## SUBJECT 4

Let the sequence of real numbers $\left(x_{n}\right)_{n \geq 1}$, given by $x_{1}=\alpha$, with $\alpha \in \mathrm{R}^{*}, \alpha \neq 2$ and $x_{n+1}=\frac{x_{n}^{2}+2 x_{n}+2}{x_{n}^{2}+1},(\forall) n \geq 1$. Calculate:
a) $\lim _{n \rightarrow \infty} x_{n}$,
b) $\lim _{n \rightarrow \infty}\left(x_{n}-1\right)^{n}$.

Prof. Dorin Arventiev

## Note:

Working time: 3 hours.

All problems are mandatory.

Each problem is graded from 0 to 7.

No points are given ex officio.

## MATHEMATICS OLYMPIAD

Local stage - Constanța, 21.02.2016

11th Grade

## Grading and marking rubric

No points are given ex officio.
## SUBJECT 4

a) $x_{n+1}-2=\frac{2 x_{n}-x_{n}^{2}}{x_{n}^{2}+1}=\frac{x_{n}}{x_{n}^{2}+1}\left(2-x_{n}\right)$,

$\Rightarrow\left|x_{n+1}-2\right|=\left|x_{n}-2\right| \cdot\left|\frac{x_{n}}{x_{n}^{2}+1}\right| \leq \frac{1}{2}\left|x_{n}-2\right|$

$1 p$

$\Rightarrow\left|x_{n}-2\right| \leq \frac{1}{2^{n-1}}\left|x_{1}-2\right| \rightarrow 0 \Rightarrow \lim _{n \rightarrow \infty} x_{n}=2$

$1 p$

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

$1 p$

$\left|n\left(x_{n}-2\right)\right| \leq \frac{n}{2^{n-1}}\left|x_{1}-2\right| \rightarrow 0 \Rightarrow n\left(x_{n}-2\right) \rightarrow 0$

$2 p$

If $\alpha \in R^{*} \backslash\{1\}, \mathrm{x}_{n} \notin\{0,2\} \quad(\forall) n \in N^{*}$.

Note: Any other correct solution, different from the one in the grading rubric, will receive the maximum score.