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lim x → 0 √ 2 − x − √ 2 + x x

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प्रश्न

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

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उत्तर

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

Rationalising the numerator: 

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

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

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

= \[\frac{- 2}{2\sqrt{2}}\] 

=  \[- \frac{1}{\sqrt{2}}\] 

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अध्याय 29: Limits - Exercise 29.4 [पृष्ठ २९]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.4 | Q 26 | पृष्ठ २९

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