## Subiectul II(7p)

Rezolvati in $\mathrm{M}_{2}(\mathrm{R})$ ecuatia : $\mathrm{X}^{3}+3 \mathrm{X}^{2}+3 \mathrm{X}=\left(\begin{array}{cc}7 & -8052 \\ 0 & 7\end{array}\right)$.


Rezolvare : Adunăm $I_{2}$ in ambii membrii ai ecuatiei, notăm $\mathrm{Y}=\mathrm{X}+I_{2}$ şi obţinem ecuaţia

$Y^{3}=\left(\begin{array}{cc}8 & -8052 \\ 0 & 8\end{array}\right)=A$ ( se ţine cont că $\mathrm{X} \cdot I_{2}=I_{2} \cdot X=X$ ) .............................. $2 \mathrm{p}$

![](https://cdn.mathpix.com/cropped/2024_06_07_10cd630d9bafd5609b0eg-1.jpg?height=106&width=1582&top_left_y=1672&top_left_x=243)

![](https://cdn.mathpix.com/cropped/2024_06_07_10cd630d9bafd5609b0eg-1.jpg?height=111&width=1568&top_left_y=1778&top_left_x=241)

Gaseste $\mathrm{Y}=\left(\begin{array}{cc}2 & -2013 \\ 0 & 2\end{array}\right)$ si apoi $\mathrm{X}=\left(\begin{array}{cc}1 & -2013 \\ 0 & 1\end{array}\right)$

## $\underline{\text { Subiectul III(7p) }}$

Fie sirul $\left(\mathrm{x}_{\mathrm{n}}\right)_{\mathrm{n} \in \mathrm{N}}$ definit prin $\mathrm{x}_{\mathrm{n}+1}=\mathrm{x}_{\mathrm{n}}+\mathrm{e}^{-\mathrm{x}_{\mathrm{n}}} \forall \mathrm{n} \in \mathrm{N}$ si $\mathrm{x}_{0} \in \mathbf{R}$. Sa se arate ca

$$
\lim _{\mathrm{n} \rightarrow \infty} \frac{\ln n}{\mathrm{x}_{\mathrm{n}}}=1
$$

Rezolvare: Monotonia : $x_{n+1}-x_{n}=e^{-x_{n}}>0, x_{n}-$ strict crescator

Marginirea: Presupunem $x_{n}$ marginit superior, atunci din Teorema lui Weierstrass obtinem $x_{n}$ convergent, si trecand la limita in relatia de recurenta se obtine $\mathrm{l}=\mathrm{l}+e^{-l}$ cu $l \in R$, deci $\lim _{n \rightarrow \infty} x_{n}=+\infty$ $2 \mathrm{p}$

Aplicand repetat Lema Cesaro-Stoltz obtinem $\lim _{n \rightarrow \infty} \frac{\ln n}{x_{n}}=\lim _{n \rightarrow \infty} \frac{\ln \left(\frac{n+1}{n}\right)}{x_{n+1}-x_{n}}=\lim _{n \rightarrow \infty} e^{x_{n}}$.

$\ln \left(\frac{n+1}{n}\right)=\lim _{n \rightarrow \infty} \frac{e^{x_{n}}}{n} \cdot \lim _{n \rightarrow \infty} n \cdot \ln \left(\frac{n+1}{n}\right)=\lim _{n \rightarrow \infty} \frac{e^{x_{n+1}-e^{x_{n}}}}{n+1-n}=\lim _{n \rightarrow \infty} e^{x_{n}} \cdot\left(e^{x_{n+1}-x_{n}}-1\right)$ $=\lim _{n \rightarrow \infty} \frac{\left(e^{x_{n+1}-x_{n}}-1\right)}{x_{n+1}-x_{n}}=1$ $.4 \mathrm{p}$

