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प्रश्न
\[\lim_{x \to \infty} \left\{ \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right\}^\frac{3x - 2}{3x + 2}\]
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उत्तर
\[\lim_{x \to \infty} \left( \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right)^\left( \frac{3x - 2}{3x + 2} \right) \]
\[ = \lim_{x \to \infty} \left[ 1 + \frac{x^2 + 2x + 3}{2 x^2 + x + 5} - 1 \right]\left( {}^\frac{3x - 2}{3x + 2} \right)\]
\[ = \lim_{x \to \infty} \left[ 1 + \frac{\left( x^2 + 2x + 3 \right) - \left( 2 x^2 + x + 5 \right)}{2 x^2 + x + 5} \right]^\left( \frac{3x - 2}{3x + 2} \right) \]
\[ = \lim_{x \to \infty} \left[ 1 + \frac{\left( - x^2 + x - 2 \right)}{2 x^2 + x + 5} \right]^\left( \frac{3x - 2}{3x + 2} \right) \]
\[ = e^\lim_{x \to \infty} \left( \frac{- x^2 + x - 2}{2 x^2 + x + 5} \right) \times \left( \frac{3x - 2}{3x + 2} \right) \]
\[ = e^\lim_{x \to \infty} \left( \frac{- 1 + \frac{1}{x} - \frac{2}{x^2}}{2 + \frac{1}{x} + \frac{5}{x^2}} \right) \times \left( \frac{3 - \frac{2}{x}}{3 + \frac{2}{x}} \right) \]
\[ = e^{- \frac{1}{2} \times 1} \]
\[ = \frac{1}{\sqrt{e}}\]
\[\]
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