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I have this function which approaches zero in discrete steps:

$$\frac{1}{2^{int(x)}}$$

My question is that although this function shows asymptotic behaviour in that it approaches $$y=0$$ does it still have an asymptote even though it isn't continuous?

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    What do you mean by "have an asymptote"?2012-03-19
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    What is the definition of asymptote you are using? In the ones I know, continuity is not involved.2012-03-19
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    Is that supposed to be ${1\over2^{int(x)}}$? If so, can you edit?2012-03-19
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    Is this about Big-Oh notation huh?2012-03-19
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    @GerryMyerson Fixed, sorry!2012-03-19

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