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lim x → 1 √ 3 + x − √ 5 − x x 2 − 1

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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: 

\[\lim_{x \to 1} \left[ \frac{\left( \sqrt{3 + x} - \sqrt{5 - x} \right) \left( \sqrt{3 + x} + \sqrt{5 - x} \right)}{\left( x - 1 \right) \left( x + 1 \right) \left( \sqrt{3 + x} + \sqrt{5 - x} \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{\left( 3 + x \right) - \left( 5 - x \right)}{\left( x - 1 \right) \left( x + 1 \right) \left( \sqrt{3 + x} + \sqrt{5 - x} \right)} \right]\]
\[ = \lim_{x \to 1} \frac{2\left( x - 1 \right)}{\left( x - 1 \right) \left( x + 1 \right) \left( \sqrt{3 + x} + \sqrt{5 - x} \right)}\]
\[ = \frac{2}{\left( 1 + 1 \right) \left( \sqrt{3 + 1} + \sqrt{5 - 1} \right)}\]
\[ = \frac{2}{2 \times \left( 2 + 2 \right)}\]
\[ = \frac{1}{4}\]

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Chapter 29: Limits - Exercise 29.4 [Page 29]

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R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.4 | Q 27 | Page 29

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