A) \[\sqrt{3}\]
B) \[1/\sqrt{3}\]
C) \[-\sqrt{3}\]
D) \[-1/\sqrt{3}\]
Correct Answer: B
Solution :
Using L?Hospital?s rule, \[\underset{x\to \pi /6}{\mathop{\lim }}\,\frac{3\cos x+\sqrt{3}\sin x}{6}=\frac{3.\frac{\sqrt{3}}{2}+\sqrt{3}.\frac{1}{2}}{6}=\frac{1}{\sqrt{3}}\].
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