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$$a,b \gt 0$$ $$\mathop {\lim }\limits_{x \to 0 } \frac{a}{x}\left[ {\frac{x}{b}} \right]$$

So, I know that if x is $x \in \mathbb{Z}$ then the limit is $a\over [b]$
I couldn't figure out the solution for $x \notin \mathbb{Z}$

By the way, $[x]$ notation is equal to floor(x).

I'll be glad for a direction here

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If $x<b$, then $\lfloor x/b \rfloor=0$. – Ian Mateus Dec 17 '13 at 19:12
OK, that's a good one, but what about the case when $x \ge b$? – AndrePoole Dec 17 '13 at 19:15
@AndrePoole: you are making $x$ tend to $0$, hence ultimately you have $x<b$. – zarathustra Dec 17 '13 at 19:16
up vote 5 down vote accepted


Assume that $b>0$, prove that $$\mathop {\lim }\limits_{x \to 0^+ } \frac{a}{x}\left[ {\frac{x}{b}} \right]=0$$ and $$\mathop {\lim }\limits_{x \to 0^{-} } \frac{a}{x}\left[ {\frac{x}{b}} \right]=\infty$$

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Oh, that's easy actually. Got it now. Thanks – AndrePoole Dec 17 '13 at 19:17
I'm such a n00b :) – AndrePoole Dec 17 '13 at 19:18
@AndrePoole You're welcome. – user63181 Dec 17 '13 at 19:18
@AndrePoole. So were we all. – Rick Decker Dec 17 '13 at 20:40
I see now what is going inside the limit now. :+) – Babak S. Dec 18 '13 at 6:00

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