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.
 A: 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.
A: We will use Stolz theorem. Many times before applying this theorem it is a good idea to make the denominator linear.
We do so by observing we can calculate $\sqrt L= \lim _n \frac{x_n^{3/2}}{n}$ instead.
\begin{align*}
\lim _n \frac{x_n^{3/2}}{n} &=^{\text{ST}} \lim _n x_{n+1}^{3/2}- x_n^{3/2} \\
&= \lim _n \frac{x_{n+1}^{3}- x_n^{3}}{  x_{n+1}^{3/2}+ x_n^{3/2}}\\
&= \lim _n \frac{3 x_n^{3/2}}{  x_{n+1}^{3/2}+ x_n^{3/2}}\text{,  see Vortuoz's answer}\\
&= \lim _n \frac{3 x_n^{3/2}}{  ( x_n +\frac{1}{\sqrt{x_n}} )^{3/2}+ x_n^{3/2}} \\
&= \lim _n \frac{3 }{  ( 1 +\frac{1}{\sqrt{x_n^3}} )^{3/2}+ 1} \\
&=\frac{3}{2}
\end{align*}
In the second to last line we used that $\lim _n x_n=\infty$.
Therefore $$L=\frac{9}{4}.$$
A: Just for reference, we can prove that $(x_n)$ has the following asymptotic expansion: There exists a constant $C$, depending only on $x_0$, such that
$$ x_n^{3/2} = \frac{3}{2}n + \frac{1}{4}\log n + C + \mathcal{O}\left( \frac{\log n}{n} \right) \tag{1} $$
where the implicit bound for $\mathcal{O}$ possibly depends on $x_0$. Here are some explanations:


*

*Continuum analogue of the problem $y'(t) = y(t)^{-1/2}$ has the solution $y(t)^{3/2} = \frac{3}{2}t + \text{const}$. This suggests that $x_n^{3/2} \approx \frac{3}{2}n$.

*From the previous observation, we ''linearize'' $x_n$ by considering $y_n := x_n^{3/2}$. This new sequence satisfies the following recurrence relation:
$$ y_{n+1} = y_n \left( 1 + \frac{1}{y_n} \right)^{3/2} = y_n + \frac{3}{2} + \frac{3}{8y_n} + \mathcal{O}(y_n^{-2}). $$
As in @clark's answer, this immediately yields the asymptotics $y_n \sim \frac{3}{2}n$. Moreover, this type of recurrence relation is well-understood. See my blog posting, for instance.
