Limit of the composition $f \circ f$

Question

Say, for example, that $\lim_{x \to -1}f(x) = 0 = \lim_{x \to -2} f(x)$. How can the limit of $\lim_{x \to -1} f(f(x)) \neq \lim_{x \to -2} f(f(x))$? Wouldn't they all be $f(0)$?

Answer 1

Here's an alternative example. I'm going to switch to $-1$ and $+1$ as the two points, for convenience.

Let $$f(x) = \begin{cases} -2(x+1)^2, & x<0 \\ 3(x-1)^2, & x>0 \end{cases}.$$ (You can define $f(0)$ to be anything you want, or leave it undefined.) Note that $\lim\limits_{x\to 0^+} f(x) = 3$ and $\lim\limits_{x\to 0^-} f(x)=-2$.

Then $\lim\limits_{x\to -1} f(f(x)) = -2$ and $\lim\limits_{x\to +1} f(f(x)) = 3$.

You have to actually think through the limit definitions here. No shortcuts!

Answer 2

Consider $f(x)=\frac{(x+1)^2(x+2)^2}x$. Then $\lim_{x\to-1}f(f(x))=-\infty$ while $\lim_{x\to2}f(f(x))=+\infty$.