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प्रश्न
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
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उत्तर
\[\lim_{x \to \pi} \left[ \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2} \right]\]
\[\text{ Rationalising the numerator, we get }: \]
\[ \lim_{x \to \pi} \left[ \frac{\left( \sqrt{2 + \cos x} - 1 \right) \times \left( \sqrt{2 + \cos x} + 1 \right)}{\left( \pi - x \right)^2 \left( \sqrt{2 + \cos x} + 1 \right)} \right]\]
\[ = \lim_{x \to \pi} \left[ \frac{2 + \cos x - 1}{\left( \pi - x \right)^2 \left( \sqrt{2 + \cos x} + 1 \right)} \right]\]
\[ = \lim_{x \to \pi} \left[ \frac{1 + \cos x}{\left( \pi - x \right)^2 \left[ \sqrt{2 + \cos x} + 1 \right]} \right]\]
Let x = π \[-\]h
when x → π, then h → 0
\[ = \lim_{h \to 0} \left[ \frac{1 - \cos h}{h^2 \left[ \sqrt{2 - \cos h} + 1 \right]} \right] \left\{ \because \cos \left( \pi - \theta \right) = - \cos \theta \right\}\]
\[ = \lim_{h \to 0} \left[ \frac{2 \sin^2 \left( \frac{h}{2} \right)}{4 \times \frac{h^2}{4}\left[ \sqrt{2 - \cos h} + 1 \right]} \right]\]
\[ = \frac{1}{2} \lim_{h \to 0} \left[ \left( \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right)^2 \times \frac{1}{\left[ \sqrt{2 - \cos h} + 1 \right]} \right]\]
\[ = \frac{1}{2} \times 1 \times \frac{1}{\left( \sqrt{2 - \cos 0} + 1 \right)}\]
\[ = \frac{1}{2} \times \frac{1}{\left( \sqrt{1} + 1 \right)}\]
\[ = \frac{1}{2 \times 2}\]
\[ = \frac{1}{4}\]
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