$x$ 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?

  • 4
    $\begingroup$ f(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. $\endgroup$ – Inazuma Apr 26 '16 at 7:17
  • $\begingroup$ $f\equiv 1$ 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 $0$. $\endgroup$ – Math1000 Apr 26 '16 at 8:19

No. In order for a function $f(x)$ to be continuous at $x=a$, it must meet three conditions:

  1. $f(a)$ is defined.

  2. $\displaystyle\lim_{x\to a}f(x)$ exists.

  3. $\displaystyle\lim_{x\to a}f(x)=f(a)$.

Your function does not meet the first criterion. Hence, $f(x)$ is not continuous at $0$.

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