Limit of a composite function that is not continuous Below is a link to the graph that I'm working with.
Graph of function f
From this graph, based on my understanding:
$$\lim_{x\rightarrow 2} f(x)$$
The limit does not exist. This is because as $x$ approaches $2$, $f(x)$ is approaching multiple values (specifically $-2$ and $2$).
The next question is where I'm finding this confusing, which is to compute the following limit:
$$\lim_{x\rightarrow -3} f(f(x))$$
As $x$ approaches $-3$ (but $x \neq -3$), $f(x)$ is close to $2$. I know that for composition of limits, we can only "sub in" the output of the inner function, which is $2$ in this case, as the input of the outer function when at least 1 of following conditions are satisfied:
Consider $g\circ f(x)$

*

*$f(x) \neq 2$ for $x$ on an interval centered at $-3$, except maybe at $-3$.

*$g$ continuous at $2$.

Based on what I see on the graph, $f(x)=2$ on the interval around $x=-3$, so the first condition fails. Also, $g$, which I think is the outer $f$ function, is not continuous at $x=2$. So both conditions have failed. Based on this, we would not be able to "sub in" the inner function into the outer function legitimately.
So I tried to work backwards from the answer, which is $-1$. The only thing I could come up with was that as $x$ approaches $-3$, $f(x)=2$. It is not close to $2$, it is exactly $2$. Since it is exactly $2$ and not approaching $2$, we can just sub in the output $2$ into the outer function $f$, and $f(2)=-1$. I'm a bit skeptical that this kind of answer would be correct, but I've run out of ideas. Has my reasoning made sense so far? Perhaps I may be missing out on a key idea?
Any assistance would be greatly appreciated!
EDIT:
Although I'm not 100% sure, I think the answer has to do with the idea that $\lim_{x\rightarrow -3} f(f(x))$ isn't actually a composition of limits at all. The prerequisites for composition of limits are:

*

*$\lim_{x\to a}f(x)=L$

*$\lim_{y\to L}g(y)=M$
$\lim_{x\to a}f(x)=L$ means that when $x$ is close to $a$ (but $x\neq a$), $f(x)$ is close to $L$. In our case of $x\to -3$, when $x$ is close to $-3$ (but $x\neq -3$), $f(x)=2$ and is not close to 2. So when we use the output of (1) (which is not a limit anymore by the definition) as the input for (2), there is no need to obey at least one of the two conditions because this is not a composition of limits. We can simply compute and get the final answer of $-1$.
 A: Since you mention you are interested in the reasoning behind the answers, lets go into that first. To be more general, let $A,B,C\subseteq\mathbb{R}$ be non-empty and lets say $f:A\to B$ and $g:B\to C$ are two functions. Given some $a\in A$, we can say that $$\lim_{x\to a}g(f(a))=g\left(\lim_{x\to a}f(x)\right)$$ provided that $g$ is continuous at $\lim\limits_{x\to a}f(x)$.
In your case, we had two points of interest: $x=-3$ and $x=0$. Both of these cases fail to satisfy the above conditions, so we have to resort to another method, that being left and right sided limits.
In the case where $x=-3$, we can approach $x$ from the left and the right; approaching from the left, graphically, tells us that $\lim\limits_{x\to -3^-}f(f(x))=-1$ and approaching from the right tells us that $\lim\limits_{x\to -3^+}f(f(x))=-1$. This means $\lim\limits_{x\to -3}f(f(x))=-1$.
Now, in the case where $x=0$, approaching from the left of zero yields $\lim\limits_{x\to 0^-}f(f(x))=-2$ and approaching from the right similarly yields $\lim\limits_{x\to 0^+}f(f(x))=-2$.
Since you don't need to be completely rigorous in your justification, to compute, for example, the left sided limits, simply move your finger near the left of $-3$ and approximate the out put of $f$ where you finger is. For example, lets say your finger is near $x'=-3.5$. Then $f(-3.5)=2$ and $f(f(-3.5))=f(2)=-1$. From the right, lets say your finger is near $x'=-2.5$. Then $f(f(-2.5))=f(2)=-1$. Since these limits agree, then you can be sure your answer is  $-1$. This is not a very precise answer, but the reasoning behind it is left and right sided limits if you're more interested. Trying to provide epsilon-delta proofs would be a good way to strengthen your understanding of why this works.
