1
$\begingroup$

$x$ 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?

$\endgroup$
  • 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
5
$\begingroup$

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 $0$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.