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प्रश्न
\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\]
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उत्तर
\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x\left( x^2 - 3x + 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{- 2\left( 2x - 3 \right)}{x\left( x^2 - 2x - x + 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x\left\{ x\left( x - 2 \right) - 1\left( x - 2 \right) \right\}} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x\left( x - 1 \right)\left( x - 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x\left( x - 1 \right) - 2\left( 2x - 3 \right)}{x\left( x - 1 \right)\left( x + 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x^2 - x - 4x + 6}{x\left( x - 1 \right)\left( x + 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x^2 - 5x + 6}{x\left( x - 1 \right)\left( x + 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x^2 - 2x - 3x + 6}{x\left( x - 1 \right)\left( x - 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x\left( x - 2 \right) - 3\left( x - 2 \right)}{x\left( x - 1 \right)\left( x - 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{\left( x - 3 \right)\left( x - 2 \right)}{x\left( x - 1 \right)\left( x - 2 \right)} \right]\]
\[ = \frac{2 - 3}{2\left( 2 - 1 \right)}\]
\[ = - \frac{1}{2}\]
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