Question

$\underset{h\to 0}{lim}\frac{2(\sqrt{3}\sin (\frac{\pi }{6}+h)-\cos (\frac{\pi }{6}+h))}{\sqrt{3}h(\sqrt{3}\cos h-\sin h)}$ is equal to

• A. $\frac{4}{3}$
• B. $-\frac{4}{3}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

Answer: (A)

$\frac{4}{3}$

$\underset{h\to 0}{lim}\frac{2(\sqrt{3}\sin (\frac{\pi }{6}+h)-\cos (\frac{\pi }{6}+h))}{\sqrt{3}h(\sqrt{3}\cos h-\sin h)}$

$=\underset{h\to 0}{lim}\frac{2(\frac{\sqrt{3}}{2}\sin (\frac{\pi }{6}+h)-\frac{1}{2}\cos (\frac{\pi }{6}+h))}{\sqrt{3}h(\frac{\sqrt{3}}{2}\cos h-\frac{1}{2}sinh)}$

$=\underset{h\to 0}{lim}-\frac{2\cos (\frac{\pi }{2}+h)}{\sqrt{3}hsin(\frac{\pi }{3}-h)}$

$=\underset{h\to 0}{lim}\frac{2\sin h}{\sqrt{3}h\sin (\frac{\pi }{3}-h)}$

$=\underset{h\to 0}{lim}\frac{2}{\sqrt{3}\sin (\frac{\pi }{3}-h)}$

$=\frac{2\times 2}{\sqrt{3}\sqrt{3}}$

$=\frac{4}{3}$

So,the correct option is (A).