# Proving the limit $\lim\limits_{x\to \infty }\left(\sqrt{x}\left(\sqrt[x]{x}-1\right)\right)=0$

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 $$\lim\limits_{n\rightarrow\infty}\sqrt n \cdot x_n=0$$

• What tools are allowed to you? This is easily solved using the asymptotic expansion $$x^{1/x}=1+\mathcal{O}\left(\frac{\log x}{x}\right).$$ Mar 19 '19 at 21:48
• Unfortunately I have not been exposed to asymptotic expansions Mar 19 '19 at 21:50

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 $$0$$.

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 $$\frac{1}{n_x} \geq \frac{1}{x} > \frac{1}{n_x+1} \Rightarrow \\ n_x^{\frac{1}{n_x+1}} and finally $$\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.

L'Hopital's Rule works.

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