fun-lim

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



1) $-\frac{2}{3}$
2) $-\frac{3}{4}$
3) $-2\sqrt{3}$
4) $\frac{4}{3}$
Solution: Explanation: No Explanation
Limits

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