Limit $ lim_{n rightarrow infty}sqrt{(1+frac{1}{n}+frac{1}{n^3})^{5n+1}} $












1















What's the limit of this?
$$
lim_{n rightarrow infty}sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}
$$




Can't find a way to solve it, I tried to use something looks like $(1+1/n)^n$ but didn't succeed.



Many thanks!










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  • Start with logarithms; then Taylor series
    – Claude Leibovici
    Dec 5 '18 at 12:04










  • Is answer is 1?
    – Cloud JR
    Dec 5 '18 at 12:10
















1















What's the limit of this?
$$
lim_{n rightarrow infty}sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}
$$




Can't find a way to solve it, I tried to use something looks like $(1+1/n)^n$ but didn't succeed.



Many thanks!










share|cite|improve this question
























  • Start with logarithms; then Taylor series
    – Claude Leibovici
    Dec 5 '18 at 12:04










  • Is answer is 1?
    – Cloud JR
    Dec 5 '18 at 12:10














1












1








1








What's the limit of this?
$$
lim_{n rightarrow infty}sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}
$$




Can't find a way to solve it, I tried to use something looks like $(1+1/n)^n$ but didn't succeed.



Many thanks!










share|cite|improve this question
















What's the limit of this?
$$
lim_{n rightarrow infty}sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}
$$




Can't find a way to solve it, I tried to use something looks like $(1+1/n)^n$ but didn't succeed.



Many thanks!







calculus limits exponential-function






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edited Dec 5 '18 at 19:33









amWhy

192k28225439




192k28225439










asked Dec 5 '18 at 12:00









Omer GaflaOmer Gafla

91




91












  • Start with logarithms; then Taylor series
    – Claude Leibovici
    Dec 5 '18 at 12:04










  • Is answer is 1?
    – Cloud JR
    Dec 5 '18 at 12:10


















  • Start with logarithms; then Taylor series
    – Claude Leibovici
    Dec 5 '18 at 12:04










  • Is answer is 1?
    – Cloud JR
    Dec 5 '18 at 12:10
















Start with logarithms; then Taylor series
– Claude Leibovici
Dec 5 '18 at 12:04




Start with logarithms; then Taylor series
– Claude Leibovici
Dec 5 '18 at 12:04












Is answer is 1?
– Cloud JR
Dec 5 '18 at 12:10




Is answer is 1?
– Cloud JR
Dec 5 '18 at 12:10










8 Answers
8






active

oldest

votes


















6














Let
$$
a_n = left(1 + frac 1n + frac 1{n^3}right)^n.
$$

To use that limit, we squeeze it like so:
$$
left(1 + frac 1nright)^n leqslant a_n leqslant left(1 + frac {n^2+1}{n^3}right)^{n^3/(n^2+1)},
$$

so $lim a_n = mathrm e$. Then the original limit is $mathrm e^{5/2}$ by breaking the expression into
$$
sqrt {a_n ^5 cdot left(1+frac 1n + frac 1{n^3}right)},
$$

and the expression in the () has the limit $1$.



UPDATE



The proof is actually not complete. If only the limit $lim (1+frac 1n)^n$ is allowed to use, then we need to justify
$$
lim_n left(1+frac {n^2+1}{n^3}right)^{n^3/(n^2+1)} = mathrm e.
$$

Then we still need to squeeze this. Let this sequence be $b_n$, then
$$
b_n = left(1 + frac 1{frac {n^3}{n^2+1}}right)^{n^3/(n^2+1)},
$$

note that
$$
frac {n^2+1}{n^3} leqslant frac {n^2+n}{n^3} = frac {n+1}{n^2} leqslant frac {n+1}{n^2-1} = frac 1{n-1},
$$

and also
$$
frac {n^3}{n^2+1} leqslant frac {n^3+n}{n^2+1} = n,
$$

so
$$
left(1+frac 1n right)^n leqslant a_n leqslant b_n leqslant left(1+frac 1{n-1}right)^n = left(1+frac 1{n-1}right)^{n-1}cdot left(1 +frac 1{n-1}right),
$$

then take the limit $nto infty$, we have
$$
lim a_n = lim b_n =mathrm e.
$$



Remark



This seems complicated, so actually you could show that
$$
lim_n left(1 + frac 1 {f(n)}right)^{f(n)} =mathrm e
$$

where $lim f(n) = +infty$ after you learning the limit of functions, i.e. the well known
$$
lim_{xto +infty} left(1+frac 1xright)^x = mathrm e
$$

to make things a bit easier by invoking the relationship between limits of sequences and functions. This answer is a demo when there is some restrictions on the tools you are allowed to use.






share|cite|improve this answer































    4














    You may use the fact:





    • $(1+x_n)^{large frac{1}{x_n}} stackrel{n to infty}{longrightarrow} e$ for any sequence $(x_n)$ with $0<x_n stackrel{n to infty}{longrightarrow} 0$


    It follows



    $$begin{eqnarray*} sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}
    & = & sqrt{left(1 + frac{n^2+1}{n^3}right)^{largefrac{n^3}{n^2+1}cdot underbrace{large frac{(5n+1)(n^2+1)}{n^3}}_{large stackrel{n to infty}{longrightarrow} 5}}} \
    & stackrel{n to infty}{longrightarrow} & sqrt{e^5}
    end{eqnarray*}$$






    share|cite|improve this answer























    • I add large for powers in a square root for a good look! If you feel uncomfortable to this , I'm sorry and kindly edit again!
      – Chinnapparaj R
      Dec 5 '18 at 12:22



















    3














    Short answer: (correct but not 100% rigorous, see @xbh)



    The term $1/n^3$ is negligible compared to $1/n$, and $1$ is negligible compared to $5n$. Then your limit is the same as that of



    $$left(left(1+frac1nright)^{1/n}right)^{5/2}.$$





    With a little more trickery, you can write



    $$frac1m=frac1n+frac1{n^3}$$ and it is not a big deal to show



    $$n=m+o(m).$$



    Now you have



    $$sqrt{left(1+frac1mright)^{5m+o(m)+1}}$$ where only the first exponent brings a non-unit contribution.






