Question

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

• A. $0$
• B. $1$
• C. $\frac{2}{3}$
• D. $4$

Answer 1

The correct option is D $4$

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

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

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

$=2.\underset{x\to 0}{lim}\frac{(\cos x-1)(\sin x-\cos x-1)}{\cos x(1-\cos x)}$

$=2.\underset{x\to 0}{lim}\frac{1-\sin x+\cos x}{\cos x}=2\times 2=4$

Hence option ${}^{\prime }{D}^{\prime }$ is the answer.