Question

$\underset{x\to \frac{\pi }{6}}{lim}\frac{\sin (x-\frac{\pi }{6})}{\sqrt{3-2cosx}}$ is equal to :

• A. $0$
• B. $\frac{1}{(\sqrt{3}-2)}$
• C. $1$
• D. $\infty$

Answer 1

Answer: (A)

$0$

${lim}_{x\to \frac{\pi }{6}}\frac{\sin (x-\frac{\pi }{6})}{\sqrt{3-2\cos x}}$

at $\frac{\pi }{6}$ we get the numerator as $\sin (x-\frac{\pi }{6})=0$

and the denominator as $\sqrt{3-2\cos x}=\sqrt{3-\sqrt{3}}$

So,it is not an indeterminant form as we have $\frac{0}{\sqrt{3-\sqrt{3}}}$

Hence the answer will be zero.