    share|cite|improve this answer



















    • 3




      Still more rigorous than the average engineer would do :-)
      – Carl Witthoft
      Dec 5 '18 at 13:51



















    2














    Hint:



    $$lim_{ntoinfty}left(1+dfrac{n^2+1}{n^3}right)^{(5n+1)/2}$$



    $$=left(lim_{ntoinfty}left(1+dfrac{n^2+1}{n^3}right)^{n^3/(n^2+1)}right)^{lim_{ntoinfty}dfrac{(5n+1)(n^2+1)}{2n^3}}$$



    As $lim_{ntoinfty}dfrac{n^2+1}{n^3}=0,$ the inner limit converges to $e$



    $lim_{ntoinfty}dfrac{(5n+1)(n^2+1)}{2n^3}=lim_{ntoinfty}dfrac{left(5+dfrac1nright)left(1+dfrac1{n^2}right)}2=?$






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      1














      Rewrite:
      $$sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}=expleft(tfrac{5n+1}{2}logleft(1+tfrac{1}{n}+tfrac{1}{n^3}right)right)$$
      and $log(1+x) approx x$ for small $x$, so:
      $$lim_{n to infty}sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}=lim_{n to infty}expleft(tfrac{5n+1}{2}left(tfrac{1}{n}+tfrac{1}{n^3}right)right)=e^{5/2}$$






      share|cite|improve this answer





























        0














        It's $$e^{5/2} .$$ I've been involved in a few posts in which the query is like that of this one, recently, the subject being one of my favourite ones.



        It scarcely matters about the square-root, really: that doesn't affect the limiting process itself here atall. A crucial point though is that in a limit such as this, the terms of higher order than linear in $1/n$ become negligible. This can be shown in various ways, such as by analysing the terms in a generalised expansion combinatorially. I have discussed this elsewhere; and I think a repetition of any more than a little of it would be a tad redundant here. But I'll say that it's a bit like the way the terms beyond the linear ones fade out in the derivation of the derivative of a function - but that here it's 'cast' in a multiplicative form.



        Also the constant (with respect to $n$) in the exponent becomes negligible: this can be seen by noting that leaving it in would be the same as leaving a finite power of the bracket multiplying the whole thing; And the limit of a finite power of it as $ntoinfty$ is certainly 1.



        So we are left with $$lim_{ntoinfty}left(1+{1over n}right)^{5nover2}$$ This is now just the standard limit form of the exponential function $$lim_{ntoinfty}left(1+{xover n}right)^n$$$$=$$$$lim_{ntoinfty}left(1+{1over n}right)^{nx}$$$$=$$$$lim_{ntoinfty}left(1+{sqrt{x}over n}right)^{nsqrt{x}}, $$ or whatever.



        The reason I have written it several times like that is just to emphasise that factors can be transferred at will between the term in the bracket & the exponent. That the exponential function of $x$ is identical with this limt is a standard result - showing that it is is beyond the ambit of this answer, really. Or is that actually what you're asking for?






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          0














          Because I can :-), in R



          >    whatlim <- function(n) {   
          term = sqrt( (1 + 1/n + 1/n^3)^(5*n+1) )
          delta = exp(2.5) - term
          return(delta) }

          > converge = NULL
          > for (jj in 0:10) {
          converge[jj+1] = whatlim(10^jj) }

          > converge
          [1] -1.481751e+01 5.525765e-01 8.732760e-02 9.095940e-03 9.132772e-04
          [6] 9.136407e-05 9.139366e-06 8.958993e-07 2.765578e-07 -2.510824e-06
          [11] -2.519047e-06


          Where pretty obviously the last two terms suffer from floating-point precision limit.






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            0














            Let's assume the well known result $$lim_{ntoinfty} left(1+frac{1}{n}right)^n=etag{1}$$ and the lemma of Thomas Andrews:




            Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$ as $nto infty $.




            Consider the sequence $$a_n=dfrac{1+dfrac{1}{n}+dfrac {1}{n^3}}{1+dfrac{1}{n}}$$ and check that $n(a_n-1) to 1$ so that $a_n^nto 1$ and using $(1)$ we get $$lim_{nto infty} left(1+frac {1}{n}+frac{1}{n^3}right)^n=etag{2}$$ and then we get $$lim_{ntoinfty} left(1+frac {1}{n}+frac {1}{n^3}right)^{5n+1}=e^5$$ and taking square roots we get the answer as $e^{5/2}$.






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              8 Answers
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              8 Answers
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              active

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              active

              oldest

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              6














              Let
              $$
              a_n = left(1 + frac 1n + frac 1{n^3}right)^n.
              $$

              To use that limit, we squeeze it like so:
              $$
              left(1 + frac 1nright)^n leqslant a_n leqslant left(1 + frac {n^2+1}{n^3}right)^{n^3/(n^2+1)},
              $$

              so $lim a_n = mathrm e$. Then the original limit is $mathrm e^{5/2}$ by breaking the expression into
              $$
              sqrt {a_n ^5 cdot left(1+frac 1n + frac 1{n^3}right)},
              $$

              and the expression in the () has the limit $1$.



