Lim2h→03sinπ6+h-cosπ6+h3.h3 cosh-sinh=

2 Lth→03sinπ6+h-cosπ6+h3 h3 cosh-sin h =2Lth→032sinπ6+h-12 cosπ6+h3 h32 cosh-12 sin h =2Lth→0sinπ6+h-π6 3 h sin π3-h =2 Lth→0 sin hh.13 sin π3-h =23 32=43

Lim2h→03sinπ6+h-cosπ6+h3.h3 cosh-sinh=

2 Lth→03sinπ6+h-cosπ6+h3 h3 cosh-sin h =2Lth→032sinπ6+h-12 cosπ6+h3 h32 cosh-12 sin h =2Lth→0sinπ6+h-π6 3 h sin π3-h =2 Lth→0 sin hh.13 sin π3-h =23 32=43

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$\underset{h \to 0}{Lim 2} [\frac{\sqrt{3} \sin (\frac{π}{6} + h) - \cos (\frac{π}{6} + h)}{\sqrt{3} . h (\sqrt{3} \cosh - \sinh)}] =$

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$\frac{- 2}{\sqrt{3}}$

$\frac{- 4}{3}$

$\frac{2}{3}$

$\frac{4}{3}$

answer is D.

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Detailed Solution

$2 \underset{h \to 0}{Lt} \frac{\sqrt{3} \sin (\frac{π}{6} + h) - \cos (\frac{π}{6} + h)}{\sqrt{3} h (\sqrt{3} \cosh - \sin h)} = 2 \underset{h \to 0}{Lt} \frac{\frac{\sqrt{3}}{2} \sin (\frac{π}{6} + h) - \frac{1}{2} \cos (\frac{π}{6} + h)}{\sqrt{3} h (\frac{\sqrt{3}}{2} \cosh - \frac{1}{2} \sin h)} = 2 \underset{h \to 0}{Lt} \frac{\sin (\frac{π}{6} + h - \frac{π}{6})}{\sqrt{3} h \sin (\frac{π}{3} - h)} = 2 \underset{h \to 0}{Lt} \frac{\sin h}{h} . \frac{1}{\sqrt{3} \sin (\frac{π}{3} - h)} = \frac{2}{\sqrt{3} \frac{\sqrt{3}}{2}} = \frac{4}{3}$