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.

Rewrite the limit as
$$L = \lim_{x \to +\infty} \frac{x^{\frac{x^2 + 1}{x}}}{\left(\frac{x^2 + 1}{x}\right)^x} = \lim_{x \to +\infty} \frac{x^{\frac{2x^2 + 1}{x}}}{(x^2 + 1)^x} = \lim_{x \to +\infty} e^\ell,$$
where
$$\require{cancel}\ell = \frac{2x^2 + 1}x\ln x - x\ln(x^2 + 1) = \cancel{2x\ln x} + \frac1x\ln x - \cancel{2x\ln x} - x\ln\left(1 + \frac1{x^2}\right).$$
Since $\ell \to 0$ and $\exp(x)$ is continuous, we conclude that the limit is $1$.
