)|kn\
            end{array}
            $



            Therefore,
            to make
            $|sqrt[n]{n^k}-1|
            le epsilon$
            ,
            choose $n$ large enough
            so that
            $|(sqrt[n]{n}-1)|
            le frac{epsilon}{kn}
            $
            .






            share|cite|improve this answer












            You know that
            $sqrt[n]{n}
            to 1$
            .



            Also
            $sqrt[n]{n^k}
            =(sqrt[n]{n})^k
            $

            so



            $begin{array}\
            sqrt[n]{n^k}-1
            &=(sqrt[n]{n})^k-1\
            &=(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j\
            text{so}\
            |sqrt[n]{n^k}-1|
            &=|(sqrt[n]{n}-1)sum_{j=0}^{k-1} sqrt[n]{n}^j|\
            &=|(sqrt[n]{n}-1)||sum_{j=0}^{k-1} sqrt[n]{n}^j|\
            &le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^j|\
            &le|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |sqrt[n]{n}^n|\
            &=|(sqrt[n]{n}-1)|sum_{j=0}^{k-1} |n|\
            &=|(sqrt[n]{n}-1)|kn\
            end{array}
            $



            Therefore,
            to make
            $|sqrt[n]{n^k}-1|
            le epsilon$
            ,
            choose $n$ large enough
            so that
            $|(sqrt[n]{n}-1)|
            le frac{epsilon}{kn}
            $
            .







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 21 at 22:02









            marty cohen

            71.7k546124




            71.7k546124






















                up vote
                2
                down vote













                You can if you know four things.



                1) If $f$ is continuous and $a_n to M$ and for all $a_n$ that $f(a_n)$ is defined and $f(M)$ is defined then $a_nto M$ means $f(a_n) to f(M)$.



                But you DO have to prove this sometime. I'm sure your text has proven that somewhere.



                And



                2) you need to know that $sqrt[n]{n} to 1$.



                But you claim you have already shown that.



                And 3) that $sqrt[n]{n^3} = (sqrt[n]{n})^3$.



                Which is basic. It follows that as for all $M > 0$ and $nin mathbb N$ there is a unique $k = sqrt[n]{M}$ so that $k^n = M$. And as $(k^j)^n= (k^n)^j = M^j; for j in mathbb N$ it follows that $(sqrt[n]{M})^j= sqrt[n]{M^3}$.



                And finally you need to know 4) that $()^3: mathbb R to mathbb R$ is continuous.



                Which is .... basic. But you should have proven that sometime.



                So $lim sqrt[n]{n^3} = lim (sqrt[n]{n})^3 = (lim sqrt[n]{n})^3 = 1^3 =1$.






                share|cite|improve this answer

























                  up vote
                  2
                  down vote













                  You can if you know four things.



                  1) If $f$ is continuous and $a_n to M$ and for all $a_n$ that $f(a_n)$ is defined and $f(M)$ is defined then $a_nto M$ means $f(a_n) to f(M)$.



                  But you DO have to prove this sometime. I'm sure your text has proven that somewhere.



                  And



                  2) you need to know that $sqrt[n]{n} to 1$.



                  But you claim you have already shown that.



                  And 3) that $sqrt[n]{n^3} = (sqrt[n]{n})^3$.



                  Which is basic. It follows that as for all $M > 0$ and $nin mathbb N$ there is a unique $k = sqrt[n]{M}$ so that $k^n = M$. And as $(k^j)^n= (k^n)^j = M^j; for j in mathbb N$ it follows that $(sqrt[n]{M})^j= sqrt[n]{M^3}$.



                  And finally you need to know 4) that $()^3: mathbb R to mathbb R$ is continuous.



                  Which is .... basic. But you should have proven that sometime.



                  So $lim sqrt[n]{n^3} = lim (sqrt[n]{n})^3 = (lim sqrt[n]{n})^3 = 1^3 =1$.






                  share|cite|improve this answer























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    You can if you know four things.



                    1) If $f$ is continuous and $a_n to M$ and for all $a_n$ that $f(a_n)$ is defined and $f(M)$ is defined then $a_nto M$ means $f(a_n) to f(M)$.



