Question

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

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

Answer 1

Answer: (C)

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