Xrfgtjtk

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.

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

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

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.

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 $e$

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

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:
$$lim_{n to infty}sqrt{left(1+frac{1}{n}+frac{1}{n^3}right)^{5n+1}}=lim_{n to infty}expleft(tfrac{5n+1}{2}left(tfrac{1}{n}+tfrac{1}{n^3}right)right)=e^{5/2}$$

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?

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.

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 $e^{5/2}$.