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प्रश्न
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
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उत्तर
\[\lim_{x \to 1} \left[ \frac{\left( x^3 - 1 \right) - x\left( x^2 + x - 2 \right)}{\left( x^2 + x - 2 \right)\left( x^3 - 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{\left( x^3 - 1 \right) - x^3 - x^2 + 2x}{\left( x^2 + x - 2 \right)\left( x^3 - 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- x^2 + 2x - 1}{\left( x^2 + x - 2 \right)\left( x - 1 \right)\left( x^2 + x + 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- \left( x^2 - 2x + 1 \right)}{\left( x^2 + x - 2 \right)\left( x - 1 \right)\left( x^2 + x + 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- \left( x - 1 \right)^2}{\left( x^2 + x - 2 \right)\left( x - 1 \right)\left( x^2 + x + 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- \left( x - 1 \right)}{\left( x^2 + 2x - x - 2 \right)\left( x^2 + x + 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- \left( x - 1 \right)}{\left\{ x\left( x + 2 \right) - 1\left( x + 2 \right) \right\}\left( x^2 + x + 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- \left( x - 1 \right)}{\left( x - 1 \right)\left( x + 2 \right)\left( x^2 + x + 1 \right)} \right]\]
\[ = \frac{- 1}{\left( 1 + 2 \right)\left( 1 + 1 + 1 \right)}\]
\[ = \frac{- 1}{9}\]
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