Continuity of composition of two-variables and one-variable functions

Question

I'm asked to proof the following:
Let $ f: R^n \to R$ be continouos,
and $\ g: R \to R$ as well.
Proof that $\ h := g \circ f $ is continouos.
Seems to me that it follows straight from the standard definition of continuity, that is:
a function $f$ is continouos if $$\lim_{x\to x_0}{f(x)}=f(x_0) $$ I reasoned as followed.
$\lim_{(x,y)\to(x_0,y_0)}{f(x,y)}=f(x_0,y_0)$
$\lim_{z\to z_0}{g(z)}=g(z_0)$ and, composing the functions,
$\lim_{f(x,y)\to(f(x_0,y_0)} {g[f(x,y)]}=g[f(x_0, y_0]$, but since $f(x,y) \in R$, $f(x_0, y_0)\in R $, $g$ is continouos, then $g \circ f$ is continouos. I've just started trying to tackle proofs and I guess my reasoning is pretty naive and I'm missing the point here, but I can't even see why it should be wrong (if it is). Thanks for your attention.

Answer 1

You should keep limits out of this argument. It is a basic fact of analysis that the composition of two continuous functions is again continuous whereas there is some small print concerning nested limits.

Assume $f$ is continuous at $x_0$, and $g$ is continuous at $y_0:=f(x_0)$. We have to prove that $h:=g\circ f$ is continuous at $x_0$.

Let $g(y_0)=:z_0$. Then $h(x_0)=z_0$ as well. Let a "tolerance" $\epsilon>0$ be given. Since $g$ is continuous at $y_0$ there is an "allowance" $\delta>0$ such that $$|y-y_0|<\delta\quad\Rightarrow\quad |g(y)-z_0|<\epsilon\ .\tag{1}$$ Now $f$ is continuous at $x_0$. This means that the values $f(x)$ are very near $f(x_0)=y_0$ when $x$ is sufficiently near $x_0$. To be precise: There is a $\delta'>0$ such that $$|x-x_0|<\delta'\quad\Rightarrow\quad |f(x)-y_0|<\delta\ ,$$ whereby the "tolerance" $\delta$ in this second backstep is the $\delta$ obtained in the first backstep. The new "allowance" $\delta'$ is the quantity we are after: Using $(1)$ we now have the chain $$|x-x_0|<\delta'\quad\Rightarrow\quad |f(x)-y_0|<\delta\quad\Rightarrow\quad|g\bigl(f(x)\bigr)-z_0\bigr|<\epsilon\ .$$ As $\epsilon>0$ was arbitrary this proves the continuity of $h=g\circ f$ at $x_0$.