$\underset{x\to 0}{\lim }\frac{x}{a}\left[\frac{b}{x}\right]\left(a\ne 0\right)$  =     , $where \left[·\right] denotes integral part of x$ 

$\begin{array}{l}\frac{b}{x}-1<\left[\frac{b}{x}\right]\le \frac{b}{x} \left(\sin ce x-1<\left[ x\right] \le x \right)\\ \frac{x}{a}\left(\frac{b}{x}-1\right)<\frac{x}{a}\left[\frac{b}{x}\right]\le \frac{x}{a}\frac{b}{x}\\ \frac{b}{a}-\frac{x}{a}<\frac{x}{a}\left[\frac{b}{x}\right]<\frac{b}{a}\\ \underset{x\to 0}{\lim }\left(\frac{b}{a}-\frac{x}{a}\right)<\underset{x\to 0}{\lim }\frac{x}{a}\left[\frac{b}{x}\right]<\underset{x\to 0}{\lim }\frac{b}{a}\\ \frac{b}{a}<\underset{x\to 0}{\lim }\frac{x}{a}\left[\frac{b}{x}\right]<\frac{b}{a} \left(by Sandwitch theorem\right)\\ \therefore \underset{x\to 0}{\lim }\frac{x}{a}\left[\frac{b}{x}\right]=\frac{b}{a}\end{array}$