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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 $x_n=\sqrt[n]n-1$, $n=\left(1+x_n\right)^n\geq 1+\binom{n}{3}x_n^3$. Therefore $x_n^3\leq 12n^{-2},$ if $n\geq 4$ and $\sqrt n\cdot x_n\leq 3n^{-1/6}$. Hence $\lim\limits_{n\rightarrow\infty}\sqrt n \cdot x_n=0$

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  • $\begingroup$ What tools are allowed to you? This is easily solved using the asymptotic expansion $$ x^{1/x}=1+\mathcal{O}\left(\frac{\log x}{x}\right).$$ $\endgroup$ Mar 19 '19 at 21:48
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    $\begingroup$ Unfortunately I have not been exposed to asymptotic expansions $\endgroup$
    – New2Math
    Mar 19 '19 at 21:50
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One has $$\sqrt{x}\left( \sqrt[x]{x}-1\right) = \sqrt{x} \left(e^{\ln(x)/x}-1 \right) = \frac{\ln(x)}{\sqrt{x}} \frac{e^{\ln(x)/x}-1 }{\frac{\ln(x)}{x}} \quad \quad (1)$$

Note that when $x$ tends to $+\infty$, $$\frac{\ln(x)}{x} \rightarrow 0$$

so by definition of the derivative, $$\frac{e^{\ln(x)/x}-1 }{\frac{\ln(x)}{x}} \rightarrow \exp'(0) = 1$$

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

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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.

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L'Hopital's Rule works.

$$\lim_{x \to \infty} \sqrt{x}(\sqrt[x]{x}-1)= \lim_{x \to \infty} \frac{e^{\ln x \over x}-1}{\frac{1}{\sqrt x}} = \lim_{x \to \infty} \frac{\frac{1}{x^2}(1+{\ln x \over x^2})e^{\frac{\ln x}{x}}}{-\frac{1}{2} x^{-3/2}} = \lim_{x \to \infty}\frac{1}{-(\sqrt{x}/2)}=0.$$

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