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प्रश्न
\[\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}\]
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उत्तर
\[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6} \right]\] It is of the form \[\frac{0}{0}\]
⇒ \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( x^2 + x - 2 \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( x^2 + 2x - x - 2 \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( x\left( x + 2 \right) - 1\left( x + 2 \right) \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( x - 1 \right)\left( x + 2 \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( \left( \sqrt{x} \right)^2 - 1^2 \right)\left( x + 2 \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( \sqrt{x} + 1 \right)\left( \sqrt{x} - 1 \right)\left( x + 2 \right)} \right]\]
= \[\frac{- 1}{3\left( 2 \right) \times 3}\]
= \[\frac{- 1}{18}\]
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