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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"? –  Antonio Vargas Mar 19 '12 at 2:15
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What is the definition of asymptote you are using? In the ones I know, continuity is not involved. –  Arturo Magidin Mar 19 '12 at 2:15
    
Is that supposed to be ${1\over2^{int(x)}}$? If so, can you edit? –  Gerry Myerson Mar 19 '12 at 2:27
    
Is this about Big-Oh notation huh? –  FiniteA Mar 19 '12 at 3:51
    
@GerryMyerson Fixed, sorry! –  Charlie Somerville Mar 19 '12 at 7:29

1 Answer 1

One usually says that a function $f:(a,\infty)\to\mathbb R$ has horizontal asymptote $y=c$ if $$\lim_{x\to \infty} f(x)=c \tag1$$ The function $f(x)=2^{-\lfloor x\rfloor }$ satisfies (1) with $c=0$ and therefore has horizontal asymptote $y=0$.

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