I need your help with evaluating this limit:

limn→∞sinsin…sin⏟n compositionsn,

i.e. we apply the sin function n times.

Thank you.

**Answer**

The first sine is in I1=[−1,1] hence the nth term of the sequence is in the interval In defined recursively by I1=[−1,1] and In+1=sin(In). One sees that In=[−xn,xn] where x1=1 and xn+1=sin(xn). The sine function is such that 0≤sin(x)≤x for every nonnegative x hence (xn) is nonincreasing and bounded below by zero hence it converges to a limit ℓ. The sine function is continuous hence ℓ=sin(ℓ). The only fixed point of the sine function is zero hence ℓ=0. This proves that xn→0, that the sequence (In) is nonincreasing and that its intersection is reduced to the point zero and finally, that the sequence considered in the post converges to zero.

**Edit:** The argument above shows that for **every** sequence (zn), the sequence (sn) defined by sn=sinsin⋯sin(zn) (the sine function being iterated n times to define sn) converges to zero. In other words, there is nothing particular about the choice zn=n.

**Attribution***Source : Link , Question Author : Community , Answer Author : Did*