Ltx→0Sin⁡2x+2Sin2⁡x−2Sin⁡xCos⁡x−Cos2⁡x=

# $\underset{x\to 0}{\mathrm{Lt}}\frac{\mathrm{Sin}2x+2{\mathrm{Sin}}^{2}x-2\mathrm{Sin}x}{\mathrm{Cos}x-{\mathrm{Cos}}^{2}x}=$

1. A

0

2. B

1

3. C

$\frac{2}{3}$

4. D

4

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### Solution:

Given limit is

$=\underset{x\to 0}{\mathrm{Lt}}\frac{\mathrm{Sin}2x+2{\mathrm{Sin}}^{2}x-2\mathrm{Sin}x}{\mathrm{Cos}x-{\mathrm{Cos}}^{2}x}$

$=\underset{x\to 0}{lim} \frac{2\mathrm{sin}x\mathrm{cos}x+2-2{\mathrm{cos}}^{2}x-2\mathrm{sin}x}{\mathrm{cos}x-{\mathrm{cos}}^{2}x}$

$=2\cdot \underset{x\to 0}{lim} \frac{\mathrm{sin}x\mathrm{cos}x-\mathrm{sin}x-\left({\mathrm{cos}}^{2}x-1\right)}{\mathrm{cos}x-{\mathrm{cos}}^{2}x}$

$=2\cdot \underset{x\to 0}{lim} \frac{\mathrm{sin}x\left(\mathrm{cos}x-1\right)-\left(\mathrm{cos}x-1\right)\left(\mathrm{cos}x+1\right)}{\mathrm{cos}x-{\mathrm{cos}}^{2}x}$

$=2\cdot \underset{x\to 0}{lim} \frac{\left(\mathrm{cos}x-1\right)\left(\mathrm{sin}x-\mathrm{cos}x-1\right)}{\mathrm{cos}x\left(1-\mathrm{cos}x\right)}$

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