Question

$\displaystyle\lim _{x \rightarrow \pi / 3} \frac{\cos \left(x+\frac{\pi}{6}\right)}{(1-2 \cos x)^{2 / 3}}=$

• A. 1
• B. -1
• C. 0
• D. None of these

Answer 1

Answer: (C)

$\displaystyle\lim _{x \rightarrow \pi / 3} \frac{\cos (x+\pi / 6)}{(1-2 \cos x)^{2 / 3}}$
$=\displaystyle\lim _{z \rightarrow 0} \frac{\cos (\pi / 2+z)}{[1-2 \cos (\pi / 3+z)]^{2 / 3}}$
[putting $x-\pi / 3=z$ ]
$=\displaystyle\lim _{z \rightarrow 0} \frac{-\sin z}{(1-\cos z+\sqrt{3} \sin z)^{2 / 3}}$
$=\displaystyle\lim _{z \rightarrow 0} \frac{-2 \sin \left(\frac{z}{2}\right) \cos \left(\frac{z}{2}\right)}{\left.2 \sin \left(\frac{z}{2}\right)\right]^{2 / 3}\left[\sin \left(\frac{z}{2}\right)+\sqrt{3} \cos \left(\frac{z}{2}\right)\right]^{2 / 3}}$
$=\displaystyle\lim _{z \rightarrow 0} \frac{-2^{1 / 3}\left[\sin \left(\frac{z}{2}\right)\right]^{1 / 3} \cos \left(\frac{z}{2}\right)}{\left[\sin \left(\frac{z}{2}\right)+\sqrt{3} \cos \left(\frac{z}{2}\right)\right]^{2 / 3}=\frac{-2^{1 / 3} \cdot 0 \cdot 1}{(\sqrt{3})^{2 / 3}}=0}$