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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