Somemathhere

Let $f$ be twice continuously differentiable function defined on $\mathbb{R}$, such that $f(x)f''(x)=1$ for all $x\ge 0$, $f(0)=1$ and $f'(0)=0$. Find $$\lim_{x\to\infty} \frac{f(x)}{x\sqrt{\ln(x)}}.$$

The first two limits of the following problem were proposed at VJIMC, 2005, Category I. The exact value of the last limit was proposed by a user at https://artofproblemsolving.com , where you can also see his solution along other lines.  Let $(x_n)_{n\ge 2}$ be a sequence of real numbers, such that $x_2>0$ and for every $n\ge 2$ holds  $$x_{n+1}=-1+\sqrt[n]{1+nx_n}.$$ Prove consecutively that  $$1)\ \lim_{n\to\infty}x_n=0,\  \ \ 2)\ \lim_{n\to\infty}nx_n=0,\  \ \ 3)\ \lim_{n\to\infty}n^2x_n=4.$$ $\textbf{Proof.}$  $\textbf{1)}$ Clearly all the elements of the sequence are positive. The inequality $-1+\sqrt[n]{1+nx_n}<x_n$ is equivalent to $(1+nx_n)<(1+x_n)^n$, which is seen to be true after expanding, since all the summands on the right are positive. This shows that the sequence is strictly decreasing. Hence $$0<x_{n+1}=-1+\sqrt[n]{1+nx_n}\le -1+\sqrt[n]{1+nx_2},$$ and since the right hand side clearly tends to $0$ we obtain $\displaystyle\lim_...