_ h 0 2 [ 3 ( 6 + h ) - ( 6 + h ) ] 3 h( 3 h - h) = - Vidyadip

MCQ

$\mathop {\lim }\limits_{h \to 0} \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)}} = $

A
$ - \frac{2}{3}$
B
$ - \frac{3}{4}$
C
$ - 2\sqrt 3 $
✓
$\frac{4}{3}$

✓

Answer

Correct option: D.

$\frac{4}{3}$

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

$ = \mathop {\lim }\limits_{h \to 0} \,\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)}}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{4}{{\sqrt 3 }}.\frac{{\sin \,h}}{h}.\frac{1}{{(\sqrt 3 \,\cos \,h - \sin \,h)}} = \frac{4}{3}$.

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