Function and sequence I'm working on this exercice but I'm not sure about my answers and the mathematical thinking. I did 1) 2) 4) without problems but I doubt on the other.
Let $f$ be a function:
$$f :\; ]0,+\infty[ \rightarrow \mathbb{R} \\
                x\mapsto \frac{x}{2}+\frac{1}{x}$$
1) Study the variations of $f$.
2) Show that the equation $f(x) =x$ admits a unique solution.
Then we consider the  sequence defined by :
$$u_0=1$$ $$u_{n+1}=f(u_n)$$ for any natural number.
3) Explain why the sequence is defined.
We studied the variations of $f$ and with the limits we can see that $u_n\gt 0$.
4) Prove that $$u_n \gt\sqrt2 $$
5) Compare $u_0$ and $u_1$ and say if the sequence is monotonic.
We compute $u_0\gt u_1$. According to the variations of $f$, I would say that the sequence is not monotonic but I'm not sure.
6) Is $\{u_n\}$ convergent? What is the limit of this sequence?
I'm not sure of the way to answer to this question.
Thank you for your help
 A: 3) To show the sequence $\{u_n\}$ is defined we need to show $f(u_n)$ is defined for all $n\geq 0$. That is, that $u_n$ is in the domain of $f$. You have done that by showing $u_n\gt 0$.
5)
\begin{eqnarray*}
u_{n+1} -u_n &=& f(u_n)-u_n \\
&=& \dfrac{u_n}{2} + \dfrac{1}{u_n} - u_n \\
&=& \dfrac{1}{u_n} - \dfrac{u_n}{2}.
\end{eqnarray*}
For $n\geq 1,\; u_n\gt\sqrt{2}\;\;$ so that $\;\dfrac{1}{u_n} \lt \dfrac{1}{\sqrt2}\;$ and $\;\dfrac{u_n}{2} \gt \dfrac{1}{\sqrt{2}}$.
Therefore, $\dfrac{1}{u_n} - \dfrac{u_n}{2} \lt 0,\;$ so for $n\geq 1$ we have $u_n\gt u_{n+1}$.
But $u_0 = 1 \lt \dfrac{3}{2} = u_1.\;$ So $\{u_n\}$ is not motonotic.
$$\\$$
6) We showed $\{u_n\}_{n=1}^{\infty}$ is decreasing and is also bounded below by $\sqrt{2}$. So this sequence has a limit. Let's call it $L$.
For $n\geq 1,\; u_{n+1} = \dfrac{u_n}{2} + \dfrac{1}{u_n}$. Since $\{u_n\}$ converges, we can take the limit as $n\to\infty$ of both sides of this equation, giving,
\begin{eqnarray*}
L &=& \dfrac{L}{2} - \dfrac{1}{L} \\
\therefore\quad L^2 &=& 2 \\
\therefore\quad L &=& \sqrt{2} \qquad\text{since we know $u_n\gt 0$.}
\end{eqnarray*}
