Functions Question 119
Question: $ \underset{h\to 0}{\mathop{\lim }},\frac{2
[ \sqrt{3}\sin ( \frac{\pi }{6}+h )-\cos ( \frac{\pi }{6}+h ) ]}{\sqrt{3}h(\sqrt{3}\cos h-\sin h)}= $ [BIT Ranchi 1987]
Options:
A) $ -\frac{2}{3} $
B) $ -\frac{3}{4} $
C) $ -2\sqrt{3} $
D) $ \frac{4}{3} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{h\to 0}{\mathop{\lim }}\frac{2,[ \sqrt{3}\sin ,( \frac{\pi }{6}+h )-\cos ,( \frac{\pi }{6}+h ) ]}{\sqrt{3},h,(\sqrt{3},\cos ,h-\sin ,h)} $ $ =\underset{h\to 0}{\mathop{\lim }}\frac{\frac{4}{\sqrt{3}},[ \frac{\sqrt{3}}{2}\sin ,( \frac{\pi }{6}+h )-\frac{1}{2}\cos ,( \frac{\pi }{6}+h ) ]}{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} $ .
BETA
