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