Finding the limit of a composite function graphically 
Question :
$$\lim_{x\to 1^+}f(1-f(2-x))$$
I'm having trouble understanding how to evaluate composite limits graphically when they aren't continuous. Could someone help me understand how to in a simpler way instead of explicitly stating the formula
 A: $$\lim_{x\to 1^+}f(1-f(2-x))$$
$$=\lim_{\epsilon\to 0^+}f(1-f(2-1-\epsilon))$$
$$=\lim_{\epsilon\to 0^+}f(1-f(1-\epsilon))$$
$$=\lim_{\epsilon\to 0^+}f(1+1-\epsilon)$$
$$=\lim_{\epsilon\to 0^+}f(2-\epsilon)=1$$
A: Here,
lim(x->1+) f(1-f(2-x))
=lim(h->0+) f(1-f(2-1-h))
                 (Substituting x=1+h).
=lim(h->0+) f(1-f(1-h)).
Note from graph that  for x belonging to (0,1),         f(x)=-x.
Thus f(1-h)=-1+h.
Thus required limit
=lim(h->0+) f(1+1-h)
=lim(h->0+) f(2-h)
=1( from graph).
A: Note that as $x\to 1^{+}$, the variable $t=2-x\to 1^{-}$ and hence $u=f(t) \to -1^{+}$ (from graph). And then $z=1-u\to 2^{-}$ and thus finally $v=f(z) \to 1$ (from graph).
The expression $f(1-f(2-x))$ has been handled in a step by step manner by use of different variables starting with $t=2-x$ followed by introduction of $u, z, v$.
Also note that it is not necessary to guess the formula for $f(x) $ from graph (in this easy example it can be done because the graph consists of straight line segments). What is important is to notice that the graph gives information about left and right hand limits of $f$ at various points of interest like $-1,0,1,2$ and that is all we need to solve the problem here. 
