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