Can we calculate this limit analytically? What will be the limit of the following for $x=0$?
$\sin(\cos(\tan(\sin(\cos(\tan(\sin(\cos(\tan( ... \mbox{infinite times} ... \sin(\cos(\tan(x)))))))...)$
On a calculator it seems to approach 0.72063..., but how do I get the answer analytically?
 A: As requested, this is a comment turned answer which shows why the sequences always 
converges and independent of initial starting value of $x$.
Let $f(x) = \sin(\cos(\tan(x)))$. 
After one iteration, it is clear $f$ maps every $x$ into
some value inside the interval $[-1,1]$.
If one plot $f(x)$ over $[-1,1]$, one find:


*

*$f( [-1, 1] ) \subset [0,1]$.

*$f(x)$ is strictly monotonic decreasing over $[0,1]$.


The $1^{st}$ observation tells us we only need to study the behavior of $f^{\circ n}(x)$
for $x \in [0,1]$.
The $2^{nd}$ observation tell us $f^{\circ 2} = f \circ f$ is strictly monotonic
increasing over $[0,1]$. This implies for any $x \in [0,1]$, the even and odd sub-sequences
$f^{\circ 2n}(x)$ and $f^{\circ 2n+1}(x)$ are monotonic in $n$. Since both of them are
bounded by $0$ and $1$ and $f^{\circ 2}$ is continuous, these two sub-sequences converge
to some fixed points of $f^{\circ 2}$. 
If one make another plot of $f^{\circ 2}$ over $[0,1]$, one will notice $f^{\circ 2}$ has
a unique fixed point (the same one as the fixed point of $f$). This implies the limit of the even and odd sub-sequences coincides and equal to the unique fixed point of $f$.
From this, we can conclude:
Start from any $x \in \mathbb{R}$, the sequence $f^{\circ n}(x)$ always converges to
a unique $y \in [0,1]$ which satisfies the equation $y = f(y) = \sin(\cos(\tan y)))$. 
