The solution were so messy that there was no way that I could came up with it on my own. Although the binomial Expansion seems reasonable the rest seems so forced because the solution required an induction for an estimate. I also don't know how to pick just the Right estimates to get where I want is there Maybe a rule or a method?
Do you Maybe have another solution than:
For $x_n=\sqrt[n]n-1$, $n=\left(1+x_n\right)^n\geq 1+\binom{n}{3}x_n^3$. Therefore $x_n^3\leq 12n^{-2},$ if $n\geq 4$ and $\sqrt n\cdot x_n\leq 3n^{-1/6}$. Hence $\lim\limits_{n\rightarrow\infty}\sqrt n \cdot x_n=0$
That would work if we had to deal with $n$'s, but we have $x$'s instead, which typically means dealing with limits of functions, not just a particular sequence. But, since $x\rightarrow\infty$ (positive infinite, otherwise we are in trouble defining $x^{\frac{1}{x}}$) we can assume $x>0$ and thus, we can apply floor functions, i.e. $n_x = \left \lfloor x \right \rfloor$, where $n_x \in\mathbb{N}$ $$n_x \leq x < n_x +1$$ or $$\frac{1}{n_x} \geq \frac{1}{x} > \frac{1}{n_x+1} \Rightarrow \\ n_x^{\frac{1}{n_x+1}}<n_x^{\frac{1}{x}} \leq x^{\frac{1}{x}} < (n_x +1)^{\frac{1}{x}}\leq (n_x +1)^{\frac{1}{n_x}} $$ and finally $$\sqrt{n_x}\left(n_x^{\frac{1}{n_x+1}}-1\right)< \sqrt{x}\left(x^{\frac{1}{x}}-1\right) < \sqrt{n_x +1}\left((n_x +1)^{\frac{1}{n_x}}-1\right)$$ Obviously $n_x\rightarrow\infty$ when $x\rightarrow\infty$. Now you can apply the binomial trick you mentioned and squeeze.