# Finding limits of composition functions of a piecewise?

Having insane amounts of trouble doing this. Here's a graph of $f(x)$:

How am I to calculate $\lim_{x \to 0} f(f(x))$ , $\lim_{x \to 3} f(f(x))$ , $\lim_{x \to 0} f(1+x^2)$. One that is even more confusing for me is $\lim_{x \to 0} f((1+x)^2)$. Is there some law I'm missing that is preventing me from calculating these limits? I can't seem to grab the intuition or any idea whatsoever of how to get about solving these.

• I think your best approach is to work through it step by step. If $x\to 0$, then since $f$ is continuous at $0$, we just have $\lim_{x\to 0}f(x)=f(0)=1$. Okay, so we have $$\lim_{x\to 0}f(f(x))=\lim_{x\to 1}f(x).$$ Can you generalise this approach? – Ian Coley Oct 4 '14 at 22:25
• go step by step and this theorem might help you proofwiki.org/wiki/Limit_of_Composite_Function – sha Oct 4 '14 at 22:27
• @IanColey does that mean that limit does not exist?? – Backslash Oct 4 '14 at 22:39
• @IanColey You are wrong. For the first limit you have to be aware of the fact that in a neighbourhood of $f(0)$ the function $f$ is constant, hence the limit is just $f(f(0)) = f(1) = -2$. – Crostul Oct 4 '14 at 22:50
• @Qasim Have you worked this out yet? – Vincenzo Oliva Oct 30 '14 at 21:32

As for the first limit, note that for $|x| < 1 , f(f(x)) = f(1) = -2$. So the first limit is $$\lim_{x \to 0} f(f(x)) = -2$$
The second limit does not exist, since (sufficiently near to $3$) for $x>3$ you have $f (f(x)) = 1$ while for $x<3$ you have $f(f(x)) > \frac{3}{2}$.
The third limit is just $$\lim_{x \to 0} f(1+x^2) = \lim_{h \to 1^+} f(h) = 2$$
• I just used the definition of limit. The intuition is "see how $f(f(x))$ behaves locally". Moreover this function $f$ has the property that all right-hand and left-hand limits exist, so you can easily compute the limit on the two sides and see wether these two coincide. – Crostul Oct 5 '14 at 7:38