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

# The value of $\underset{x\to 0}{lim} \left(\left[\frac{11x}{\mathrm{sin}x}\right]+\left[\frac{21\mathrm{sin}x}{x}\right]\right)$where is the greatest integer less than or equal tois

1. A

32

2. B

31

3. C

11

4. D

21

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### Solution:

Since  and $\underset{x\to 0+}{lim} \frac{x}{\mathrm{sin}x}=1$

so   Thus

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

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

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

Similarly

as  Thus

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

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