Ltx→0Sin⁡2x+2Sin2⁡x−2Sin⁡xCos⁡x−Cos2⁡x= - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Given limit is

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

$= lim_{x \to 0} \frac{2 \sin x \cos x + 2 - 2 \cos^{2} x - 2 \sin x}{\cos x - \cos^{2} x}$

$= 2 \cdot lim_{x \to 0} \frac{\sin x \cos x - \sin x - (\cos^{2} x - 1)}{\cos x - \cos^{2} x}$

$= 2 \cdot lim_{x \to 0} \frac{\sin x (\cos x - 1) - (\cos x - 1) (\cos x + 1)}{\cos x - \cos^{2} x}$

$= 2 \cdot lim_{x \to 0} \frac{(\cos x - 1) (\sin x - \cos x - 1)}{\cos x (1 - \cos x)}$

$= 2 \cdot lim_{x \to 0} \frac{1 - \sin x + \cos x}{\cos x} = 2 \times 2 = 4$

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