Lim x - 1-5 x-4-x-x2-1

Lim_x → 1√(5x 4) √(x)/x^2 1 Rationalising the numenator =lim_x → 1(√(5x 4) √(x))/(x 1)(x+1)×(√(5x 4)+√(x))/(√(5x 4)+√(x)) =lim_x → 1((5 4) x)/(x 1)(x+1)(√(5x 4) ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

lim x → 1√5 x...

Question

${lim}_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{2} - 1}$

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Solution

${lim}_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{2} - 1}$

Rationalising the numenator

$= {lim}_{x \to 1} \frac{(\sqrt{5 x - 4} - \sqrt{x})}{(x - 1) (x + 1)} \times \frac{(\sqrt{5 x - 4} + \sqrt{x})}{(\sqrt{5 x - 4} + \sqrt{x})}$

$= {lim}_{x \to 1} \frac{((5 - 4) - x)}{(x - 1) (x + 1) (\sqrt{5 x - 4} + \sqrt{x})}$

$= {lim}_{x \to 1} \frac{4 (x - 1)}{(x - 1) (x + 1) (\sqrt{5 x - 4} + \sqrt{x})}$

$= {lim}_{x \to 1} \frac{4}{(x + 1) (\sqrt{5 x - 4} + \sqrt{x})}$

$= \frac{4}{(1 + 1) (\sqrt{5 - 4} + \sqrt{1})}$

$= \frac{4}{2 (1 + 1)}$

$= \frac{4}{4} = 1$

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= ______

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limx→1√5x−4−√xx2−1

limx→1√5x−4−√xx2−1 Rationalising the numenator =limx→1(√5x−4−√x)(x−1)(x+1)×(√5x−4+√x)(√5x−4+√x) =limx→1((5−4)−x)(x−1)(x+1)(√5x−4+√x) =limx→14(x−1)(x−1)(x+1)(√5x−4+√x) =limx→14(x+1)(√5x−4+√x) =4(1+1)(√5−4+√1) =42(1+1) =44=1

${lim}_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{2} - 1}$