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If I have a removable discontinuity, do i have any kind of asymptote?

I originally thought no, but this confused me a bit:

http://www.purplemath.com/modules/asymtote4.htm

Close to the bottom, it says that the function (with a removable discontinuity) has both a vertical asymptote at x = 2 and a slant asymptote. Is this correct?

Jon
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  • Yes, but a graph can have a vertical asymptote as well as a slant asymptote. There is nothing wrong with that... – imranfat Aug 28 '14 at 19:01
  • That's not what I am asking. Can a graph have BOTH a vertical asymptote at x = 2 AND a removable discontinuity at x = 2? I've been told no by my teacher, so that's what I am not understanding. – Jon Aug 28 '14 at 19:04
  • A function $f$ is said to have a vertical asymptote at $a$ if, and only if, $\lim \limits_{x\to a}(f(x))=\pm \infty$ and it is said to have a removable discontinuity at $a$ if, and only if, $\lim \limits_{x\to a}(f(x))\in \mathbb R$, so your teacher is right by virtue of the definitions. – Git Gud Aug 28 '14 at 19:08

1 Answers1

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The answer given is incorrect, there is no vertical asymptote for that example. It has a removable discontinuity at $x=2$.

Paul Sundheim
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  • Thank you, I thought so as well. – Jon Aug 28 '14 at 19:11
  • THe problem lies with purple math. It first suggest that there is a vertical asymptote and then it says that it actually is a removable disc. I can now see where the confusion comes from. The graph has no vertical asymptote at all. So I concure with Paul (+1) – imranfat Aug 28 '14 at 19:26