The value of $\underset{x\to 0}{lim} \left\{\left[\frac{100x}{\sin x}\right]+\left[\frac{99\sin x}{x}\right]\right\},$ where $[·]$ represents the greatest integer function, is

We know that

$\begin{array}{l}x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}>\sin x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ for all }x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}>0\\ \text{ and, }x<\sin x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ for all }x<0\end{array}$

$\begin{array}{l}\Rightarrow \frac{x}{\sin x}>1\text{ and }\frac{\sin x}{x}<1\text{ for all }x\ne 0\\ \Rightarrow \frac{100x}{\sin x}>100\text{ and }\frac{99\sin x}{x}\ne 0\text{ for all }x\ne 0\\ \therefore \underset{x\to 0}{lim} \left(\left[\frac{100x}{\sin x}\right]+\left[\frac{99\sin x}{x}\right]\right)=\underset{x\to 0}{lim} (100+98)=198\end{array}$