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

Consider the sequence of functions, $ f_0 (x) = x, f_1 (x) = \sin (x),
f_2 (x) = \sin{(\sin (x)) }.$ For what values $ 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 $ x
$, this sequence converges to $ 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 $ x$ such that $ f_1 (x) = f_2 (x) =
\cdots = f_n (x) = \cdots = L$, and in that case, the limit would be $ L $. What are
the fixed points of the $ sin $ composition? Only $ 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.

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