A) ½
B) 1/4
C) 2
D) 4
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{2\sin 4x\cos 2x}{2\sin x\cos 4x}=\underset{x\to 0}{\mathop{\lim }}\,4\left( \frac{\sin 4x}{4x} \right)\,\left( \frac{x}{\sin x} \right)\frac{\cos 2x}{\cos 4x}=4\]. Aliter : \[\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\frac{2\,\,\sin 2x}{2x}+\frac{6\,\,\sin 6x}{6x}}{\frac{5\,\,\sin 5x}{5x}-\frac{3\,\,\sin 3x}{3x}}=\frac{2+6}{5-3}=4.\]
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