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  $x>\sin x,$ for $x>0$  and  $\underset{x\to 0+}{\lim }\frac{x}{\sin x}=1$
so $\frac{11x}{\sin x}\to 11$ as $x\to 0+$ but $\frac{11x}{\sin x}>11.$Thus
$\left[\frac{11x}{\sin x}\right]=11$ for value $x\to 0+.$  Similarly
$\frac{21\sin x}{x}\to 21$ as$x\to 0+$ but$20\le \frac{21\sin x}{x}<21$ $\left[\frac{21\sin x}{x}\right]=20.$  Hence
$\underset{x\to 0+}{\lim }\text{}\left(\left[\frac{11x}{\sin x}\right]+\left[\frac{21\sin x}{x}\right]\right)=31$
Similarly  $x<\sin x$ for $x<0$ , so  $\left[\frac{11x}{\sin x}\right]=11$
as $x\to 0-$  and $\left[21\frac{\sin x}{x}\right]=20$ as  $x\to 0-.$  Thus

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