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प्रश्न
\[\lim_{x \to 1} \left\{ \frac{x^3 + 2 x^2 + x + 1}{x^2 + 2x + 3} \right\}^\frac{1 - \cos \left( x - 1 \right)}{\left( x - 1 \right)^2}\]
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उत्तर
\[ = \lim_{x \to 1} \left\{ 1 + \frac{x^3 + 2 x^2 + x + 1}{x^2 + 2x + 3} - 1 \right\}^\frac{1 - \cos \left( x - 1 \right)}{\left( x - 1 \right)^2} \]
\[ = e^\lim_{x \to 1} \left\{ \frac{x^3 + 2 x^2 + x + 1 - x^2 - 2x - 3}{x^2 + 2x + 3} \right\} \times \left[ \frac{1 - \cos \left( x - 1 \right)}{\left( x - 1 \right)^2} \right] \]
\[ = e^\lim_{x \to 1} \left\{ \frac{x^3 + x^2 - x - 2}{x^2 + 2x + 3} \right\} \times \left[ \frac{2 \sin^2 \frac{\left( x - 1 \right)}{2}}{4 \times \frac{\left( x - 1 \right)^2}{4}} \right] \]
\[ = e^\left( \frac{1 + 1 - 1 - 2}{1 + 2 + 3} \right) \times \frac{1}{2} \]
\[ = e^{- \frac{1}{12}}\]
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