Prove the convergence of sequence and find its limit Let $f : \mathbb{R} \rightarrow \mathbb{R}$ with:
$$f(x) = \frac{1}{x^2}e^{\frac{1}{x}}$$
We consider the sequence $x_n$, having $x_0 \in \left ( 0, \frac{1}{2} \right )$ and $x_{n+1} = f(\frac{1}{x_n})$, for any $n \in \mathbb{N}$.
Prove that $x_n$ is convergent and find its limit.
So far, I've only found that $x_n > 0$ $\forall n \in \mathbb{N}$. I have no idea what I should do next.
Thank you in advance.
 A: Outline:


*

*First, $f$ is unnecessarily confusing, because of the reciprocals. Write instead $g\colon(0,\infty)\to\mathbb{R}$ with $g(x) = x^2 e^x$ (i.e., $f(x) = g(\frac{1}{x})$: your recurrence relation is now
$$
x_{n+1} = g(x_n) \qquad n\in\mathbb{N}.
$$

*We now want to prove that $(x_n)_n$ is a decreasing and positive sequence. By monotone convergence, it will converge.


*

*Prove that $x\mapsto \frac{g(x)}{x} = x e^x$ is increasing on $(0,\infty)$, and bounded between $0$ and $1$ on $(0,1/2]$. Why? Because then $\frac{x_{n+1}}{x_n} = \frac{g(x_n)}{x_n}$ will be, by induction, in $(0,1)$.

*Use the above to get that indeed $(x_n)_n$ is a decreasing and positive sequence.

*Conclude by monotone convergence.


*Now that you showed convergence, you know there exists $\ell\in[0,1/2]$ such that $x_n \xrightarrow[n\to\infty]{} \ell$. By continuity of the function $g$, this $\ell$ must then satisfy
$$
\ell = g(\ell)
$$
i.e.
$\ell^2 e^\ell = \ell$. The solutions are either $0$, or values $\ell$ such that $\ell e^\ell = 1$. But by the above,  $x e^x \in (0,1)$ for all $x\in (0,1/2]$, so there is no solution there... Hence, the only possible limit is $0$.
