Tricky limit as $n$ tends to infinity of an expression involving a bunch of roots In this question, @user513057 asked how to prove that for $n$ large enough,
$$
(n+1)\cdot ((n+1)!)^{\frac{1}{n+1}} -n\cdot (n!)^\frac{1}{n}< n+1
$$
In the answer by @Von Neumann, the latter rewrote this inequality as 
$$
[2 \pi (n+1)]^{1/(2(n+1))}\frac{n+1}{e} - \frac{n^2}{e(n+1)}(2\pi n)^{\frac1{2n}} < 1
$$
Then he argued, that one can replace $n+1$ by $n$ if $n$ is large enough. I don't see how to formally justify this though. 
In order to do this, I would like to determine the limit 
$$
\lim_{x\to\infty} [2 \pi (x+1)]^{1/(2(x+1))}\frac{x+1}{e} - \frac{x^2}{e(x+1)}(2\pi x)^{\frac{1}{2x}}.
$$
Wolfram Alpha says that this limit equals $\frac2e$. I don't see any way of proving this though.
I tried computing the derivative of the above function in order to show that it is decreasing, but that didn't lead to any results. I also tried computing the difference of the $n+1$-st and the $n$-th term, but that didn't work either.

EDIT: This is equivalent to proving that
  $$
\lim_{x\to\infty} \frac1x\left((x+1)\cdot\frac{x+1}e\cdot \sqrt[2(x+1)]{2\pi(x+1)}-x\cdot\frac{x}e\cdot \sqrt[2x]{2\pi x}\right)=\frac2e
$$
  or equivalently that
  $$
\lim_{x\to\infty} \frac1x\left((x+1)^2\cdot \sqrt[2(x+1)]{2\pi(x+1)}-x^2\cdot \sqrt[2x]{2\pi x}\right)=2
$$
  or rewritten again that 
  $$
\lim_{x\to\infty} \frac{(x+1)^2}x\cdot \sqrt[2(x+1)]{2\pi(x+1)}-x\cdot \sqrt[2x]{2\pi x}=2
$$

Wolfram Alpha query for the last limit.

EDIT 2:
  Using l'Hospital, one could also prove that 
  \begin{multline}
\frac12 (4x-\ln(2\pi (x+1))+5) (2\pi (x+1))^{1/(2 x+2)}\\-2 (2\pi x)^{1/(2x)}(x-(1/4)\ln(2\pi x)+\frac14)
\end{multline}
  converges to $2$.

Third Wolfram Alpha query
 A: I'll be using the form
$$
L=\lim_{x \rightarrow\infty}\left(\frac{(x+1)^2}{x}\left(2\pi(x+1)\right)^{\frac{1}{2(x+1)}}-x\left(2\pi x\right)^{\frac{1}{2x}}\right).
$$
We'll look at the function
$$
f(x)=(2\pi x)^{1/2x}=\exp\left(\frac{1}{2x}\log(2\pi x)\right).
$$
First note that $\lim_{x \rightarrow\infty}f(x)=1$, so we can simplify the original limit:
\begin{align}
L&=\lim_{x \rightarrow\infty}\frac{(x+1)^2f(x+1)-x^2f(x)}{x}\\
&=\lim_{x \rightarrow\infty}\color{orange}{\frac{x^2f(x+1)-x^2f(x)}x}
+\color{blue}{\frac{2xf(x+1)}x}
+\color{green}{\frac{f(x+1)}{x}}\\
&=\color{blue}2+\color{green}0+\color{orange}{\lim_{x \rightarrow\infty}x(f(x+1)-f(x))}\\
&=2+\lim_{x \rightarrow\infty}x(f(x+1)-f(x)).
\end{align}
Now use Lagrange's theorem to write $f(x+1)-f(x)=f'(\xi(x))$ for some $\xi\in(x,x+1)$. To make use of this, we evaluate the limit of $g(x)=xf'(x)$:
$$
\lim_{x \rightarrow\infty}g(x)=\lim_{x \rightarrow\infty}xf'(x)=\lim_{x \rightarrow\infty}f(x)\frac{1-\log(2\pi x)}{2x}=0
$$
Since $\xi(x)\in(x,x+1)$ we have:
$$
\lim_{x \rightarrow\infty}x(f(x+1)-f(x))=\lim_{x \rightarrow\infty}\xi(x)f'(\xi(x))+(x-\xi(x))f'(\xi(x))=\lim_{x \rightarrow\infty}g(\xi(x))=0
$$
Recall that $\lim_{x \rightarrow\infty}\xi(x)=\infty$; the term with $x-\xi(x)$ disappears as this expression is bounded and $f'(\xi(x))$ vanishes in the limit. The very last equality holds because of $\lim_{x \rightarrow\infty}g(x)=0$ and $\lim_{x \rightarrow\infty}\xi(x)=\infty$. This establishes $L=2$.

EDIT: Here's an approach without Lagrange's theorem. We'll again look at the limit
$$
\lim_{x \rightarrow\infty}x(f(x+1)-f(x))
$$
Calculating $f''$, we find
$$
f''(x)=f(x)\left(\left(\frac{1-\log{2\pi x}}{2x^2}\right)^2+\frac{2\log{2\pi x}-3}{2x^3}\right)
$$
In particular, for large $x:f''(x)>0$. This means the function is convex and because $f'<0$ for large $x$, we can estimate
$$
\vert f(x+1)-f(x)\vert < |f'(x)|\cdot 1=\vert f'(x)\vert
$$
But since $\lim_{x \rightarrow\infty}xf'(x)=0$, this implies $\lim_{x \rightarrow\infty}x(f(x+1)-f(x))=0$.
