We know that if $f(x)$ and $g(x)$ are continuous in a domain then $f(x)/g(x)$ is continuous in the domain except for those elements in the domain for which $g(x) = 0$. If I take two functions $f(x) = x$ and $g(x) = 1/x$, then $f(x)$ is continuous for all $R$ and $g(x)$ is continuous for $R-\{0\}$ but if I take $f(x)/g(x)$ which is $x/1/x$ or $x^2$ then it is continuous for all $R$ even when $0$ was not in the domain of $g(x)$. Is this correct?
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2Not exactly because $\frac{x}{\frac{1}{x}} = x^2$ only when $x \neq 0$, so at $x=0$, $\frac{f(x)}{g(x)}$ is undefined – Osama Ghani Jul 20 '18 at 07:46
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To elaborate, $f(x)/g(x)$ is undefined at $0$ but has a removable discontinuity. So if it were defined at $0$, the discontinuity could be removed. – Jam Jul 20 '18 at 07:47
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A function is not just defined by its mapping rule but also by the domain and codomain!
As mentioned in the comments, you have to check the domain carefully. If you have two functions $f:D\to \mathbb R$ and $g:E\to\mathbb R$ then $h:=\frac{f}{g}$ is defined on $$ \{x\in D\cap E~:~g(x)\neq 0\}. $$ In your case $D=\mathbb R$ and $E=\mathbb R\setminus \{0\}$. Because $g(x)\neq 0$ holds for all $x\in E$, you get $D\cap E=E$ as your domain of the quotient $h$. Further $h$ is continuous at each point of $E$ as composition of continuous functions.
Nevertheless, there are a lot of examples, where the quotient $h$ can be continuously extended to a larger domain $F\supset E$. In your case, you can extend $h$ continuously to whole $\mathbb R$. But keep in mind, that the extension is formally not the same as the quotient $h$.
Mundron Schmidt
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