              UPDATE



              The proof is actually not complete. If only the limit $lim (1+frac 1n)^n$ is allowed to use, then we need to justify
              $$
              lim_n left(1+frac {n^2+1}{n^3}right)^{n^3/(n^2+1)} = mathrm e.
              $$

              Then we still need to squeeze this. Let this sequence be $b_n$, then
              $$
              b_n = left(1 + frac 1{frac {n^3}{n^2+1}}right)^{n^3/(n^2+1)},
              $$

              note that
              $$
              frac {n^2+1}{n^3} leqslant frac {n^2+n}{n^3} = frac {n+1}{n^2} leqslant frac {n+1}{n^2-1} = frac 1{n-1},
              $$

              and also
              $$
              frac {n^3}{n^2+1} leqslant frac {n^3+n}{n^2+1} = n,
              $$

              so
              $$
              left(1+frac 1n right)^n leqslant a_n leqslant b_n leqslant left(1+frac 1{n-1}right)^n = left(1+frac 1{n-1}right)^{n-1}cdot left(1 +frac 1{n-1}right),
              $$

              then take the limit $nto infty$, we have
              $$
              lim a_n = lim b_n =mathrm e.
              $$



              Remark



              This seems complicated, so actually you could show that
              $$
              lim_n left(1 + frac 1 {f(n)}right)^{f(n)} =mathrm e
              $$

              where $lim f(n) = +infty$ after you learning the limit of functions, i.e. the well known
              $$
              lim_{xto +infty} left(1+frac 1xright)^x = mathrm e
              $$

              to make things a bit easier by invoking the relationship between limits of sequences and functions. This answer is a demo when there is some restrictions on the tools you are allowed to use.






              share|cite|improve this answer




























                6














                Let
                $$
                a_n = left(1 + frac 1n + frac 1{n^3}right)^n.
                $$

                To use that limit, we squeeze it like so:
                $$
                left(1 + frac 1nright)^n leqslant a_n leqslant left(1 + frac {n^2+1}{n^3}right)^{n^3/(n^2+1)},
                $$

                so $lim a_n = mathrm e$. Then the original limit is $mathrm e^{5/2}$ by breaking the expression into
                $$
                sqrt {a_n ^5 cdot left(1+frac 1n + frac 1{n^3}right)},
                $$

                and the expression in the () has the limit $1$.



                UPDATE



                The proof is actually not complete. If only the limit $lim (1+frac 1n)^n$ is allowed to use, then we need to justify
                $$
                lim_n left(1+frac {n^2+1}{n^3}right)^{n^3/(n^2+1)} = mathrm e.
                $$

                Then we still need to squeeze this. Let this sequence be $b_n$, then
                $$
                b_n = left(1 + frac 1{frac {n^3}{n^2+1}}right)^{n^3/(n^2+1)},
                $$

                note that
                $$
                frac {n^2+1}{n^3} leqslant frac {n^2+n}{n^3} = frac {n+1}{n^2} leqslant frac {n+1}{n^2-1} = frac 1{n-1},
                $$

                and also
                $$
                frac {n^3}{n^2+1} leqslant frac {n^3+n}{n^2+1} = n,
                $$

                so
                $$
                left(1+frac 1n right)^n leqslant a_n leqslant b_n leqslant left(1+frac 1{n-1}right)^n = left(1+frac 1{n-1}right)^{n-1}cdot left(1 +frac 1{n-1}right),
                $$

                then take the limit $nto infty$, we have
                $$
                lim a_n = lim b_n =mathrm e.
                $$



                Remark



                This seems complicated, so actually you could show that
                $$
                lim_n left(1 + frac 1 {f(n)}right)^{f(n)} =mathrm e
                $$

                where $lim f(n) = +infty$ after you learning the limit of functions, i.e. the well known
                $$
                lim_{xto +infty} left(1+frac 1xright)^x = mathrm e
                $$

                to make things a bit easier by invoking the relationship between limits of sequences and functions. This answer is a demo when there is some restrictions on the tools you are allowed to use.






                share|cite|improve this answer


























                  6












                  6








                  6






                  Let
                  $$
                  a_n = left(1 + frac 1n + frac 1{n^3}right)^n.
                  $$

                  To use that limit, we squeeze it like so:
                  $$
                  left(1 + frac 1nright)^n leqslant a_n leqslant left(1 + frac {n^2+1}{n^3}right)^{n^3/(n^2+1)},
                  $$

                  so $lim a_n = mathrm e$. Then the original limit is $mathrm e^{5/2}$ by breaking the expression into
                  $$
                  sqrt {a_n ^5 cdot left(1+frac 1n + frac 1{n^3}right)},
                  $$

                  and the expression in the () has the limit $1$.



                  UPDATE



                  The proof is actually not complete. If only the limit $lim (1+frac 1n)^n$ is allowed to use, then we need to justify
                  $$
                  lim_n left(1+frac {n^2+1}{n^3}right)^{n^3/(n^2+1)} = mathrm e.
                  $$

                  Then we still need to squeeze this. Let this sequence be $b_n$, then
                  $$
                  b_n = left(1 + frac 1{frac {n^3}{n^2+1}}right)^{n^3/(n^2+1)},
                  $$

                  note that
                  $$
                  frac {n^2+1}{n^3} leqslant frac {n^2+n}{n^3} = frac {n+1}{n^2} leqslant frac {n+1}{n^2-1} = frac 1{n-1},
                  $$

                  and also
                  $$
                  frac {n^3}{n^2+1} leqslant frac {n^3+n}{n^2+1} = n,
                  $$

                  so
                  $$
                  left(1+frac 1n right)^n leqslant a_n leqslant b_n leqslant left(1+frac 1{n-1}right)^n = left(1+frac 1{n-1}right)^{n-1}cdot left(1 +frac 1{n-1}right),
                  $$

                  then take the limit $nto infty$, we have
                  $$
                  lim a_n = lim b_n =mathrm e.
                  $$



                  Remark



                  This seems complicated, so actually you could show that
                  $$
                  lim_n left(1 + frac 1 {f(n)}right)^{f(n)} =mathrm e
                  $$

                  where $lim f(n) = +infty$ after you learning the limit of functions, i.e. the well known
                  $$
                  lim_{xto +infty} left(1+frac 1xright)^x = mathrm e
                  $$

                  to make things a bit easier by invoking the relationship between limits of sequences and functions. This answer is a demo when there is some restrictions on the tools you are allowed to use.






                  share|cite|improve this answer














                  Let
                  $$
                  a_n = left(1 + frac 1n + frac 1{n^3}right)^n.
                  $$

                  To use that limit, we squeeze it like so:
                  $$
                  left(1 + frac 1nright)^n leqslant a_n leqslant left(1 + frac {n^2+1}{n^3}right)^{n^3/(n^2+1)},
                  $$

                  so $lim a_n = mathrm e$. Then the original limit is $mathrm e^{5/2}$ by breaking the expression into
                  $$
                  sqrt {a_n ^5 cdot left(1+frac 1n + frac 1{n^3}right)},
                  $$

                  and the expression in the () has the limit $1$.



