$x$ is divided by $x$. Thus, $f(x)=1$ when $x \neq 0$.
However, at $0$ can we consider $f(x)$ as $1$?
More specifically, do we have to define a rational function as a reduced form?
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4f(x) is not defined at x = 0 so f(x) is not anything. It simply just 'does not exist'. For a more humorous explanation, go ask Siri to divide 0 by 0. – Inazuma Apr 26 '16 at 07:17
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$f\equiv 1$ on $\mathbb R\setminus{0}$ so $\lim_{x\to 0} f(x)=1$. But $f(0)$ is undefined, so it is meaningless to speak of the continuity of $f$ at $0$. – Math1000 Apr 26 '16 at 08:19
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No. In order for a function $f(x)$ to be continuous at $x=a$, it must meet three conditions:
$f(a)$ is defined.
$\displaystyle\lim_{x\to a}f(x)$ exists.
$\displaystyle\lim_{x\to a}f(x)=f(a)$.
Your function does not meet the first criterion. Hence, $f(x)$ is not continuous at $0$.
Christopher Carl Heckman
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Not at all, cf. http://math.stackexchange.com/questions/1482787/can-we-talk-about-the-continuity-discontinuity-of-a-function-at-a-point-which-is/1482900#1482900 – Michael Hoppe Apr 26 '16 at 09:05