Consider the following recurrence
$L_0= 100$, $L_{k+1}=[ L_k^{1/2}]L_k,$
where $[x]=\max\{m\in \mathbb {Z} \mid m\leq x\} .$
I have to prove that
$$\displaystyle \lim_{k\rightarrow \infty}\frac {\log L_k}{ \left(\frac {3}{2}\right)^k \log L_0}=1. $$
My answer:
If $L_k^{1/2}$ is a integer number, then
$L_1= L_0^{3/2} $, $L_2= L_0^{ (3/2)^2} $,..., $L_k= L_0^{(3/2)^k} $.
The result is obtained by applying logarithm. How to control the fractional part in the general case?
In the general case (that is, always), the formula does not hold. It is just that 100 is a big number, and a square at that, which makes the error too small to be visible with the naked eye. Try $L_0=2\text{ or }3$, that would make it stand out more prominently.
Really, your limit is just a product of terms of the form $\dfrac{\log L_{k+1}}{{3\over2}\log L_k}$. Every such term is no greater than 1. Calculate a few terms (up to $L_2$, if you insists on having $L_0=100$) and you'll find out that at least some of them are definitely smaller than 1. Just how could the product end up being 1?