                    But you DO have to prove this sometime. I'm sure your text has proven that somewhere.



                    And



                    2) you need to know that $sqrt[n]{n} to 1$.



                    But you claim you have already shown that.



                    And 3) that $sqrt[n]{n^3} = (sqrt[n]{n})^3$.



                    Which is basic. It follows that as for all $M > 0$ and $nin mathbb N$ there is a unique $k = sqrt[n]{M}$ so that $k^n = M$. And as $(k^j)^n= (k^n)^j = M^j; for j in mathbb N$ it follows that $(sqrt[n]{M})^j= sqrt[n]{M^3}$.



                    And finally you need to know 4) that $()^3: mathbb R to mathbb R$ is continuous.



                    Which is .... basic. But you should have proven that sometime.



                    So $lim sqrt[n]{n^3} = lim (sqrt[n]{n})^3 = (lim sqrt[n]{n})^3 = 1^3 =1$.






                    share|cite|improve this answer












                    You can if you know four things.



                    1) If $f$ is continuous and $a_n to M$ and for all $a_n$ that $f(a_n)$ is defined and $f(M)$ is defined then $a_nto M$ means $f(a_n) to f(M)$.



                    But you DO have to prove this sometime. I'm sure your text has proven that somewhere.



                    And



                    2) you need to know that $sqrt[n]{n} to 1$.



                    But you claim you have already shown that.



                    And 3) that $sqrt[n]{n^3} = (sqrt[n]{n})^3$.



                    Which is basic. It follows that as for all $M > 0$ and $nin mathbb N$ there is a unique $k = sqrt[n]{M}$ so that $k^n = M$. And as $(k^j)^n= (k^n)^j = M^j; for j in mathbb N$ it follows that $(sqrt[n]{M})^j= sqrt[n]{M^3}$.



                    And finally you need to know 4) that $()^3: mathbb R to mathbb R$ is continuous.



                    Which is .... basic. But you should have proven that sometime.



                    So $lim sqrt[n]{n^3} = lim (sqrt[n]{n})^3 = (lim sqrt[n]{n})^3 = 1^3 =1$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 21 at 22:18









                    fleablood

                    67.4k22684




                    67.4k22684






















                        up vote
                        2
                        down vote













                        Yes, because the cubic root function is continuous, so that you can swap the limit and the root.





                        Continuity at $1$ is ensured by the fact that



                        $$|sqrt[3]{1+delta}-1|=left|frac{delta}{(sqrt[3]{1+delta})^2+sqrt[3]{1+delta}+1}right|<fracdelta3<epsilon$$ holds with $delta<3epsilon$.






                        share|cite|improve this answer



























                          up vote
                          2
                          down vote













                          Yes, because the cubic root function is continuous, so that you can swap the limit and the root.





                          Continuity at $1$ is ensured by the fact that



                          $$|sqrt[3]{1+delta}-1|=left|frac{delta}{(sqrt[3]{1+delta})^2+sqrt[3]{1+delta}+1}right|<fracdelta3<epsilon$$ holds with $delta<3epsilon$.






                          share|cite|improve this answer

























                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            Yes, because the cubic root function is continuous, so that you can swap the limit and the root.





                            Continuity at $1$ is ensured by the fact that



                            $$|sqrt[3]{1+delta}-1|=left|frac{delta}{(sqrt[3]{1+delta})^2+sqrt[3]{1+delta}+1}right|<fracdelta3<epsilon$$ holds with $delta<3epsilon$.






                            share|cite|improve this answer














                            Yes, because the cubic root function is continuous, so that you can swap the limit and the root.





                            Continuity at $1$ is ensured by the fact that



                            $$|sqrt[3]{1+delta}-1|=left|frac{delta}{(sqrt[3]{1+delta})^2+sqrt[3]{1+delta}+1}right|<fracdelta3<epsilon$$ holds with $delta<3epsilon$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 22 at 8:07

























                            answered Nov 21 at 21:52









                            Yves Daoust

                            123k668219




                            123k668219















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