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