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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 $x_n=sqrt[n]n-1$, $n=left(1+x_nright)^ngeq 1+binomn3x_n^3$. Therefore $x_n^3leq 12n^-2,$ if $ngeq 4$ and $sqrt ncdot x_nleq 3n^-1/6$. Hence $limlimits_nrightarrowinftysqrt n cdot x_n=0$

One has
$$sqrtxleft( sqrt[x]x-1right) = sqrtx left(e^ln(x)/x-1 right) = fracln(x)sqrtx frace^ln(x)/x-1 fracln(x)x quad quad (1)$$

Note that when $x$ tends to $+infty$,
$$fracln(x)x rightarrow 0$$

so by definition of the derivative,
$$frace^ln(x)/x-1 fracln(x)x rightarrow exp'(0) = 1$$

Finally, you get from $(1)$ that the limit is $0$.

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 $xrightarrowinfty$ (positive infinite, otherwise we are in trouble defining $x^frac1x$) we can assume $x>0$ and thus, we can apply floor functions, i.e. $n_x = left lfloor x right rfloor$, where $n_x inmathbbN$
$$n_x leq x < n_x +1$$
or
$$frac1n_x geq frac1x > frac1n_x+1 Rightarrow \
n_x^frac1n_x+1<n_x^frac1x leq x^frac1x < (n_x +1)^frac1xleq (n_x +1)^frac1n_x $$
and finally
$$sqrtn_xleft(n_x^frac1n_x+1-1right)< sqrtxleft(x^frac1x-1right) < sqrtn_x +1left((n_x +1)^frac1n_x-1right)$$
Obviously $n_xrightarrowinfty$ when $xrightarrowinfty$.
Now you can apply the binomial trick you mentioned and squeeze.

L'Hopital's Rule works.

$$lim_x to infty sqrtx(sqrt[x]x-1)= lim_x to infty frace^ln x over x-1frac1sqrt x = lim_x to infty fracfrac1x^2(1+ln x over x^2)e^fracln xx-frac12 x^-3/2 = lim_x to inftyfrac1-(sqrtx/2)=0.$$