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