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Lim X → 0 √ a + X − √ a X √ a 2 + a X - Mathematics

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प्रश्न

\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]

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उत्तर

\[\lim_{x \to 0} \left[ \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}} \right]\] It is of the form \[\frac{0}{0}\] 

Rationalising the numerator: 

\[\lim_{x \to 0} \left[ \frac{\left( \sqrt{a + x} - \sqrt{a} \right) \times \left( \sqrt{a + x} + \sqrt{a} \right)}{\left( x\sqrt{a^2 + ax} \right) \times \left( \sqrt{a + x} + \sqrt{a} \right)} \right]\] 

=  \[\lim_{x \to 0} \left[ \frac{a + x - a}{x\left( \sqrt{a^2 + ax} \right) \left( \sqrt{a + x} + \sqrt{a} \right)} \right]\] 

= \[\frac{1}{\left( \sqrt{a^2} \right) \left( \sqrt{a} + \sqrt{a} \right)}\] 

=  \[\frac{1}{2a\sqrt{a}}\] 

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पाठ 29: Limits - Exercise 29.4 [पृष्ठ २८]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.4 | Q 16 | पृष्ठ २८

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