$\displaystyle lim_{x \to 0^{+}} \frac{b}{x}*[\frac{x}{a}] =$

(a > 0 , b > 0)

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- Jan 31st 2010, 10:44 AMgilyosLimit Problem
$\displaystyle lim_{x \to 0^{+}} \frac{b}{x}*[\frac{x}{a}] =$

(a > 0 , b > 0) - Jan 31st 2010, 10:47 AMMath Major
I must be reading your question wrong, because it looks like you have

$\displaystyle \mathop{\lim}\limits_{x \to 0+} \frac{b}{x} * \frac{x}{a} $

In which case those x's will cancel and the limit is just

$\displaystyle \frac{b}{a} $ - Jan 31st 2010, 11:57 AMgilyos
No , it's with [ ]

- Jan 31st 2010, 12:20 PMHallsofIvy
Which is NOT standard notation. I would guess that you mean the floor function which would be [x]

**without**the upper serifs. - Jan 31st 2010, 12:33 PMPlato
Which is the greatest integer function or the floor function.

We know that $\displaystyle a>0$ so it follows that $\displaystyle \lim _{x \to 0^ + } \frac{x}{a} = 0$.

That being the case at some point, $\displaystyle \frac{x}{a}<1 $ so that means $\displaystyle \left\lfloor {\frac{x}{a}} \right\rfloor = 0$.

Now what can you say about this limit? - Jan 31st 2010, 12:45 PMgilyos
$\displaystyle \frac{x}{a}-1 <= [\frac{x}{a}] <= \frac{x}{a} $

$\displaystyle (\frac{x}{a}-1)*\frac{b}{x} <= [\frac{x}{a}] <= \frac{x}{a}*\frac{b}{x} = \frac{b}{a} $

but , what I doing now ??? - Jan 31st 2010, 12:55 PMPlato
That has absolutely nothing to do with this question.

Do you understand that $\displaystyle \lim _{x \to 0^ + } \frac{x}{a} = 0$ means that $\displaystyle \left( {\exists \delta > 0} \right)\left[ {0 < x < \delta \Rightarrow \quad 0<\frac{x}{a} < 1} \right]$

Which means in turn that $\displaystyle \left( {\exists \delta > 0} \right)\left[ {0 < x < \delta \Rightarrow \quad \left\lfloor {\frac{x}{a}} \right\rfloor = 0} \right]$

Which implies $\displaystyle \left( {\exists \delta > 0} \right)\left[ {0 < x < \delta \Rightarrow \quad \frac{b}{x}\left\lfloor {\frac{x}{a}} \right\rfloor = 0} \right]$.

Note is does not matter what $\displaystyle b$ is.