Why is true? $\mathop {\lim }\limits_{x \to 0} \frac{x}{a}\left[ {\frac{b}{x}} \right] = \frac{b}{a}$ $$\begin{array}{l}a,b > 0\\\mathop {\lim }\limits_{x \to 0} \frac{x}{a}\left[ {\frac{b}{x}} \right] = \frac{b}{a}\\\end{array}
$$
I asked already a similar question, but I'm still not sure what makes it true.
As $x$ is decreasing to $0$, $x \over a$ is converging to $0$ but $\left[ {\frac{b}{x}} \right]$ is converging to $\infty$, So we left with $0*\infty$ which isn't helpful.  
How can you solve it?

We have,
$\displaystyle\mathtt{\lim_{x\to\,0}\dfrac{x}{a}\left[\dfrac{b}{x}\right]}$
$\displaystyle=\mathtt{\lim_{x\to\,0}\dfrac{x}{a}\left(\dfrac{b}{x}-\left\{\dfrac{b}{x}\right\}\right)}$
$\displaystyle=\mathtt{\lim_{x\to\,0}\dfrac{x}{a}\cdot\dfrac{b}{x}-\dfrac{x}{a}\cdot\left\{\dfrac{b}{x}\right\}}$
$\displaystyle=\mathtt{\lim_{x\to\,0}\dfrac{b}{a}-\lim_{x\to\,0}\dfrac{x}{a}\cdot\left\{\dfrac{b}{x}\right\}}$
$\displaystyle=\mathtt{\dfrac{b}{a}-\lim_{x\to\,0}\dfrac{x}{a}\cdot\left\{\dfrac{b}{x}\right\}}$
The limit $\,\,\displaystyle\lim_{x\to\,0}\dfrac{x}{a}\cdot\left\{\dfrac{b}{x}\right\}=0$
because the term $\dfrac{x}{a}$ is $0$ , if we put $x=0$ and the term $\left\{\dfrac{b}{x}\right\}\in[0,1)\,\,\,\,\,\because\{y\}\in[0,1)$
So, $0\times[0,1)=0$
Hence, the required limit is $\dfrac{b}{a}$
