# Thread: Limit Problem

1. ## Limit Problem

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

(a > 0 , b > 0)

2. I must be reading your question wrong, because it looks like you have

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

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

$\frac{b}{a}$

3. No , it's with [ ]

4. Which is NOT standard notation. I would guess that you mean the floor function which would be [x] without the upper serifs.

5. Originally Posted by gilyos
No , it's with [ ]
Which is the greatest integer function or the floor function.
We know that $a>0$ so it follows that $\lim _{x \to 0^ + } \frac{x}{a} = 0$.
That being the case at some point, $\frac{x}{a}<1$ so that means $\left\lfloor {\frac{x}{a}} \right\rfloor = 0$.

6. $\frac{x}{a}-1 <= [\frac{x}{a}] <= \frac{x}{a}$

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

but , what I doing now ???

7. Originally Posted by gilyos
$\frac{x}{a}-1 <= [\frac{x}{a}] <= \frac{x}{a}$
$(\frac{x}{a}-1)*\frac{b}{x} <= [\frac{x}{a}] <= \frac{x}{a}*\frac{b}{x} = \frac{b}{a}$
but , what I doing now ???
That has absolutely nothing to do with this question.
Do you understand that $\lim _{x \to 0^ + } \frac{x}{a} = 0$ means that $\left( {\exists \delta > 0} \right)\left[ {0 < x < \delta \Rightarrow \quad 0<\frac{x}{a} < 1} \right]$
Which means in turn that $\left( {\exists \delta > 0} \right)\left[ {0 < x < \delta \Rightarrow \quad \left\lfloor {\frac{x}{a}} \right\rfloor = 0} \right]$
Which implies $\left( {\exists \delta > 0} \right)\left[ {0 < x < \delta \Rightarrow \quad \frac{b}{x}\left\lfloor {\frac{x}{a}} \right\rfloor = 0} \right]$.
Note is does not matter what $b$ is.