\[\underset{h\to 0}{\mathop{\lim }}\,\frac{2\left[ \sqrt{3}\sin \left( \frac{\pi }{6}+h \right)-\cos \left( \frac{\pi }{6}+h \right) \right]}{\sqrt{3}h(\sqrt{3}\cos h-\sin h)}=\] [BIT Ranchi 1987]

\[\underset{h\to 0}{\mathop{\lim }}\,\frac{2\left[ \sqrt{3}\sin \left(

1.	\[\underset{h\to 0}{\mathop{\lim }}\,\frac{2\left[ \sqrt{3}\sin \left( \frac{\pi }{6}+h \right)-\cos \left( \frac{\pi }{6}+h \right) \right]}{\sqrt{3}h(\sqrt{3}\cos h-\sin h)}=\] [BIT Ranchi 1987]
A.	\[-\frac{2}{3}\]
B.	\[-\frac{3}{4}\]
C.	\[-2\sqrt{3}\]
D.	\[\frac{4}{3}\]
Answer» E.

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\[\underset{h\to 0}{\mathop{\lim }}\,\frac{2\left[ \sqrt{3}\sin \left( \frac{\pi }{6}+h \right)-\cos \left( \frac{\pi }{6}+h \right) \right]}{\sqrt{3}h(\sqrt{3}\cos h-\sin h)}=\] [BIT Ranchi 1987]

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