<s>[INST] Let $\langle x_n\rangle$ be a recursive relation. Find $\lim_{n\to\infty}\frac {x_n}{n^2}.$ Let $\langle x_n\rangle$ be a recursive relation given by $$x_{n+1}=x_n+a+\sqrt {b^2+4ax_n}, n\geq0, x_0 =0$$ and $a$ and $b$ are fixed positive integers. Find  $$\lim_{n\to\infty}\frac {x_n}{n^2}.$$
 [/INST] Clearly, $\lim_{n\to \infty} x_n = \infty$. We have
\begin{align}
\sqrt{x_{n+1}} - \sqrt{x_n} &= \sqrt{x_n + a + \sqrt{b^2 + 4ax_n}} - \sqrt{x_n}\\[6pt]
&= \frac{a + \sqrt{b^2 + 4ax_n}}{\sqrt{x_n + a + \sqrt{b^2 + 4ax_n}} + \sqrt{x_n}}\\[6pt]
&\to \sqrt{a} \quad \mathrm{as}\quad n \to \infty.
\end{align}
By the Stolz-Cesaro theorem, we have
$$\lim_{n\to \infty} \frac{\sqrt{x_n}}{n} = \sqrt{a}$$
and thus
$$\lim_{n\to \infty} \frac{x_n}{n^2} = a.$$
We are done.
</s>