## Subject IV. (10 points)

Calculate the limit of the sequence $\left(x_{n}\right)_{n \geq 0}$ which satisfies the relation $x_{n+1}=\sqrt{x_{n}+45}-\sqrt{x_{n}+5}$, with $x_{0} \geq-5$.

Prof. Eugen Jecan, National College Andrei Mureşanu Dej

All problems are mandatory. 10 points are awarded by default.

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Effective working time - 3 hours.

## Grading Scale for Grade XI (OLM 2015 - Local Stage)

## Official $10 p$
## Subject IV. (10 points)

From the recurrence relation, it follows that $x_{n}>0$, for any $n \geq 1$.

We can write: $\left|x_{n+1}-4\right|=\left|\left(\sqrt{x_{n}+45}-7\right)-\left(\sqrt{x_{n}+5}-3\right)\right| \leq\left|\sqrt{x_{n}+45}-7\right|+\left|\sqrt{x_{n}+5}-3\right|=$

$=\frac{\left|x_{n}-4\right|}{\sqrt{x_{n}+45}+7}+\frac{\left|x_{n}-4\right|}{\sqrt{x_{n}+5}+3}=\left|x_{n}-4\left(\frac{1}{\sqrt{x_{n}+45}+7}+\frac{1}{\sqrt{x_{n}+5}+3}\right) \leq\right| x_{n}-4\left(\frac{1}{7}+\frac{1}{3}\right)=\frac{10}{21}\left|x_{n}-4\right|$

Using the obtained inequality, we have successively: $\quad\left|x_{n+1}-4\right| \leq \frac{10}{21}\left|x_{n}-4\right| \leq\left(\frac{10}{21}\right)^{2}\left|x_{n-1}-4\right| \leq \cdots \leq\left(\frac{10}{21}\right)^{n+1}\left|x_{0}-4\right|$

Using the majorization criterion, from $\left|x_{n}-4\right| \leq\left(\frac{10}{21}\right)^{n}\left|x_{0}-4\right|$, we obtain $\lim _{n \rightarrow \infty} x_{n}=4$.