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Lim X → 1 X 2 − √ X √ X − 1 - Mathematics

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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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अध्याय 29: Limits - Exercise 29.4 [पृष्ठ ३०]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.4 | Q 29 | पृष्ठ ३०

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