We are given, $\mathop {\lim }\limits_{x \to \frac{\pi }{3}} \frac{{\sqrt {1 – \cos 6x} }}{{\sqrt 2 \frac{\pi }{3} – x}}$.
Using the identity, $\cos 6x = 1 – 2{\sin ^2}3x$ to get,
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\sqrt {1 – \cos 6x} }}{{\sqrt 2 \left( {\frac{\pi }{3} – x} \right)}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\sqrt {1 – \left( {1 – 2{{\sin }^2}3x} \right)} }}{{\sqrt 2 \left( {\frac{\pi }{3} – x} \right)}}$
$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\sqrt 2 |\sin 3x|}}{{\sqrt 2 \left( {\frac{\pi }{3} – x} \right)}}$
Using the identity $\sin 3x = \sin (\pi – 3x) = \sin 3\left( {\frac{\pi }{3} – x} \right)$ to get,
$\mathop {\lim }\limits_{x \to \frac{ – }{2}} \frac{{\sqrt 2 |\sin 3x|}}{{\sqrt 2 \left( {\frac{\pi }{3} – x} \right)}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\sin 3\left( {\frac{\pi }{3} – x} \right)}}{{\left( {\frac{\pi }{3} – x} \right)}}$
$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{3\sin 3\left( {\frac{\pi }{3} – x} \right)}}{{3\left( {\frac{\pi }{3} – x} \right)}}$
$ = 3\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\sin (\pi – 3x)}}{{(\pi – 3x)}}$
Applying, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$ and substituting the limit value to get,
$3\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\sin (\pi – 3x)}}{{(\pi – 3x)}} = 3.1$
$ = 3$
Therefore, $\mathop {\lim }\limits_{x \to \frac{7}{2}} \frac{{\sqrt {1 – \cos 6x} }}{{\sqrt 2 \left( {\frac{\pi }{3} – x} \right)}} = 3$.