$f$ is a smooth function from a manifold to itself. So is $f\circ f$, and $f\circ f\circ f$ and so on...
If this sequence is extended forever, and supposing that it converges to some function, need the function it converges to be smooth as well? (To be clear, by smooth, I mean $f$ is a diffeomorphism.)
I am tempted to deploy an argument like, "every step of the way is smooth, so the limit must be as well." However, there are sequences of rational numbers that converge, but not to rational numbers. (In fact, this is one way to create the reals.) So I will not deploy that argument, but instead ask you what the truth is!
No. For a counterexample, take $f\colon\Bbb R\to\Bbb R$ defined by $f(x)=x^{1/3}$. Then, writing $f^{\circ 2}=f\circ f$, $f^{\circ3}=f\circ f\circ f$, and so on, $$ \lim_{n\to\infty} f^{\circ n}(x) = \lim_{n\to\infty} x^{1/3^n} = \mathop{\rm sign}(x), $$ which is discontinuous at $x=0$.
In general, a limit of any sequence of smooth functions need not be smooth, so such a heuristic is probably a liability.