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Since the domain of $f(x)$ is $(-\infty, 1) \cup (1, \infty)$ is there any point discontinuity in $f(x)= \frac{x^2-1}{x-1}$?

Thomas Andrews
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UndergroundVault
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    you realise $$\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1}$$ – Chinny84 Apr 24 '15 at 14:59
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    Although $\lim_{x\to 1}f(1)$ exists, $f(1)$ is not defined, so there is discontinuity at $x = 1$. – James Pak Apr 24 '15 at 15:06
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    No, it is not defined, so it is not discontinuous at $x=1$. If you defined $f(1)=3$, then $f$ would be discontinuous at $1$. @JamesPak – Thomas Andrews Apr 24 '15 at 15:09
  • @ThomasAndrews "No, it is not defined, so it is not discontinuous at $x=1$" I don't quite get it. Continuity exists if and only if $\lim_{x \to a}f(x)$ exists and $f(a)$ is defined and $\lim_{x \to a}f(x)=f(a)$. – James Pak Apr 24 '15 at 15:27
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    But it is not called a discontinuity unless $f$ is defined at the point. @JamesPak – Thomas Andrews Apr 24 '15 at 15:51
  • @ThomasAndrews Oh I see. There is continuity or discontinuity only if $f(a)$ is defined. – James Pak Apr 24 '15 at 16:05

1 Answers1

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Rational functions are continuous in their domain.

ajotatxe
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