I was looking at the limits of composite functions and I came across this question which I do not really understand:
Let $g:\mathbb R\to\mathbb R$ be defined by $g(x)=\begin{cases}0,&\ \text{ if }x=1\\ 2,&\ \text{ if }x\ne1\end{cases}$
Let $f:\mathbb R\to\mathbb R$ be defined by $f(x)=x+1$, for all $x\in\mathbb R$
Show that $\lim_{x\to0}g(f(x))\ne g(f(0)).$
I am guessing that the answer is really trivial and that I am just misunderstanding something, but a simple explanation would be really helpful. Thank you.
You can calculate $g(f(0))=g(1)=0$ directly.
As for the limit, for any $x\ne0$, you have $x+1\ne1$, so $g(f(x))=g(x+1)=2$. Thus $$\lim_{x\to0}g(f(x))=2.$$
So, as your question mentions, $\lim_{x\to0}g(f(x))\ne g(f(0)).$