                  UPDATE



                  The proof is actually not complete. If only the limit $lim (1+frac 1n)^n$ is allowed to use, then we need to justify
                  $$
                  lim_n left(1+frac {n^2+1}{n^3}right)^{n^3/(n^2+1)} = mathrm e.
                  $$

                  Then we still need to squeeze this. Let this sequence be $b_n$, then
                  $$
                  b_n = left(1 + frac 1{frac {n^3}{n^2+1}}right)^{n^3/(n^2+1)},
                  $$

                  note that
                  $$
                  frac {n^2+1}{n^3} leqslant frac {n^2+n}{n^3} = frac {n+1}{n^2} leqslant frac {n+1}{n^2-1} = frac 1{n-1},
                  $$

                  and also
                  $$
                  frac {n^3}{n^2+1} leqslant frac {n^3+n}{n^2+1} = n,
                  $$

                  so
                  $$
                  left(1+frac 1n right)^n leqslant a_n leqslant b_n leqslant left(1+frac 1{n-1}right)^n = left(1+frac 1{n-1}right)^{n-1}cdot left(1 +frac 1{n-1}right),
                  $$

                  then take the limit $nto infty$, we have
                  $$
                  lim a_n = lim b_n =mathrm e.
                  $$



                  Remark



                  This seems complicated, so actually you could show that
                  $$
                  lim_n left(1 + frac 1 {f(n)}right)^{f(n)} =mathrm e
                  $$

                  where $lim f(n) = +infty$ after you learning the limit of functions, i.e. the well known
                  $$
                  lim_{xto +infty} left(1+frac 1xright)^x = mathrm e
                  $$

                  to make things a bit easier by invoking the relationship between limits of sequences and functions. This answer is a demo when there is some restrictions on the tools you are allowed to use.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 6 '18 at 4:54

























                  answered Dec 5 '18 at 12:21









                  xbhxbh

                  5,8281522




                  5,8281522























                      4














                      You may use the fact:





                      • $(1+x_n)^{large frac{1}{x_n}} stackrel{n to infty}{longrightarrow} e$ for any sequence $(x_n)$ with $0<x_n stackrel{n to infty}{longrightarrow} 0$


                      It follows



                      $$begin{eqnarray*} sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}
                      & = & sqrt{left(1 + frac{n^2+1}{n^3}right)^{largefrac{n^3}{n^2+1}cdot underbrace{large frac{(5n+1)(n^2+1)}{n^3}}_{large stackrel{n to infty}{longrightarrow} 5}}} \
                      & stackrel{n to infty}{longrightarrow} & sqrt{e^5}
                      end{eqnarray*}$$






                      share|cite|improve this answer























                      • I add large for powers in a square root for a good look! If you feel uncomfortable to this , I'm sorry and kindly edit again!
                        – Chinnapparaj R
                        Dec 5 '18 at 12:22
















                      4














                      You may use the fact:





                      • $(1+x_n)^{large frac{1}{x_n}} stackrel{n to infty}{longrightarrow} e$ for any sequence $(x_n)$ with $0<x_n stackrel{n to infty}{longrightarrow} 0$


                      It follows



                      $$begin{eqnarray*} sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}
                      & = & sqrt{left(1 + frac{n^2+1}{n^3}right)^{largefrac{n^3}{n^2+1}cdot underbrace{large frac{(5n+1)(n^2+1)}{n^3}}_{large stackrel{n to infty}{longrightarrow} 5}}} \
                      & stackrel{n to infty}{longrightarrow} & sqrt{e^5}
                      end{eqnarray*}$$






                      share|cite|improve this answer























                      • I add large for powers in a square root for a good look! If you feel uncomfortable to this , I'm sorry and kindly edit again!
                        – Chinnapparaj R
                        Dec 5 '18 at 12:22














                      4












                      4








                      4






                      You may use the fact:





                      • $(1+x_n)^{large frac{1}{x_n}} stackrel{n to infty}{longrightarrow} e$ for any sequence $(x_n)$ with $0<x_n stackrel{n to infty}{longrightarrow} 0$


                      It follows



                      $$begin{eqnarray*} sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}
                      & = & sqrt{left(1 + frac{n^2+1}{n^3}right)^{largefrac{n^3}{n^2+1}cdot underbrace{large frac{(5n+1)(n^2+1)}{n^3}}_{large stackrel{n to infty}{longrightarrow} 5}}} \
                      & stackrel{n to infty}{longrightarrow} & sqrt{e^5}
                      end{eqnarray*}$$






                      share|cite|improve this answer














                      You may use the fact:





                      • $(1+x_n)^{large frac{1}{x_n}} stackrel{n to infty}{longrightarrow} e$ for any sequence $(x_n)$ with $0<x_n stackrel{n to infty}{longrightarrow} 0$


                      It follows



                      $$begin{eqnarray*} sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}
                      & = & sqrt{left(1 + frac{n^2+1}{n^3}right)^{largefrac{n^3}{n^2+1}cdot underbrace{large frac{(5n+1)(n^2+1)}{n^3}}_{large stackrel{n to infty}{longrightarrow} 5}}} \
                      & stackrel{n to infty}{longrightarrow} & sqrt{e^5}
                      end{eqnarray*}$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 5 '18 at 12:21









