The value of $\underset{x\to 0}{lim} \left(\left[\frac{11x}{\sin x}\right]+\left[\frac{21\sin x}{x}\right]\right)$

where$\left[x\right]$ is the greatest integer less than or equal to$x$is

Since $x>\sin x,\text{ for }x>0$ and $\underset{x\to 0+}{lim} \frac{x}{\sin x}=1$

so  $\frac{11x}{\sin x}\to 11\text{ as }x\to 0+\text{ but }\frac{11x}{\sin x}>11$ Thus

$\left[\frac{11x}{\sin x}\right]=11$ for value $x\to 0+$ Similarly

$\frac{21\sin x}{x}\to 21\text{ as }x\to 0+\text{ but }20\le \frac{21\sin x}{x}<21$

$\left[\frac{21\sin x}{x}\right]=20$ Hence.

$\underset{x\to 0+}{lim} \left(\left[\frac{11x}{\sin x}\right]+\left[\frac{21\sin x}{x}\right]\right)=31$

Similarly $x<\sin x\text{ for }x<0,\text{ so }\left[\frac{11x}{\sin x}\right]=11$

as $x\to \mid 0-\text{ and }\left[21\frac{\sin x}{x}\right]=20\text{ as }x\to 0-$ Thus

$\underset{x\to 0-}{lim} \left(\left[\frac{11x}{\sin x}\right]+\left[\frac{21\sin x}{x}\right]\right)=31$