# Show that: $\lim_{n\rightarrow\infty}e^{\frac{\log x}{\log\log xn-\log\log n}-\log\left(n\right)}=\sqrt{x}$.

I want to show that:

$$\lim_{n\rightarrow\infty}e^{\frac{\log x}{\log\log xn-\log\log n}-\log\left(n\right)}=\sqrt{x}$$

I looked it up on Wolfram Alpha, and it says:

$$\lim_{n\rightarrow\infty}e^{\frac{\log x}{\log\log xn-\log\log n}-\log\left(n\right)}=1$$

I got confused because it didn't match my computation results, suggesting that:

$$\lim_{n\rightarrow\infty}e^{\frac{\log x}{\log\log xn-\log\log n}-\log\left(n\right)}=\sqrt{x}$$

However, I did try WA for some values of $$x$$, and it gave the right value:

$$\lim_{n\rightarrow\infty}e^{\frac{\log2}{\log\log2n-\log\log n}-\log\left(n\right)}=\sqrt{2}$$

$$\lim_{n\rightarrow\infty}e^{\frac{\log7}{\log\log7n-\log\log n}-\log\left(n\right)}=\sqrt{7}$$

$$\lim_{n\rightarrow\infty}e^{\frac{\log31}{\log\log31n-\log\log n}-\log\left(n\right)}=\sqrt{31}$$

What is going on here? And how can I show the limit is $$\sqrt{x}$$?

Yes, you are correct. If $$x>0$$ then the limit is $$\sqrt{x}$$. Note that as $$n\to +\infty$$, \begin{align}\log(\log(nx))&=\log\left(\log(n)\left(1+\frac{\log(x)}{\log(n)}\right)\right)\\ &=\log(\log(n))+\frac{\log(x)}{\log(n)}-\frac{1}{2}\frac{\log^2(x)}{\log^2(n)}+o(1/\log^2(n)). \end{align} Hence \begin{align}\frac{\log(x)}{\log(\log(nx))-\log(\log(n))}-\log\left(n\right) &= \frac{\log(n)}{1-\frac{1}{2}\frac{\log(x)}{\log(n)}+o(1/\log(n))}-\log\left(n\right)\\&=\log(n)\left(1+\frac{1}{2}\frac{\log(x)}{\log(n)}+o(1/\log(n))\right)-\log\left(n\right)\\ &=\log(\sqrt{x})+o(1) \end{align} and the result follows.
Add Assumptions->{x>0} to your Limit command to get the desired $$\sqrt{x}$$.