3. Let the sequence $\left(x_{n}\right)_{n \geq 0}$ be given by $x_{0}=1, x_{1}=\frac{1}{2}$ and $x_{n+2}=x_{n+1} \cdot x_{n}^{2}, \forall n \geq 0$. Find the general term $x_{n}$ of the sequence and then calculate $\lim _{n \rightarrow \infty} x_{n}$.
Solution. It is first shown inductively that $x_{n}>0, \forall n \geq 0$, then the recurrence relation is logarithmized and we obtain: $\ln x_{n+2}=\ln x_{n+1}+2 \ln x_{n}, \forall n \geq 0$.

Let $\quad \ln x_{n}=y_{n}, \quad \forall n \geq 0 \Rightarrow y_{n+2}=y_{n+1}+2 y_{n} \Leftrightarrow y_{n+2}-y_{n+1}-2 y_{n}=0, \quad \forall n \geq 0 \quad \mathrm{with}$ $y_{0}=\ln 1=0 \quad$ and $y_{1}=\ln \frac{1}{2}=-\ln 2$. The characteristic equation of the sequence has solutions -1 and 2, so $y_{n}=c_{1} \cdot(-1)^{n}+c_{2} \cdot 2^{n}, \forall n \geq 0$. For $n=0$ and $n=1$, we have $c_{1}+c_{2}=0$ and

$$
\begin{gathered}
-c_{1}+2 c_{2}=-\ln 2 \Rightarrow 3 c_{2}=-\ln 2 \Rightarrow c_{2}=-\frac{1}{3} \ln 2 \Rightarrow c_{1}=\frac{1}{3} \ln 2 \Rightarrow \\
y_{n}=\frac{1}{3}\left[(-1)^{n}-2^{n}\right] \ln 2, \quad \forall n \geq 0 \Leftrightarrow y_{n}=\frac{(-1)^{n}-2^{n}}{3} \cdot \ln 2 \Leftrightarrow y_{n}=\ln 2^{\frac{(-1)^{n}-2^{n}}{3}}, \quad \forall n \geq 0
\end{gathered}
$$

But $x_{n}=e^{y_{n}} \Rightarrow x_{n}=2^{\frac{(-1)^{n}-2^{n}}{3}}, \forall n \geq 0$ and $\lim _{n \rightarrow \infty} x_{n}=2^{-\infty}=0$.