Vtyjkyuk

I have the following limit

$$lim_nrightarrow+inftyleft(fracnn+1right)^n$$

It leads to an indeterminate $[1^infty]$, this means I should look into the limit of $e$. I rework the function a little bit:

$$left(fracnn+1right)^n=left(fracn+1-1n+1right)^n=left(1-frac1n+1right)^n$$

This is similar but not identical to

$$left(1+frac1nright)^nrightarrow e$$

The major difference I notice is the $n+1$ in the denominator. Is the limit considered to approach $e^-1$ because for $nrightarrowinfty$ we have $n+1sim n$?

Remark that

$$left(frac nn+1right)^n=left(1+frac1nright)^-n$$ and by continuity of the reciprocal function, the limit of the reciprocal is the reciprocal of the limit.

This shows that in general $n+1sim n$, unless there is some cancellation. (More specifically, $n+1-nnsim n-n$.)

You can write $$left(fracnn+1right)^n=left(left(1-frac1n+1right)^n+1right)^fracnn+1=a_n^fracnn+1$$where $lim_ntoinftya_n=e^-1$ and $lim_ntoinftyfracnn+1=1$.

Then: $$lim_ntoinftya_n^fracnn+1=left(e^-1right)^1=e^-1$$

Note that:
$$lim_ntoinftyleft(1+frac1nright)^n=lim_nto-inftyleft(1+frac1nright)^n=lim_nto+inftyleft(1+frac1nright)^n=e.$$
So, one way:
$$lim_nrightarrow+inftyleft(fracnn+1right)^n=lim_nrightarrow+inftyleft(frac1fracn+1nright)^n=\
lim_nrightarrow+inftyfrac1left(1+frac1nright)^n=frac1lim_limitsnrightarrow+inftyleft(1+frac1nright)^n=frac1e.$$
or the other way:
$$lim_nto+inftyleft(1-frac1n+1right)^n=lim_nto+inftyleft(left[1+frac1-(n+1)right]^-(n+1)right)^fracn-(n+1)=\
e^lim_limitsnto+infty fracn-(n+1)=e^-1=frac1e.$$

$$L=lim_nrightarrow+inftyleft(fracnn+1right)^n$$
$$log L=lim_nrightarrow+inftylog left(fracnn+1right)^n$$
$$log L=lim_nrightarrow+inftynlog left(fracnn+1right)$$
$$log L=lim_nrightarrow+inftyfraclog left(fracnn+1right)frac1n$$

Apply LHopital's rule to the RHS

$$log L=lim_nrightarrow+inftyfracfrac1n^2+nfrac-1n^2$$
$$log L=-1lim_nrightarrow+inftyfracn^2n^2+n=-1$$

$$L=e^-1$$

Recall that $forall ain mathbbR$

$$left(1+frac1nright)^nrightarrow eimplies left(1+fracanright)^nrightarrow e^a$$

therefore from your step

$$left(fracnn+1right)^n=left(fracn+1-1n+1right)^n=left(1-frac1n+1right)^n=left(1+frac-1n+1right)^n=fracleft(1+frac-1n+1right)^n+1left(1+frac-1n+1right)to frace^-11=e^-1$$