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प्रश्न
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
उत्तर
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{5 + \cos \left( \pi - h \right)} - 2}{\left( \pi - \left( \pi - h \right) \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{5 - \cos h} - 2}{h^2}\]
\[\text{ PRationalising the numerator, we get }: \]
\[ \lim_{h \to 0} \frac{\left( \sqrt{5 - \cos h} - 2 \right) \left( \sqrt{5 - \cos h} + 2 \right)}{h^2 \left( \sqrt{5 - \cos h} + 2 \right)}\]
\[ = \lim_{h \to 0} \frac{5 - \cos h - 4}{h^2 \left( \sqrt{5 - \cos h} + 2 \right)}\]
\[ = \lim_{h \to 0} \frac{1 - \cos h}{h^2 \left[ \sqrt{5 - \cos h} + 2 \right]}\]
\[ = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}}{4\left( \frac{h^2}{4} \right) \left( \sqrt{5 - \cos h} + 2 \right)}\]
\[ = \frac{1}{2 \left( \sqrt{5 - 1} + 2 \right)}\]
\[ = \frac{1}{2\left( 4 \right)}\]
\[ = \frac{1}{8}\]
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