                      Chinnapparaj R

                      5,2481828




                      5,2481828










                      answered Dec 5 '18 at 12:17









                      trancelocationtrancelocation

                      9,4501622




                      9,4501622












                      • I add large for powers in a square root for a good look! If you feel uncomfortable to this , I'm sorry and kindly edit again!
                        – Chinnapparaj R
                        Dec 5 '18 at 12:22


















                      • I add large for powers in a square root for a good look! If you feel uncomfortable to this , I'm sorry and kindly edit again!
                        – Chinnapparaj R
                        Dec 5 '18 at 12:22
















                      I add large for powers in a square root for a good look! If you feel uncomfortable to this , I'm sorry and kindly edit again!
                      – Chinnapparaj R
                      Dec 5 '18 at 12:22




                      I add large for powers in a square root for a good look! If you feel uncomfortable to this , I'm sorry and kindly edit again!
                      – Chinnapparaj R
                      Dec 5 '18 at 12:22











                      3














                      Short answer: (correct but not 100% rigorous, see @xbh)



                      The term $1/n^3$ is negligible compared to $1/n$, and $1$ is negligible compared to $5n$. Then your limit is the same as that of



                      $$left(left(1+frac1nright)^{1/n}right)^{5/2}.$$





                      With a little more trickery, you can write



                      $$frac1m=frac1n+frac1{n^3}$$ and it is not a big deal to show



                      $$n=m+o(m).$$



                      Now you have



                      $$sqrt{left(1+frac1mright)^{5m+o(m)+1}}$$ where only the first exponent brings a non-unit contribution.






                      share|cite|improve this answer



















                      • 3




                        Still more rigorous than the average engineer would do :-)
                        – Carl Witthoft
                        Dec 5 '18 at 13:51
















                      3














                      Short answer: (correct but not 100% rigorous, see @xbh)



                      The term $1/n^3$ is negligible compared to $1/n$, and $1$ is negligible compared to $5n$. Then your limit is the same as that of



                      $$left(left(1+frac1nright)^{1/n}right)^{5/2}.$$





                      With a little more trickery, you can write



                      $$frac1m=frac1n+frac1{n^3}$$ and it is not a big deal to show



                      $$n=m+o(m).$$



                      Now you have



                      $$sqrt{left(1+frac1mright)^{5m+o(m)+1}}$$ where only the first exponent brings a non-unit contribution.






                      share|cite|improve this answer



















                      • 3




                        Still more rigorous than the average engineer would do :-)
                        – Carl Witthoft
                        Dec 5 '18 at 13:51














                      3












                      3








                      3






                      Short answer: (correct but not 100% rigorous, see @xbh)



                      The term $1/n^3$ is negligible compared to $1/n$, and $1$ is negligible compared to $5n$. Then your limit is the same as that of



                      $$left(left(1+frac1nright)^{1/n}right)^{5/2}.$$





                      With a little more trickery, you can write



                      $$frac1m=frac1n+frac1{n^3}$$ and it is not a big deal to show



                      $$n=m+o(m).$$



                      Now you have



                      $$sqrt{left(1+frac1mright)^{5m+o(m)+1}}$$ where only the first exponent brings a non-unit contribution.






                      share|cite|improve this answer














                      Short answer: (correct but not 100% rigorous, see @xbh)



                      The term $1/n^3$ is negligible compared to $1/n$, and $1$ is negligible compared to $5n$. Then your limit is the same as that of



                      $$left(left(1+frac1nright)^{1/n}right)^{5/2}.$$





                      With a little more trickery, you can write



                      $$frac1m=frac1n+frac1{n^3}$$ and it is not a big deal to show



                      $$n=m+o(m).$$



                      Now you have



                      $$sqrt{left(1+frac1mright)^{5m+o(m)+1}}$$ where only the first exponent brings a non-unit contribution.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 5 '18 at 13:44

























                      answered Dec 5 '18 at 13:36









                      Yves DaoustYves Daoust

                      124k671222




                      124k671222








                      • 3




                        Still more rigorous than the average engineer would do :-)
                        – Carl Witthoft
                        Dec 5 '18 at 13:51














                      • 3




                        Still more rigorous than the average engineer would do :-)
                        – Carl Witthoft
                        Dec 5 '18 at 13:51








                      3




                      3




                      Still more rigorous than the average engineer would do :-)
                      – Carl Witthoft
                      Dec 5 '18 at 13:51




                      Still more rigorous than the average engineer would do :-)
                      – Carl Witthoft
                      Dec 5 '18 at 13:51











                      2














                      Hint:



                      $$lim_{ntoinfty}left(1+dfrac{n^2+1}{n^3}right)^{(5n+1)/2}$$



                      $$=left(lim_{ntoinfty}left(1+dfrac{n^2+1}{n^3}right)^{n^3/(n^2+1)}right)^{lim_{ntoinfty}dfrac{(5n+1)(n^2+1)}{2n^3}}$$



                      As $lim_{ntoinfty}dfrac{n^2+1}{n^3}=0,$ the inner limit converges to $e$



                      $lim_{ntoinfty}dfrac{(5n+1)(n^2+1)}{2n^3}=lim_{ntoinfty}dfrac{left(5+dfrac1nright)left(1+dfrac1{n^2}right)}2=?$






                      share|cite|improve this answer


























                        2














                        Hint:



                        $$lim_{ntoinfty}left(1+dfrac{n^2+1}{n^3}right)^{(5n+1)/2}$$



                        $$=left(lim_{ntoinfty}left(1+dfrac{n^2+1}{n^3}right)^{n^3/(n^2+1)}right)^{lim_{ntoinfty}dfrac{(5n+1)(n^2+1)}{2n^3}}$$



                        As $lim_{ntoinfty}dfrac{n^2+1}{n^3}=0,$ the inner limit converges to $e$



                        $lim_{ntoinfty}dfrac{(5n+1)(n^2+1)}{2n^3}=lim_{ntoinfty}dfrac{left(5+dfrac1nright)left(1+dfrac1{n^2}right)}2=?$






