Understanding a theorem involving limits

Question

I am reading from the book Multidimensional real analysis vol I by Duistermaat and Kolk and am trying to understand the following theorem from it:

Theorem 1.4.2: Suppose $A\subset \mathbb R^n,B\subset \mathbb R^p, f:A\to \mathbb R^p, g:B\to \mathbb R^q, a\in\overline{A},b\in\overline{B},c\in\mathbb R^q$. Further suppose $\lim_{x\to a}f (x)=b $ and $\lim_{y\to b}g(y)=c$. Then $\lim_{x\to a}(g\circ f )(x)=c$.

My question is somewhat silly. For $\lim_{x\to a}(g\circ f )(x)$ to make sense we must have $a\in\overline{\mbox {dom }g\circ f}$, i.e. we must have $a\in\overline{A\cap f^{-1}(B)}$. However we are only given $a\in\overline{A}$. Is this an error ?

Answer 1

You're right. This seems to be an error. You can repair it by restricting yourself to $$f:A\rightarrow B.$$ Without it, $g\circ f$ may not be well defined.

Another thing. You stated $$\lim_{x\rightarrow a} (g\circ f)(x)=b.$$ Shouldn't this $b$ be a $c$?

Answer 2

If $f:[0,1[\rightarrow \mathbb R$ is defined by $f(x)=x$, then we can say $\lim_{x\to 1}f (x)=1 $, even if $1$ is not contained neither on the demain or the image of $f$. The need for $a$ to be in be in ${\overline{ f^{-1}(B)}}$ may exist but is implicitly asserted by $\lim_{x\to a}f (x)=b $.

Also, the corret result should be: $\lim_{x\to a}(g\circ f )(x)=c$.