Recursive sequence through a function 
\begin{align}f(x) &= \dfrac{e^{1/x}}{x^2}\\
x_{n+1} &= f\left( \dfrac{1}{x_n}\right)\end{align} Show that $x_n$ is convergent and that its limit is $0$. 

It is very easy to find the limit from the recurrence relation. Just plugging $l$ instead of the terms of the sequence. But how do I show it is convergent? 
I tried to evaluate the difference $x_{n+1}-x_n$ but its a messy calculation...
Also I tried using the function monotony. But I don't see how it tells me something about the function
 A: It depends on $x_0$ whether your sequence converges or not. It is easy to see, that there exists exactly one $\hat{x}>0$ such that $\hat{x}e^{\hat{x}}=1$.
If $x_0\in(0,\hat{x})$ then you get 
$$
0<x_0e^{x_0}<1 \Leftrightarrow 0<x_0^2e^{x_0}<x_0\Leftrightarrow 0<x_1=f\left(\frac1{x_0}\right)<x_0.
$$
This implies that $(x_n)_n$ is monoton decreasing and bounded from below by $0$. Define $x:=\lim_{n\to\infty} x_n<\hat{x}$ and assume $x>0$. From the definition $x_{n+1}=f\left(\frac1{x_n}\right)$ we get $x=f\left(\frac1{x}\right)$ which is a contadiction. Therefore $x=0$ holds.
If $x_0=\hat{x}$ then 
$$
\hat{x}e^{\hat{x}}=1\Leftrightarrow \hat{x}^2e^{\hat{x}}=\hat{x}\Leftrightarrow f\left(\frac1{\hat{x}}\right)=\hat{x}.
$$
This yields $x_n=\hat{x}$ for all $n$ and $x_n\to \hat{x}\neq 0$.
If $x_0>\hat{x}$ then
$$
x_0e^{x_0}>1 \Leftrightarrow x_0^2e^{x_0}>x_0\Leftrightarrow x_1=f\left(\frac1{x_0}\right)>x_0.
$$
This yields that $(x_n)_n$ is a monoton increasing sequence. Assume it converges to $x:=\lim_{n\to\infty} x_n>\hat{x}$. Like in the first case, we get a contradiction. Therefore our assumption that $(x_n)_n$ converges is false and we get that $(x_n)_n$ is an unbounded sequence.
If $x_0<0$ then $x_1>0$ and we have to consider the same cases for $x_1$ insteat of $x_0$ as above.