                        share|cite|improve this answer
























                          2












                          2








                          2






                          Hint:



                          $$lim_{ntoinfty}left(1+dfrac{n^2+1}{n^3}right)^{(5n+1)/2}$$



                          $$=left(lim_{ntoinfty}left(1+dfrac{n^2+1}{n^3}right)^{n^3/(n^2+1)}right)^{lim_{ntoinfty}dfrac{(5n+1)(n^2+1)}{2n^3}}$$



                          As $lim_{ntoinfty}dfrac{n^2+1}{n^3}=0,$ the inner limit converges to $e$



                          $lim_{ntoinfty}dfrac{(5n+1)(n^2+1)}{2n^3}=lim_{ntoinfty}dfrac{left(5+dfrac1nright)left(1+dfrac1{n^2}right)}2=?$






                          share|cite|improve this answer












                          Hint:



                          $$lim_{ntoinfty}left(1+dfrac{n^2+1}{n^3}right)^{(5n+1)/2}$$



                          $$=left(lim_{ntoinfty}left(1+dfrac{n^2+1}{n^3}right)^{n^3/(n^2+1)}right)^{lim_{ntoinfty}dfrac{(5n+1)(n^2+1)}{2n^3}}$$



                          As $lim_{ntoinfty}dfrac{n^2+1}{n^3}=0,$ the inner limit converges to $e$



                          $lim_{ntoinfty}dfrac{(5n+1)(n^2+1)}{2n^3}=lim_{ntoinfty}dfrac{left(5+dfrac1nright)left(1+dfrac1{n^2}right)}2=?$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 5 '18 at 12:17









                          lab bhattacharjeelab bhattacharjee

                          224k15156274




                          224k15156274























                              1














                              Rewrite:
                              $$sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}=expleft(tfrac{5n+1}{2}logleft(1+tfrac{1}{n}+tfrac{1}{n^3}right)right)$$
                              and $log(1+x) approx x$ for small $x$, so:
                              $$lim_{n to infty}sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}=lim_{n to infty}expleft(tfrac{5n+1}{2}left(tfrac{1}{n}+tfrac{1}{n^3}right)right)=e^{5/2}$$






                              share|cite|improve this answer


























                                1














                                Rewrite:
                                $$sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}=expleft(tfrac{5n+1}{2}logleft(1+tfrac{1}{n}+tfrac{1}{n^3}right)right)$$
                                and $log(1+x) approx x$ for small $x$, so:
                                $$lim_{n to infty}sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}=lim_{n to infty}expleft(tfrac{5n+1}{2}left(tfrac{1}{n}+tfrac{1}{n^3}right)right)=e^{5/2}$$






                                share|cite|improve this answer
























                                  1












                                  1








                                  1






                                  Rewrite:
                                  $$sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}=expleft(tfrac{5n+1}{2}logleft(1+tfrac{1}{n}+tfrac{1}{n^3}right)right)$$
                                  and $log(1+x) approx x$ for small $x$, so:
                                  $$lim_{n to infty}sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}=lim_{n to infty}expleft(tfrac{5n+1}{2}left(tfrac{1}{n}+tfrac{1}{n^3}right)right)=e^{5/2}$$






                                  share|cite|improve this answer












                                  Rewrite:
                                  $$sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}=expleft(tfrac{5n+1}{2}logleft(1+tfrac{1}{n}+tfrac{1}{n^3}right)right)$$
                                  and $log(1+x) approx x$ for small $x$, so:
                                  $$lim_{n to infty}sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}=lim_{n to infty}expleft(tfrac{5n+1}{2}left(tfrac{1}{n}+tfrac{1}{n^3}right)right)=e^{5/2}$$







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Dec 5 '18 at 12:17









                                  StackTDStackTD

                                  22.4k2049




                                  22.4k2049























                                      0














                                      It's $$e^{5/2} .$$ I've been involved in a few posts in which the query is like that of this one, recently, the subject being one of my favourite ones.



                                      It scarcely matters about the square-root, really: that doesn't affect the limiting process itself here atall. A crucial point though is that in a limit such as this, the terms of higher order than linear in $1/n$ become negligible. This can be shown in various ways, such as by analysing the terms in a generalised expansion combinatorially. I have discussed this elsewhere; and I think a repetition of any more than a little of it would be a tad redundant here. But I'll say that it's a bit like the way the terms beyond the linear ones fade out in the derivation of the derivative of a function - but that here it's 'cast' in a multiplicative form.



                                      Also the constant (with respect to $n$) in the exponent becomes negligible: this can be seen by noting that leaving it in would be the same as leaving a finite power of the bracket multiplying the whole thing; And the limit of a finite power of it as $ntoinfty$ is certainly 1.



                                      So we are left with $$lim_{ntoinfty}left(1+{1over n}right)^{5nover2}$$ This is now just the standard limit form of the exponential function $$lim_{ntoinfty}left(1+{xover n}right)^n$$$$=$$$$lim_{ntoinfty}left(1+{1over n}right)^{nx}$$$$=$$$$lim_{ntoinfty}left(1+{sqrt{x}over n}right)^{nsqrt{x}}, $$ or whatever.



                                      The reason I have written it several times like that is just to emphasise that factors can be transferred at will between the term in the bracket & the exponent. That the exponential function of $x$ is identical with this limt is a standard result - showing that it is is beyond the ambit of this answer, really. Or is that actually what you're asking for?






                                      share|cite|improve this answer


























                                        0














                                        It's $$e^{5/2} .$$ I've been involved in a few posts in which the query is like that of this one, recently, the subject being one of my favourite ones.



                                        It scarcely matters about the square-root, really: that doesn't affect the limiting process itself here atall. A crucial point though is that in a limit such as this, the terms of higher order than linear in $1/n$ become negligible. This can be shown in various ways, such as by analysing the terms in a generalised expansion combinatorially. I have discussed this elsewhere; and I think a repetition of any more than a little of it would be a tad redundant here. But I'll say that it's a bit like the way the terms beyond the linear ones fade out in the derivation of the derivative of a function - but that here it's 'cast' in a multiplicative form.



