2. Consider the sequence defined as follows: $x_{1}=1$,

$$
x_{n+1}=x_{n}+\frac{1}{2 x_{n}}
$$

for any $n$.

a) Find the limit of the sequence.

b) Calculate

$$
\lim _{n \rightarrow \infty} \frac{x_{n}}{\sqrt{n}}
$$
Solution. a) The sequence is strictly increasing. If it were bounded, it would converge to a real number $L$, and we would obtain by passing to the limit $L=L+\frac{1}{2 L}$, which is absurd, so the limit is $\infty$

b) We calculate, using Stolz-Cesaro,

$$
\lim _{n \rightarrow \infty} \frac{x_{n}^{2}}{n}=\lim _{n \rightarrow \infty} \frac{x_{n+1}^{2}-x_{n}^{2}}{n+1-n}=\lim _{n \rightarrow \infty}\left(\left(x_{n}+\frac{1}{2 x_{n}}\right)^{2}-x_{n}^{2}\right)=\lim _{n \rightarrow \infty}\left(1+\frac{1}{4 x_{n}^{2}}\right)=1
$$

so the required limit is 1