The value of limx→0 11xsin⁡x+21sin⁡xxwhere [x] is the greatest integer less than or equal to x is

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

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

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

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

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

$lim_{x \to 0 +} ([\frac{11 x}{\sin x}] + [\frac{21 \sin x}{x}]) = 31$

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

as $x \to∣ 0 - and [21 \frac{\sin x}{x}] = 20 as x \to 0 -$ Thus

$lim_{x \to 0 -} ([\frac{11 x}{\sin x}] + [\frac{21 \sin x}{x}]) = 31$

The value of $lim_{x \to 0} ([\frac{11 x}{\sin x}] + [\frac{21 \sin x}{x}])$
where $[x]$ is the greatest integer less than or equal to $x$ is