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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
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2 + \cos \left( \pi - h \right)} - 1}{\left( \pi - \left( \pi - h \right) \right)^2}\]
\[ = \lim_{h \to 0} \left( \frac{\sqrt{2 - \cos h} - 1}{h^2} \right) \times \frac{\sqrt{2 - \cos h} + 1}{\sqrt{2 - \cos h} + 1}\]
\[ = \lim_{h \to 0} \frac{2 - \cos h - 1}{h^2 \left( \sqrt{2 - \cos h} + 1 \right)}\]
\[ = \lim_{h \to 0} \frac{1 - \cos h}{h^2 \left( \sqrt{2 - \cos h} + 1 \right)}\]
\[ = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}}{4 \times \frac{h^2}{4} \times \left[ \sqrt{2 - \cos h} + 1 \right]}\]
\[ = \frac{1 \times 1^2}{2 \left( \sqrt{2 - 1} + 1 \right)}\]
\[ = \frac{1}{4}\]
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