Given limit
We need to evaluate
L = lim h → 0 2 { 3 sin ( π 6 + h ) − cos ( π 6 + h ) 3 h ( 3 cos h − sin h ) } . L=\lim_{h\to 0}2\left\{\frac{\sqrt 3\sin\left(\frac\pi6+h\right)-\cos\left(\frac\pi6+h\right)}{\sqrt 3\,h\left(\sqrt 3\cos h-\sin h\right)}\right\}. L = h → 0 lim 2 { 3 h ( 3 cos h − sin h ) 3 sin ( 6 π + h ) − cos ( 6 π + h ) } . So,
L = 2 lim h → 0 3 sin ( π 6 + h ) − cos ( π 6 + h ) 3 h ( 3 cos h − sin h ) . L=2\lim_{h\to 0}\frac{\sqrt 3\sin\left(\frac\pi6+h\right)-\cos\left(\frac\pi6+h\right)}{\sqrt 3\,h\left(\sqrt 3\cos h-\sin h\right)}. L = 2 h → 0 lim 3 h ( 3 cos h − sin h ) 3 sin ( 6 π + h ) − cos ( 6 π + h ) .
Expand the numerator
Use angle addition formulas:
sin ( π 6 + h ) = sin π 6 cos h + cos π 6 sin h = 1 2 cos h + 3 2 sin h , \sin\left(\frac\pi6+h\right)=\sin\frac\pi6\cos h+\cos\frac\pi6\sin h
=\frac12\cos h+\frac{\sqrt3}{2}\sin h, sin ( 6 π + h ) = sin 6 π cos h + cos 6 π sin h = 2 1 cos h + 2 3 sin h , cos ( π 6 + h ) = cos π 6 cos h − sin π 6 sin h = 3 2 cos h − 1 2 sin h . \cos\left(\frac\pi6+h\right)=\cos\frac\pi6\cos h-\sin\frac\pi6\sin h
=\frac{\sqrt3}{2}\cos h-\frac12\sin h. cos ( 6 π + h ) = cos 6 π cos h − sin 6 π sin h = 2 3 cos h − 2 1 sin h . Now,
3 sin ( π 6 + h ) = 3 ( 1 2 cos h + 3 2 sin h ) = 3 2 cos h + 3 2 sin h . \sqrt3\sin\left(\frac\pi6+h\right)
=\sqrt3\left(\frac12\cos h+\frac{\sqrt3}{2}\sin h\right)
=\frac{\sqrt3}{2}\cos h+\frac32\sin h. 3 sin ( 6 π + h ) = 3 ( 2 1 cos h + 2 3 sin h ) = 2 3 cos h + 2 3 sin h . Hence,
3 sin ( π 6 + h ) − cos ( π 6 + h ) = ( 3 2 cos h + 3 2 sin h ) − ( 3 2 cos h − 1 2 sin h ) . \sqrt3\sin\left(\frac\pi6+h\right)-\cos\left(\frac\pi6+h\right)
=\left(\frac{\sqrt3}{2}\cos h+\frac32\sin h\right)-\left(\frac{\sqrt3}{2}\cos h-\frac12\sin h\right). 3 sin ( 6 π + h ) − cos ( 6 π + h ) = ( 2 3 cos h + 2 3 sin h ) − ( 2 3 cos h − 2 1 sin h ) . The cosine terms cancel:
= 3 2 sin h + 1 2 sin h = 2 sin h . =\frac32\sin h+\frac12\sin h=2\sin h. = 2 3 sin h + 2 1 sin h = 2 sin h . So the limit becomes
L = 2 lim h → 0 2 sin h 3 h ( 3 cos h − sin h ) = 4 lim h → 0 sin h 3 h ( 3 cos h − sin h ) . L=2\lim_{h\to 0}\frac{2\sin h}{\sqrt3\,h\left(\sqrt3\cos h-\sin h\right)}
=4\lim_{h\to 0}\frac{\sin h}{\sqrt3\,h\left(\sqrt3\cos h-\sin h\right)}. L = 2 h → 0 lim 3 h ( 3 cos h − sin h ) 2 sin h = 4 h → 0 lim 3 h ( 3 cos h − sin h ) sin h .
Separate the standard limit
L = 4 3 lim h → 0 ( sin h h ) 1 3 cos h − sin h . L=\frac{4}{\sqrt3}\lim_{h\to 0}\left(\frac{\sin h}{h}\right)\frac{1}{\sqrt3\cos h-\sin h}. L = 3 4 h → 0 lim ( h sin h ) 3 cos h − sin h 1 . As h → 0 h\to 0 h → 0
sin h h → 1 , cos h → 1 , sin h → 0. \frac{\sin h}{h}\to 1,\qquad \cos h\to 1,\qquad \sin h\to 0. h sin h → 1 , cos h → 1 , sin h → 0. Therefore,
3 cos h − sin h → 3 . \sqrt3\cos h-\sin h\to \sqrt3. 3 cos h − sin h → 3 . Thus,
L = 4 3 ⋅ 1 ⋅ 1 3 = 4 3 . L=\frac{4}{\sqrt3}\cdot 1\cdot \frac{1}{\sqrt3}=\frac{4}{3}. L = 3 4 ⋅ 1 ⋅ 3 1 = 3 4 .
Check options
The value is
4 3 . \boxed{\frac{4}{3}}. 3 4 . So the correct option is A .