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Lim X → 0 √ 1 + X 2 − √ 1 − X 2 X - Mathematics

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प्रश्न

\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\] 

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उत्तर

\[\lim_{x \to 0} \left[ \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x} \right]\]  It is of the form\[\frac{0}{0}\] 

Rationalising the numerator: 

\[\lim_{x \to 0} \left[ \frac{\left( \sqrt{1 + x^2} - \sqrt{1 - x^2} \right)\left( \sqrt{1 + x^2} + \sqrt{1 - x^2} \right)}{\left( \sqrt{1 + x^2} + \sqrt{1 - x^2} \right)x} \right]\] 

=  \[\lim_{x \to 0} \left[ \frac{\left( 1 + x^2 \right) - \left( 1 - x^2 \right)}{x\left\{ \sqrt{1 + x^2} + \sqrt{1 - x^2} \right\}} \right]\] 

=  \[\lim_{x \to 0} \left[ \frac{2 x^2}{x\left\{ \sqrt{1 + x^2} + \sqrt{1 - x^2} \right\}} \right]\] 

=  \[\frac{2 \times 0}{\sqrt{1 + 0} + \sqrt{1 - 0}}\] 

= 0

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पाठ 29: Limits - Exercise 29.4 [पृष्ठ २९]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.4 | Q 21 | पृष्ठ २९

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