Confusing Limit With Logarithms So I am probably below the knowledge level of the average mathematician here- computer science student here studying algorithms at the graduate level and came across this peculiarity that I would appreciate some context to. Hopefully this is a softball question for you folks.
Why is
$$\lim_{x\to \infty} \frac{(\ln (\ln x))^{(\ln (\ln (x) )}}{x^2} = 0$$
But,
$$\lim_{x\to \infty} \frac{(\ln (\ln x))^{\ln (x) }}{x^2} = \infty$$
I have tried to apply L'Hopital's rule to this case which may be correct, but gives some annoying derivatives to decipher. Would appreciate a bit more reasoning as to why $x^2$ dominates the first case but not the second. To my intuition, since $ln(ln(x))$ approaches infinity similar to $ln(x)$ (albeit incredibly slowly), $x^2$ should be dominated in both cases. Wolfram seems to disagree.
 A: We can convert this limit of logs by making the substitution $x=e^{e^{y}}$ (in the spirit of Yuriy S's comment). This will of course not really change the structure of the limit, but I at least find it easier to think about the size of exponentials than the size of logarithms.
Noting that $\ln(e^{e^y})=e^y$ and $\ln(\ln(e^{e^y}))=y$, we have
$$\lim_{x\to \infty} \frac{(\ln (\ln x))^{(\ln (\ln (x) )}}{x^2} = \lim\limits_{y\to\infty}\frac{y^y}{\left(e^{e^y}\right)^2}=\lim\limits_{y\to\infty}\frac{e^{y\ln(y)}}{e^{2e^y}}=0.$$
In the exponents above, $2e^y$ easily beats $y\ln(y)$, so the limit is $0$.
On the other hand 
$$\lim_{x\to \infty} \frac{(\ln (\ln x))^{\ln (x) }}{x^2} = \lim\limits_{y\to\infty}\frac{y^{e^y}}{\left(e^{e^y}\right)^2}=\lim\limits_{y\to\infty}\frac{e^{e^y\ln(y)}}{e^{2e^y}}=\infty.$$
In the exponents above, $e^y\ln(y)$ beats $2e^y$, so the limit is infinite.
Final note: the power of $x$ in the denominator of the limit is really a red herring. Any positive power of $x$ will give the same results.
