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

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प्रश्न

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

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उत्तर

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

Rationalising the denominator:
\[\lim_{x \to 0} \left[ \frac{x}{\left( \sqrt{1 + x} - \sqrt{1 - x} \right)} \times \frac{\left( \sqrt{1 + x} + \sqrt{1 - x} \right)}{\left( \sqrt{1 + x} + \sqrt{1 - x} \right)} \right]\] 

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

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

=  \[\frac{\sqrt{1} + \sqrt{1}}{2}\] 

=  \[\frac{2}{2}\] 

= 1

 

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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 7 | पृष्ठ २८

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