A) 0
B) \[\frac{1}{2}\]
C) 1
D) 2
Correct Answer: C
Solution :
[c] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin 2x+4x}{2x+\sin 4x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{\sin 2x}{x}+4}{2+\frac{\sin 4x}{x}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2\left( \frac{\sin 2x}{2x} \right)+4}{2+4\left( \frac{\sin 4x}{4x} \right)}\] Applying limit, we get \[\frac{2+4}{2+4}=1\]
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