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Lim X → 1 √ 3 + X − √ 5 − X X 2 − 1 - Mathematics

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Question

\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\] 

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Solution

\[\lim_{x \to 1} \left[ \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1} \right]\] It is of the form \[\frac{0}{0}\] 

Rationalising the numerator: 

\[\lim_{x \to 1} \left[ \frac{\left( \sqrt{3 + x} - \sqrt{5 - x} \right)\left( \sqrt{3 + x} + \sqrt{5 - x} \right)}{\left( x - 1 \right)\left( x + 1 \right)\left( \sqrt{3 + x} + \sqrt{5 - x} \right)} \right]\] 

= \[\lim_{x \to 1} \left[ \frac{\left( 3 + x \right) - \left( 5 - x \right)}{\left( x - 1 \right)\left( x + 1 \right)\left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\] 

=  \[\lim_{x \to 1} \left[ \frac{2x - 2}{\left( x - 1 \right)\left( x + 1 \right) \left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\]

= \[\lim_{x \to 1} \left[ \frac{2\left( x - 1 \right)}{\left( x - 1 \right)\left( x + 1 \right)\left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\] 

= \[\frac{2}{2 \left( \sqrt{4} + \sqrt{4} \right)}\]

=  \[\frac{1}{4}\]

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Chapter 29: Limits - Exercise 29.4 [Page 28]

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RD Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.4 | Q 20 | Page 28

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