                                        Also the constant (with respect to $n$) in the exponent becomes negligible: this can be seen by noting that leaving it in would be the same as leaving a finite power of the bracket multiplying the whole thing; And the limit of a finite power of it as $ntoinfty$ is certainly 1.



                                        So we are left with $$lim_{ntoinfty}left(1+{1over n}right)^{5nover2}$$ This is now just the standard limit form of the exponential function $$lim_{ntoinfty}left(1+{xover n}right)^n$$$$=$$$$lim_{ntoinfty}left(1+{1over n}right)^{nx}$$$$=$$$$lim_{ntoinfty}left(1+{sqrt{x}over n}right)^{nsqrt{x}}, $$ or whatever.



                                        The reason I have written it several times like that is just to emphasise that factors can be transferred at will between the term in the bracket & the exponent. That the exponential function of $x$ is identical with this limt is a standard result - showing that it is is beyond the ambit of this answer, really. Or is that actually what you're asking for?






                                        share|cite|improve this answer
























                                          0












                                          0








                                          0






                                          It's $$e^{5/2} .$$ I've been involved in a few posts in which the query is like that of this one, recently, the subject being one of my favourite ones.



                                          It scarcely matters about the square-root, really: that doesn't affect the limiting process itself here atall. A crucial point though is that in a limit such as this, the terms of higher order than linear in $1/n$ become negligible. This can be shown in various ways, such as by analysing the terms in a generalised expansion combinatorially. I have discussed this elsewhere; and I think a repetition of any more than a little of it would be a tad redundant here. But I'll say that it's a bit like the way the terms beyond the linear ones fade out in the derivation of the derivative of a function - but that here it's 'cast' in a multiplicative form.



                                          Also the constant (with respect to $n$) in the exponent becomes negligible: this can be seen by noting that leaving it in would be the same as leaving a finite power of the bracket multiplying the whole thing; And the limit of a finite power of it as $ntoinfty$ is certainly 1.



                                          So we are left with $$lim_{ntoinfty}left(1+{1over n}right)^{5nover2}$$ This is now just the standard limit form of the exponential function $$lim_{ntoinfty}left(1+{xover n}right)^n$$$$=$$$$lim_{ntoinfty}left(1+{1over n}right)^{nx}$$$$=$$$$lim_{ntoinfty}left(1+{sqrt{x}over n}right)^{nsqrt{x}}, $$ or whatever.



                                          The reason I have written it several times like that is just to emphasise that factors can be transferred at will between the term in the bracket & the exponent. That the exponential function of $x$ is identical with this limt is a standard result - showing that it is is beyond the ambit of this answer, really. Or is that actually what you're asking for?






                                          share|cite|improve this answer












                                          It's $$e^{5/2} .$$ I've been involved in a few posts in which the query is like that of this one, recently, the subject being one of my favourite ones.



                                          It scarcely matters about the square-root, really: that doesn't affect the limiting process itself here atall. A crucial point though is that in a limit such as this, the terms of higher order than linear in $1/n$ become negligible. This can be shown in various ways, such as by analysing the terms in a generalised expansion combinatorially. I have discussed this elsewhere; and I think a repetition of any more than a little of it would be a tad redundant here. But I'll say that it's a bit like the way the terms beyond the linear ones fade out in the derivation of the derivative of a function - but that here it's 'cast' in a multiplicative form.



                                          Also the constant (with respect to $n$) in the exponent becomes negligible: this can be seen by noting that leaving it in would be the same as leaving a finite power of the bracket multiplying the whole thing; And the limit of a finite power of it as $ntoinfty$ is certainly 1.



                                          So we are left with $$lim_{ntoinfty}left(1+{1over n}right)^{5nover2}$$ This is now just the standard limit form of the exponential function $$lim_{ntoinfty}left(1+{xover n}right)^n$$$$=$$$$lim_{ntoinfty}left(1+{1over n}right)^{nx}$$$$=$$$$lim_{ntoinfty}left(1+{sqrt{x}over n}right)^{nsqrt{x}}, $$ or whatever.



                                          The reason I have written it several times like that is just to emphasise that factors can be transferred at will between the term in the bracket & the exponent. That the exponential function of $x$ is identical with this limt is a standard result - showing that it is is beyond the ambit of this answer, really. Or is that actually what you're asking for?







                                          share|cite|improve this answer












                                          share|cite|improve this answer



                                          share|cite|improve this answer










                                          answered Dec 5 '18 at 13:49









                                          AmbretteOrriseyAmbretteOrrisey

                                          55410




                                          55410























                                              0














                                              Because I can :-), in R



                                              >    whatlim <- function(n) {   
                                              term = sqrt( (1 + 1/n + 1/n^3)^(5*n+1) )
                                              delta = exp(2.5) - term
                                              return(delta) }

                                              > converge = NULL
                                              > for (jj in 0:10) {
                                              converge[jj+1] = whatlim(10^jj) }

                                              > converge
                                              [1] -1.481751e+01 5.525765e-01 8.732760e-02 9.095940e-03 9.132772e-04
                                              [6] 9.136407e-05 9.139366e-06 8.958993e-07 2.765578e-07 -2.510824e-06
                                              [11] -2.519047e-06


                                              Where pretty obviously the last two terms suffer from floating-point precision limit.






                                              share|cite|improve this answer


























                                                0














                                                Because I can :-), in R



                                                >    whatlim <- function(n) {   
                                                term = sqrt( (1 + 1/n + 1/n^3)^(5*n+1) )
                                                delta = exp(2.5) - term
                                                return(delta) }

                                                > converge = NULL
                                                > for (jj in 0:10) {
                                                converge[jj+1] = whatlim(10^jj) }

                                                > converge
                                                [1] -1.481751e+01 5.525765e-01 8.732760e-02 9.095940e-03 9.132772e-04
                                                [6] 9.136407e-05 9.139366e-06 8.958993e-07 2.765578e-07 -2.510824e-06
                                                [11] -2.519047e-06


                                                Where pretty obviously the last two terms suffer from floating-point precision limit.






