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प्रश्न
\[\lim_{x \to 1} \frac{ x^2 - \sqrt{x}}{\sqrt{x} - 1}\]
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उत्तर
\[\lim_{x \to 1} \left[ \frac{x^2 - \sqrt{x}}{\sqrt{x} - 1} \right]\]It is of the form \[\frac{0}{0}\]
\[\lim_{x \to 1} \left[ \frac{\sqrt{x}\left( x\sqrt{x} - 1 \right)}{\sqrt{x} - 1} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\sqrt{x}\left( x^{3/2} - 1 \right)}{x^\frac{1}{2} - 1} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\sqrt{x}\left( \left( \sqrt{x} \right)^3 - 1^3 \right)}{\left( \sqrt{x} - 1 \right)} \right] \left[ A^3 - B^3 = \left( A - B \right)\left( A^2 + AB + B^2 \right) \right]\]
=\[\lim_{x \to 1} \left[ \frac{\left( \sqrt{x} \right)\left( \sqrt{x} - 1 \right)\left( x + \sqrt{x} + 1 \right)}{\left( \sqrt{x} - 1 \right)} \right]\]
= 1 (1 + 1 + 1)
= 3
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