Limit of $x_n^3/n^2$ when $x_{n+1}=x_n+ 1/\sqrt {x_n}$ with $x_0 \gt 0$ 
Let $(x_n)_{n \ge 0}$  a sequence of real numbers with $x_0 \gt 0$ and
  $x_{n+1}=x_n+ \frac {1}{\sqrt {x_n}}$.
Check the existence and find $$L=\lim_{n \rightarrow \infty} \frac {x_n^3} {n^2}$$


Because $x_n$ is increasing, there is $\lim_{n \rightarrow \infty} {x_n}=l$ and cannot be finite (otherwise $l=l + \frac 1 {\sqrt {l}}$ impossible) therefore $L$ is indeterminate of $\frac {\infty}{\infty}$ form.

Using Stolz theorem one can get
$$
L = \lim_{n\to \infty} \frac{x_n^3}{n^2} = \lim_{n\to \infty}\frac{x_{n+1}^3-x_n^3}{2n+1} = \lim_{n\to \infty} \frac{3x_n^{3/2} + 3 + \frac{1}{x_n^{3/2}}}{2n+1} = \lim_{n\to \infty} \frac{3x_n^{3/2}}{2n+1}
$$
Notice that
$$
\frac{3x_n^{3/2}}{2n+1} \sim \frac{3x_n^{3/2}}{2n}
$$
That's why we get
$$
L = \frac{3}{2}\sqrt{L}
$$
and $L=0$ or $L=9/4$ or $L=\infty$ or the limit does not exist. Let's prove that the case $L=0$ is impossible. By induction we will prove $x_n^3 \ge n^2$. The base of induction is obvious. The induction step
$$
x_{n+1}^3 = x_n^3 + 3x_n^{3/2} + 3 + \frac{1}{x_n^{3/2}} \ge n^2 + 3n + 3 \ge (n+1)^2
$$
So the case $L=0$ is impossible. Similarly one can show that $L\neq \infty$. So $L = 9/4$ if someone will prove that it exists. Currently I don't know if it exists for any $x_0$ or not.
