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प्रश्न
Evaluate the following limit:
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]
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उत्तर
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]
\[ = \lim_{x \to \frac{\pi}{3}} \frac{\sqrt{2 \sin^2 3x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)} \left( 1 - \cos2\theta = 2 \sin^2 \theta \right)\]
\[ = \lim_{x \to \frac{\pi}{3}} \frac{\sqrt{2}\sin3x}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]
\[ = \lim_{x \to \frac{\pi}{3}} \frac{\sin3x}{\left( \frac{\pi}{3} - x \right)}\]
\[= \lim_{h \to 0} \frac{\sin3\left( \frac{\pi}{3} + h \right)}{\frac{\pi}{3} - \left( \frac{\pi}{3} + h \right)} \left( Put x = \frac{\pi}{3} + h \right)\]
\[ = \lim_{h \to 0} \frac{\sin\left( \pi + 3h \right)}{- h}\]
\[ = \lim_{h \to 0} \frac{- \sin3h}{- h} \left[ \sin\left( \pi + \theta \right) = - \sin\theta \right]\]
\[ = 3 \times \lim_{h \to 0} \frac{\sin3h}{3h}\]
\[ = 3 \times 1 \left( \lim_\theta \to 0 \frac{\sin\theta}{\theta} = 1 \right)\]
\[ = 3\]
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