Answer
Take these 2 functions: $$f(x)=\frac{1}{x-1}\hspace{2cm}g(x)=x+1$$
They are continuous at $x=0$ but the composite function $f(g(x))$ is not continuous at $x=0$.
Work Step by Step
Take these 2 functions: $$f(x)=\frac{1}{x-1}\hspace{2cm}g(x)=x+1$$
- At $x=0$: $$\lim_{x\to0}f(x)=\lim_{x\to0}\frac{1}{x-1}=\frac{1}{0-1}=-\frac{1}{1}=-1=f(0)$$ $$\lim_{x\to0}g(x)=\lim_{x\to0}(x+1)=0+1=1=g(0)$$
So both $f(x)$ and $g(x)$ are continuous at $x=0$.
Now consider the composite function $fog$: $$fog=f(g(x))=\frac{1}{(x+1)-1}=\frac{1}{x}$$
This function, however, is not defined at $x=0$. So since $f(g(0))$, it is not continuous at $x=0$.
This example yet does not contradict Theorem 9. Theorem 9 states explicitly that $gof$ is continuous at $c$ only if $f$ is continuous at $c$ and $g$ is continuous at $f(c)$, not just $c$ only.
In the example, even though $g(x)$ is continuous at $0$, $g(0)=1$, but $f(x)$ is not defined at $x=1$, so $f(x)$ is not continuous at $g(0)$.