                                                share|cite|improve this answer
























                                                  0












                                                  0








                                                  0






                                                  Because I can :-), in R



                                                  >    whatlim <- function(n) {   
                                                  term = sqrt( (1 + 1/n + 1/n^3)^(5*n+1) )
                                                  delta = exp(2.5) - term
                                                  return(delta) }

                                                  > converge = NULL
                                                  > for (jj in 0:10) {
                                                  converge[jj+1] = whatlim(10^jj) }

                                                  > converge
                                                  [1] -1.481751e+01 5.525765e-01 8.732760e-02 9.095940e-03 9.132772e-04
                                                  [6] 9.136407e-05 9.139366e-06 8.958993e-07 2.765578e-07 -2.510824e-06
                                                  [11] -2.519047e-06


                                                  Where pretty obviously the last two terms suffer from floating-point precision limit.






                                                  share|cite|improve this answer












                                                  Because I can :-), in R



                                                  >    whatlim <- function(n) {   
                                                  term = sqrt( (1 + 1/n + 1/n^3)^(5*n+1) )
                                                  delta = exp(2.5) - term
                                                  return(delta) }

                                                  > converge = NULL
                                                  > for (jj in 0:10) {
                                                  converge[jj+1] = whatlim(10^jj) }

                                                  > converge
                                                  [1] -1.481751e+01 5.525765e-01 8.732760e-02 9.095940e-03 9.132772e-04
                                                  [6] 9.136407e-05 9.139366e-06 8.958993e-07 2.765578e-07 -2.510824e-06
                                                  [11] -2.519047e-06


                                                  Where pretty obviously the last two terms suffer from floating-point precision limit.







                                                  share|cite|improve this answer












                                                  share|cite|improve this answer



                                                  share|cite|improve this answer










                                                  answered Dec 5 '18 at 13:55









                                                  Carl WitthoftCarl Witthoft

                                                  32618




                                                  32618























                                                      0














                                                      Let's assume the well known result $$lim_{ntoinfty} left(1+frac{1}{n}right)^n=etag{1}$$ and the lemma of Thomas Andrews:




                                                      Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$ as $nto infty $.




                                                      Consider the sequence $$a_n=dfrac{1+dfrac{1}{n}+dfrac {1}{n^3}}{1+dfrac{1}{n}}$$ and check that $n(a_n-1) to 1$ so that $a_n^nto 1$ and using $(1)$ we get $$lim_{nto infty} left(1+frac {1}{n}+frac{1}{n^3}right)^n=etag{2}$$ and then we get $$lim_{ntoinfty} left(1+frac {1}{n}+frac {1}{n^3}right)^{5n+1}=e^5$$ and taking square roots we get the answer as $e^{5/2}$.






                                                      share|cite|improve this answer




























                                                        0














                                                        Let's assume the well known result $$lim_{ntoinfty} left(1+frac{1}{n}right)^n=etag{1}$$ and the lemma of Thomas Andrews:




                                                        Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$ as $nto infty $.




                                                        Consider the sequence $$a_n=dfrac{1+dfrac{1}{n}+dfrac {1}{n^3}}{1+dfrac{1}{n}}$$ and check that $n(a_n-1) to 1$ so that $a_n^nto 1$ and using $(1)$ we get $$lim_{nto infty} left(1+frac {1}{n}+frac{1}{n^3}right)^n=etag{2}$$ and then we get $$lim_{ntoinfty} left(1+frac {1}{n}+frac {1}{n^3}right)^{5n+1}=e^5$$ and taking square roots we get the answer as $e^{5/2}$.






                                                        share|cite|improve this answer


























                                                          0












                                                          0








                                                          0






                                                          Let's assume the well known result $$lim_{ntoinfty} left(1+frac{1}{n}right)^n=etag{1}$$ and the lemma of Thomas Andrews:




                                                          Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$ as $nto infty $.




                                                          Consider the sequence $$a_n=dfrac{1+dfrac{1}{n}+dfrac {1}{n^3}}{1+dfrac{1}{n}}$$ and check that $n(a_n-1) to 1$ so that $a_n^nto 1$ and using $(1)$ we get $$lim_{nto infty} left(1+frac {1}{n}+frac{1}{n^3}right)^n=etag{2}$$ and then we get $$lim_{ntoinfty} left(1+frac {1}{n}+frac {1}{n^3}right)^{5n+1}=e^5$$ and taking square roots we get the answer as $e^{5/2}$.






                                                          share|cite|improve this answer














                                                          Let's assume the well known result $$lim_{ntoinfty} left(1+frac{1}{n}right)^n=etag{1}$$ and the lemma of Thomas Andrews:




                                                          Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$ as $nto infty $.




                                                          Consider the sequence $$a_n=dfrac{1+dfrac{1}{n}+dfrac {1}{n^3}}{1+dfrac{1}{n}}$$ and check that $n(a_n-1) to 1$ so that $a_n^nto 1$ and using $(1)$ we get $$lim_{nto infty} left(1+frac {1}{n}+frac{1}{n^3}right)^n=etag{2}$$ and then we get $$lim_{ntoinfty} left(1+frac {1}{n}+frac {1}{n^3}right)^{5n+1}=e^5$$ and taking square roots we get the answer as $e^{5/2}$.







                                                          share|cite|improve this answer














                                                          share|cite|improve this answer



                                                          share|cite|improve this answer








                                                          edited Dec 6 '18 at 5:17

























                                                          answered Dec 6 '18 at 3:59









                                                          Paramanand SinghParamanand Singh

                                                          49.1k555161




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