I am given the sequence $(x_n)_{n \ge 0}$ with the recurrence relation
$$x_{n+1}=x_n + \dfrac{2}{x_n}$$
and $x_0=1.$ I have to find the following limit:
$$\lim\limits_{n \to \infty} \dfrac{x_n}{\sqrt{n}}$$
In the first part of the problem, I had to find the limit of the sequence itself. This is what I did:
Let $$\lim\limits_{n \to \infty} x_n = a$$
My recurrence relation is:
$$x_{n+1}=x_n + \dfrac{2}{x_n}$$
If I take the limit of both sides I get:
$$\hspace{2cm} a=a+\dfrac{2}{a} \hspace{2cm} -|a$$
$$\hspace{2cm} \dfrac{2}{a}=0 \hspace{4cm}$$
Which means:
$$a=\pm \infty \hspace{1.5cm}$$
Now, since the terms of the sequence are clearly positive,
$$a= + \infty$$
Which means:
$$\hskip{6cm} \lim\limits_{n \to \infty}x_n = \infty \hskip{6cm} (1)$$
Great. I think I got this right. If not, please correct me. Now, the second part of the problem asks me to find:
$$\lim\limits_{n \to \infty} \dfrac{x_n}{\sqrt{n}}$$
And I don't know how to approach this. I can see that since we have $(1)$, this is a limit of the type $\dfrac{\infty}{\infty}$, so L'Hospital comes to mind. However I don't see any way of applying it.
Best Answer
Your first part is not rigorous. When you write let $\lim_{n\to\infty} x_n=a$ you have made an implicit assumption that the limit exists but unless this assumption is justified the approach can't be considered rigorous. Also one can't write $a=\pm\infty $.
First of all note that the sequence is consisting of positive terms and the sequence is increasing. Therefore it either tends to a limit or to $\infty $. If it tends to a limit $L$ then we must have $L\geq x_0=1$ and taking limit of the recurrence relation we get $L=L+(2/L)$ which can't hold. Thus $x_n\to\infty $.
For the next part use the hint given in comments. We have $$x_{n+1}^2=x_n^2+4+\frac{4}{x_n^2}$$ so that $$x_{n+1}^2-x_n^2\to 4$$ By Cesaro-Stolz we have $x_n^2/n\to 4$ and hence $x_n/\sqrt{n} \to 2$.