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Lim X → 1 √ 5 X − 4 − √ X X 2 − 1

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प्रश्न

\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1}\] 

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उत्तर

\[\lim_{x \to 1} \left[ \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1} \right]\] It is of the form \[\frac{0}{0}\] 

Rationalising the numerator: 

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

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

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

= \[\frac{4}{\left( \sqrt{5 - 4} + \sqrt{1} \right)\left( 1 + 1 \right)}\] 

=  \[\frac{4}{2 \times 2}\] 

= 1 

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अध्याय 29: Limits - Exercise 29.4 [पृष्ठ २८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.4 | Q 11 | पृष्ठ २८

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