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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 $\sqrt{x}$?

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2 Answers 2

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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.

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

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  • $\begingroup$ Thanks for that! It does in fact answer the “What is going on” part! $\endgroup$ Jan 17, 2020 at 9:17

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