Find $f(f(f(f(f(f(\cdots f(x)))))))$ $2018$ times 
I was given a problem to calculate $f(f(f(\dots f(x))))$ $2018$ times if $f(x)=\frac{x}{3} + 3$ and $x=4$. 

Is it even possible to be done without a computer?

We can rewrite the problem as a first-order difference equation:
\begin{align*}
    a_{n+1} &= \frac{1}{3} a_n + 3 \\ a_0 &= 4
\end{align*}
We want to know $a_{2018}$.
There are a number of standard techniques to solve these, that are covered in discrete math courses.  One trick is temporarily ignore the $+3$ at the end and find a general solution to $a_{n+1} = \frac{1}{3} a_{n}$.  It's not too hard to see that $a_{n} = \frac{1}{3^n}$ satisfies that, and so does any multiple of it.
OK, now what about the $3$?  Let's look for a constant solution to $a_{n+1} = \frac{1}{3}a_n  + 3$.  That is, we solve $a = \frac{1}{3}a + 3 $ for $a$.  We get $a = \frac{9}{2}$.
Combining the two, we have a solution $$a_n = c \cdot \left(\frac{1}{3}\right)^n + \frac{9}{2}$$
This sequence does satisfy $a_{n+1} = \frac{1}{3}a_n + 3$ for any $c$, as you can check.  We now just need to fit the initial condition $a_0 = 4$.  Substituting $n=0$ gives
$$
    4= c \cdot 1 + \frac{9}{2} \implies c = - \frac{1}{2}
$$
So our particular solution is 
$$
a_n = -\frac{1}{2} \cdot \left(\frac{1}{3}\right)^n + \frac{9}{2}
    = \frac{9}{2} - \frac{1}{2\cdot 3^n}
$$
Therefore
$$
    a_{2018} = \frac{9}{2} - \frac{1}{2\cdot 3^{2018}}
$$
This is very, very close to $\frac{9}{2} = 4.5$.
