General procedure for solving a special type of limit of a sequence What general procedure does one use to solve $\lim_{n\to\infty} a_n$,where $a_{n+1}=f(a_n),n\in\mathbb{N}$ given $a_1$?
I tried searching for articles on this but didnt find any.
My initial thought was that I simply put $a_n=a_{n-1}$ for large $n$'s and solve $a_{n}=f(a_{n-1})$, but then again im not sure of its validity since I couldnt find any reference.
 A: If you know that

*

*$a_n$ is convergent and

*$f$ is continuous

then assuming the limit of $a_n$ is $a$, you have
$$a = \lim_{n\to\infty} a_n = \lim_{n\to\infty} a_{n+1}=\lim_{n\to\infty} f(a_n) = f(\lim_{n\to\infty} a_n)=f(a)$$
which means that the only possible values of the limit are the fixed points of $f$. Note that if $f$ is not continous, then you can't make the same conclusion since the penultimate inequality does not always hold for discontinuous functions.

Also note that above, you only get candidates for $a$ (the limit). You must prove, separately, that the sequence converges and which of the candidates it converges to.
Some standard ways of doing this are

*

*By proving that the sequence is monotonic and bounded (in which case, the limit is its strict upper/lower bound if the sequence is increasing/decreasing), which may actually depend on the starting value of $a_1$ (see, for example, $f(x)=x^2$, in which case the sequence $a_n$ can converge to $0$ or $1$ or diverge, depending on $a_1$).

*By showing that $f$ is a contraction (in which case, $f$ only has one fixed point which is always the limit.

A: In general, the relevant result to deal with such problems is the fixed point theorem. This will provide sufficient conditions for the sequence to converge, but will not provide directly the limit. If the sequence converges, and $f$ is continuous the limit, in case it exists, will satisfy $L = f(L)$.
In some particular cases, you may prove convergence without using the fixed point theorem. A usual case in exercises is when you can prove that the sequence is bounded and monotonous.
