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Lim X → 1 √ 5 X − 4 − √ X X 3 − 1 - Mathematics

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Question

\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\] 

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Solution

\[\lim_{x \to 1} \left[ \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1} \right]\] It is of the form \[\frac{0}{0}\] 

Rationalising the numerator: 

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

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

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

=  \[\frac{4}{3\left( 1 + 1 \right)}\] 

= \[\frac{2}{3}\] 

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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 18 | Page 28

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