Does $ln(x) +2$ composed with itself infinitely have a limit? I was bored and playing around with my calculator, and I decided to take a number, take its natural log, add 2, and then repeat. I tried numbers as small as 5 and numbers larger than a billion, and they all seemed to end the same way. The numbers basically stopped decreasing when they get to between 1 and 4, and then stay in that interval. If $f^{{\circ}n}(x)$ is $f(x)$ composed with itself $n$ times and $f(x)=ln(x)+2$, then does this limit exist? $$\lim_{n\to{\infty}}f^{{\circ}n}(x)$$ And if it does, what happens if you replace 2 with an arbitrary constant?
By the way, the last Calc class I completed was Calc 2, if you need to know what math level I'm at.
 A: The derivative of $$f(x)=\mathrm{ln}(x)+2$$ is $1/x$, which is positive and bounded above by $1/a<1$ on the interval $[a,b]$ for any real numbers $1<a<b$. Moreover, $2 \leq f(x)$ for for all $x \geq 1$. It follows that $f(x)$ is a contraction mapping from $[a,\infty)$ to itself, when $a$ is slightly larger than $1$. Thus (by using the contraction principle) if you iterate $f$, by applying it to any number larger than $1$ over and over again, you will obtain a sequence converging to the unique fixed point of $f$ on $[1,\infty)$.
A: This is perfect question for Banach fixed point theorem.
This theorem states, that for any contraction on a complete metric space has exactly one fixed point, and defined by you iterated application of the function always converges to to fixed point.
For this case, it is enough to know that any closed interval (for example $[1, 7]$, or $[0, \infty)$) is complete metric space with respect to metric $|x - y|$. Contraction is function $T$ that satisfies $q|x - y|\geq|T(x) - T(y)|$ for any points $x, y$ for some fixed $q \in [0, 1)$.
In our case, we can choose our domain to be $[1+\varepsilon, \infty)$ for any $\varepsilon > 0$. Then, since derivative of our function is $1/x$ is bounded on the domain by $q = \frac 1 {1 + \varepsilon} < 1$, function $f(x) = \ln x + c$ is Lipshitz with constant $q$. Therefore Banach theorem can be applied: there exists unique fixed point in our domain, and any point  from it will converge to it when $f$ is applied infinitely many times.
Moreover, for any $x > 1$ we can choose $\varepsilon$ such that $x > \varepsilon + 1$, and so we have convergence on $(1, \infty)$.
One thing we should keep in mind, is that $f$ has to take values in the same complete metric space. In our case, we need to $f(x) > 1$. So any constant $c > 1$ would make any number $x_0 > 1$ converge to something. That limit will depend on $c$, as it will be fixed point of $f$, so the only solution to
$$ \ln x + c = x. $$
Others already posted what this fixed point is in your initial case $c = 2$.

Edit: when $c > 1$ is already fixed, we can extend domain on which
$f^{\circ n}(x_0)$ converges. If after first iteration we end up in $(1, \infty)$, then sequence will converge. So we can pick any $x_0$ such that $f(x_0) > 1$. We can solve this inequality
$$
\ln x_0 + c > 1 \quad \iff \quad x_0 > e^{1 - c}
$$
So in the end, any constant $c > 1$ will make sequence $f^{\circ n}(x_0)$ converge for any $x_0 > e^{1 - c}$. Note that this may still be rough condition, and other points may converge to the stable fixed point as well. In fact we could iterate procedure we just did, and obtain convergence on $(\exp(e^{1-c}-c), \infty)$ etc.
A: You're asking if there's a fixed point of the equation $x = \ln x + 2$.  It turns out that there is, at $x \approx 3.1461932206205825$.
Wolfram Alpha shows another solution at $x \approx 0.15859433956303937$, but this is an “unstable” fixed point, where iterating $x_{new} = \ln(x_{old}) + 2$ brings you away from it and towards the other solution.
