$\underset{x\to 0}{lim} \left[\frac{100\tan x\sin x}{x^2}\right]$ is (where $[·]$represents greatest integer function)

In the neighbourhood of$x = 0$, we have

$\begin{array}{r}0<\frac{\tan x\sin x}{x^2}<1\\ \Rightarrow 0<\frac{100\tan x\sin x}{.2}<100\end{array}$

Also, 

$\begin{array}{l}\underset{x\to 0}{lim} \frac{100\tan x\sin x}{x^2}=100\\ \therefore \underset{x\to 0}{lim} \left[\frac{100\tan x\sin x}{x^2}\right]=99.\end{array}$