I am studying the function $${\sqrt x(\ln x)^{(-(\ln x)^{1/k})^5}\over xe^{-(\ln x)^{6/k}}}.$$ According to Desmos, this seems to go to $0$ for $k\ge 3$, but not for $k\le 2$. I was wondering if there was a simple explanation why, but I've found it a bit difficult to manipulate all of these exponentials and logarithms.
Instead of using $1/k$, is there a more precise threshold $c$ such that replacing $1/k$ in the above by $c$ makes this whole thing go to $0$?
Thanks in advance!
Let $x=e^t$ which makes the expression to be now $$\large e^{t^{6/k}-\frac{t}{2}} \,\,t^{-t^{5/k}}$$ which is easier to explore for large values.
In order the expression goes to $0$, it must goes through a maximum first. Computing the first derivative, we need to solve $$-12 t^{6/k}+k \left(2 t^{5/k}+\color{red}{t}\right)+10 t^{5/k} \log (t)=0 \tag 1$$
Using $$0 < \color{red}{t} <t^{5/k}$$
the solution of $(1)$ are given in terms of Lambert function $$t_{(1)}^{1/k}=-\frac{5}{6} k\,\,W\left(-\frac{6}{5 e^{1/5}\, k}\right)$$ $$t_{(2)}^{1/k}=-\frac{5}{6} k\,\,W\left(-\frac{6}{5 e^{3/10}\, k}\right)$$ In the real domain, the argument of Lambert function must be larger than $\frac 1e$.
So $$t_{(1)}^{1/k}\quad \implies \qquad k > \frac{6 }{5}e^{4/5}=2.67065$$ $$t_{(2)}^{1/k}\quad \implies \qquad k > \frac{6 }{5}e^{7/10}=2.41650$$
I suppose that this explains that.