Recently, a friend of mine, Chris, posed the following question to me:

Consider the sequence of functions $latex , f_0 (x) = x, f_1 (x) = \sin (x), f_2 (x) = \sin{(\sin (x)) }.$ For what values $latex x \in {\bf R}$ does the limit of this sequence exist, and what is that limit?

After a few moments, it is relatively easy to convince oneself that for all $latex x $, this sequence converges to $latex 0 $, but a complete proof seemed tedious. Chris then told me to consider the concept of fixed points and a simple solution would arise.

If such a sequence were to converge to a limit, then it could only do so at a fixed point of that sequence, i.e. a point $latex x$ such that $latex f_1 (x) = f_2 (x) = … = f_n (x) = … = L$, and in that case, the limit would be $latex L $. What are the fixed points of the $latex sin $ composition? Only $latex 0 $! Then it takes only the simple exercise to see that the sequence does in fact have a limit for every x (one might split the cases for positive and negative angles, in which case one has a decreasing/increasing sequence that is bounded below/above for example).

A cute little exercise, I think.