How to proove that franctional part of sqrt(n) has the limes inferior (lower limit) equal to 0 and limes superior (upper limit) equal to 1?
Thanks in advance!
Using $[x]$ to mean the fractional part of $x$...
As you point out, $\underline{\lim} [ \sqrt{n} ] = 0$ because for every $n$ there is a $k \gt n$ such that $k = i^2$ (for some integer $i$) and thus $$[ \sqrt{k} ] = [ \sqrt{i^2} ] = [ i ] = 0$$
For $\overline{\lim} [ \sqrt{n} ]$ we can consider the subsequence formed only of the elements $[ \sqrt{n} ]$ where $n=k^2-1$ for some integer $k$. For each element in this sequence (for large enough $k$), $$\left[ \frac{k^2 - 1}{k} \right] \lt [ \sqrt{n} ] \lt 1$$
Then $$\lim_{k \to \infty} \left[ \frac{k^2 - 1}{k} \right] = \lim_{k \to \infty} \left(1 - \frac1k\right) < \lim_{k \to \infty} [ \sqrt{n} ] \lt 1$$
and by the squeeze theorem $ \displaystyle \lim_{k \to \infty} [ \sqrt{n} ] = 1$