To find the given limits from the graph 
The questions asked are with respect to the given graph.
1) $\lim_{x\to0} \ln(1-f(x))$
2) $\lim_{x\to1^-} \ln(2+f(x))$
3) $\lim_{x\to1} (1-f(\ln x))$
4) $\lim_{x\to1^+} f(1- f(2-x))$
I wasn't sure as to how to go about solving these composite functions and I basically just substituted the value of $x$ first to solve the inner function and then went about solving the limit which I  understand is incorrect. Is there a particular formula or way to approach these kinds of problems? Any form of help an examples to illustrate maybe one of the question will be appreciated!
 A: A calculus level answer. This method takes the following as granted:
(1) [Continuous function] If $f$ is continuous at $L$ and $\lim_{x\rightarrow c^\pm}g(x) =L$, then $\lim_{x\rightarrow c^\pm}(f\circ g)(x)=f(L)$ . 
(2) [Change of variable] Suppose $h$ is strictly decreasing and continuous at $c$. If $\lim_{u \rightarrow h(c)^-} g(u) = M$, then $\lim_{x\rightarrow c^+} g(h(x))=M$. 
Then  $\lim_{x\rightarrow 1^+} f(1-f(2-x)) = \lim_{2-x\rightarrow 1^-} f(1-f(2-x))= \lim_{u\rightarrow 1^-} f(1-f(u))$.
But $\lim_{u\rightarrow 1^-} 1-f(u) = 2$.
And $f$ is continuous at $2$. Therefore the required limit is $f(2) = 1$.

An analysis level answer: The rigorous definition of a $g(c+) = L$ is given below:

For all positive number $\epsilon >0$, there is another positive number $\delta > 0$ such that for all real number $x$ where $g(x)$ is defined, 
  $$\delta> x - c>0 \implies |g(x) - L| < \epsilon$$

Now we wish to show that $g(1+) = 1$, where $g(x) = f(1-f(2-x))$.
Pick any $\epsilon >0$. Take $\delta = \min(1, \epsilon).$ Pick any $x \in (1,1+\delta)$. We show that $|g(x)-1|< \epsilon$:
$$1< x <1+\delta \implies  1-\delta< 2-x < 1$$
Now $\delta \leq 1$, for any $t \in (1-\delta,1)$, $f(t)=-t$. Thus,
$$-1< f(2-x) < \delta-1 \implies 2-\delta< 1-f(2-x) < 2 $$
For any $t \in (2-\delta,2)$, $f(t)=t-1$. Thus,
$$1-\delta< g(x)= f(1-f(2-x)) < 1 \implies -\delta< g(x)-1 <0 \implies |g(x)-1|< \delta \leq \epsilon $$
Q.E.D.

Proof of (2): Write domain of $h$ and $g$ as $D(g)$ and $D(h)$ respectively. Also suppose $h(D(h))\subseteq D(g)$.
Pick any $\epsilon >0$. Since $\lim_{u \rightarrow h(c)^-} g(u) = M$, there is some $\delta > 0$ such that for all $u \in D(g)$,
$$\delta > h(c)-u > 0 \implies |g(u) - M|<\epsilon$$
$h$ is continuous at $c$, then $\lim_{x \rightarrow c^+} h(x) = h(c)$. There is some $\eta > 0$ such that for all $x \in D(h)$,
$$\eta > x-c > 0 \implies |h(x) - h(c)|<\delta$$
With this $\eta$, we are done: Pick any $x \in D(h)$ such that $\eta > x-c > 0$. $h(x) \in D(g)$.
$x > c \implies h(x) < h(c)$. Then $0 < h(c)-h(x) < \delta$. Therefore, $|g(h(x)) - M| < \epsilon$.
