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प्रश्न
\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to
विकल्प
\[\sqrt{3}\]
- \[\frac{1}{2}\]
\[\frac{1}{\sqrt{3}}\]
\[\sqrt{3}\]
उत्तर
\[\frac{1}{\sqrt{3}}\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\]
\[ = \lim_{h \to 0} \frac{\sin \frac{\pi}{3} - \left( \frac{\pi}{3} - h \right)\left[ \right]}{2 \cos \left( \frac{\pi}{3} - h \right) - 1}\]
\[ = \lim_{h \to 0} \frac{\sin h}{2\left[ \cos \frac{\pi}{3}\cos h + \sin \frac{\pi}{3} \sin h \right] - 1}\]
\[ = \lim_{h \to 0} \frac{\sin h}{2\left[ \frac{1}{2}\cos h + \frac{\sqrt{3}}{2} \sin h \right] - 1}\]
\[ = \lim_{h \to 0} \frac{\sin h}{\cos h + \sqrt{3} \sin h - 1}\]
\[ = \lim_{h \to 0} \frac{\sin h}{- 2 \sin^2 \frac{h}{2} + \sqrt{3} \sin h} \]
\[\text{ Dividing } N^r \text{ and } D^r \text{ by } h: \]
\[ = \lim_{h \to 0} \frac{\frac{\sin h}{h}}{- \left( 2 \times \frac{h}{4} \right) \left( \frac{\sin^2 \frac{h}{2}}{h \times \frac{h}{4}} \right) + \frac{\sqrt{3} \sin h}{h}}\]
\[ = \frac{1}{\sqrt{3}}\]
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