Finding the fixed points of a contraction Banach's fixed point theorem gives us a sufficient condition for a function in a complete metric space to have  a fixed point, namely it needs be a contraction.
I'm interested in how to calculate the limit of the sequence $x_0 = f(x), x_1 = f(x_0), \ldots, x_n = f(x_{n-1})$ for a fixed $x$. I couldn't figure out a way to do this limit with ordinary limits calculations. 
The only thing I have at my disposal is the proof of the theorem, from which we see that the sequence $x_n$ is a Cauchy sequence; from this, I'm able to say, for example, that $\left|f(f(f(x))) - f(f(f(f(x))))\right| \leq \left|f(x_0)-f(x_1)\right| ( \frac{k^3}{1-k})$, where $k$ is the contraction constant, but I can't get any further in the calculations.
My question is: how should I procede to calculate this limit exactly? If there are non-numerical (read: analytical) way to do this.
Remark: I'm interested in functions $\mathbb{R} \rightarrow \mathbb{R}$ (as it can be seen from my use of the euclidean metric in $\mathbb{R}$)

In addition to what Hans Lundmark has said about solving $x = f(x)$, you could also try writing a simple computer programme to read a number $n$ and a starting value $x_0$, and compute the result of applying f to $x_0$ $n$ times, thus delivering an approximation to the root that you are seeking. The value of $n$ may have to be fairly large in some cases. 
