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प्रश्न
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
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उत्तर
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x} \]
\[ = \lim_{x \to \infty} \left\{ 1 + \frac{3 x^2 + 1}{4 x^2 - 1} - 1 \right\}^\frac{x^3}{1 + x} \]
\[ = \lim_{x \to \infty} \left\{ 1 + \frac{3 x^2 + 1 - 4 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x} \]
\[ = \lim_{x \to \infty} \left\{ 1 + \frac{2 - x^2}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x} \]
\[ = e^\lim_{x \to \infty} \left( \frac{2 - x^2}{4 x^2 - 1} \right) \times \left( \frac{x^3}{1 + x} \right) \]
\[ = e {}^\lim_{x \to \infty} \left( \frac{- x^5 + 2 x^3}{\left( 4 x^2 - 1 \right) \left( 1 + x \right)} \right) \]
\[\text{ Dividing N^r and } D^r \text{ by } x^5 : \]
\[ = e {}^\lim_{x \to \infty} \left( \frac{- 1 + \frac{2}{x^3}}{\frac{\left( 4 x^2 - 1 \right)}{x^3} \frac{\left( 1 + x \right)}{x^2}} \right) \]
\[ = e {}^\lim_{x \to \infty} \left( \frac{- 1 + \frac{2}{x^3}}{\left( \frac{4}{x} - \frac{1}{x^3} \right) \left( \frac{1}{x^2} + \frac{1}{x} \right)} \right) \]
\[ = e^{- \frac{1}{0}} \]
\[ = e^{- \infty} = 0\]
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