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 $\sqrt{x}$?
Best Answer
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.