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 $\sqrt{x}$?

Add Assumptions->{x>0} to your Limit command to get the desired $\sqrt{x}$.
