Solving this limit: $\lim\limits_{x\to\infty}\frac{x^{x+1/x}}{(x+1/x)^x}$ $$\lim\limits_{x\to\infty}\frac{x^{x+1/x}}{(x+1/x)^x}$$
I have tried a lot of things, like: 


*

*transforming those terms to:
$$\frac{e^{(x+1/x)\ln(x)}}{e^{x\ln(x+1/x)}}$$ 

*then I tried L'Hôpital's rule but it was just getting more complex

*I also made them one, like:
$$e^{(x+1/x)\ln(x)-x\ln(x+1/x)}$$

*At last, I tried to "squeeze" them but I couldn't find the perfect function for that.


I hope that this is not a duplicate because I searched but I couldn't find a similar post.

You can rewrite the function as
$$
\left(\frac{x}{x+1/x}\right)^{\!x}x^{1/x}=
\left(\frac{x^2}{x^2+1}\right)^{\!x}x^{1/x}
$$
Now
$$
\lim_{x\to\infty}\frac{1}{x}\log x=0
$$
so
$$
\lim_{x\to\infty}x^{1/x}=1
$$
and we just need to compute
$$
\lim_{x\to\infty}x\log\frac{x^2}{x^2+1}=
-\lim_{x\to\infty}x\log\frac{x^2+1}{x^2}=
-\lim_{t\to0^+}\frac{\log(1+t^2)}{t}=
-\lim_{t\to0^+}\frac{t^2+o(t^2)}{t}=0
$$
so also
$$
\lim_{x\to\infty}\left(\frac{x^2}{x^2+1}\right)^{\!x}=1
$$
