A) 0
B) 6
C) \[\frac{1}{3}\]
D) None of these
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{1-\cos \,\,6x}{x}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{2\,\,{{\sin }^{2}}3x}{x}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x\,.\,2\,\,{{\sin }^{2}}3x}{{{x}^{2}}}=0\].
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