Lim x - 1-5 x-4-x-x3-1

Lim_x → 1√(5x 4)–√(x)/x^3 1 Rationalising the numenator =lim_x → 1(√(5x 4) √(x))/(x 1)(x^2+1+1)×(√(5x 4)+√(x))/(√(5x 4)+√(x)) =lim_x → 1(5x 4 x)/(x 1)(x^2+1+x)( ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

lim x → 1√5 x...

Question

${lim}_{x \to 1} \frac{\sqrt{5 x - 4} - - \sqrt{x}}{x^{3} - 1}$

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Solution

${lim}_{x \to 1} \frac{\sqrt{5 x - 4} - - \sqrt{x}}{x^{3} - 1}$

Rationalising the numenator

$= {lim}_{x \to 1} \frac{(\sqrt{5 x - 4} - \sqrt{x})}{(x - 1) (x^{2} + 1 + 1)} \times \frac{(\sqrt{5 x - 4} + \sqrt{x})}{(\sqrt{5 x - 4} + \sqrt{x})}$

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

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

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

$= \frac{4}{3 (1 + 1)} = \frac{4}{3 \times 2} = \frac{2}{3}$

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Similar questions

{lim}_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{3} - 1}

Q. Evaluate :

lim x \to 1 \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{3} - 1}

Q. Find the limits :

i .

lim x \to 1 [\frac{x^{2} + 1}{x + 100}]

i i .

lim x \to 2 [\frac{x^{3} - 4 x^{2} + 4 x}{x^{2} - 4}]

i i i .

lim x \to 2 [\frac{x^{2} - 4}{x^{3} - 4 x^{2} + 4 x}]

i v .

lim x \to 2 [\frac{x^{3} - 2 x^{2}}{x^{2} - 5 x + 6}]

v .

lim x \to 1 [\frac{x - 2}{x^{2} - x} - \frac{1}{x^{3} - 3 x^{2} + 2 x}]

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

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

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limx→1√5x−4−−√xx3−1

limx→1√5x−4−−√xx3−1 Rationalising the numenator =limx→1(√5x−4−√x)(x−1)(x2+1+1)×(√5x−4+√x)(√5x−4+√x) =limx→1(5x−4−x)(x−1)(x2+1+x)(√5x−4+√x) =limx→14(x−1)(x−1)(x2+x+1)(√5x−4+√x) =4(1+1+1)(√5−4+√1) =43(1+1)=43×2=23

${lim}_{x \to 1} \frac{\sqrt{5 x - 4} - - \sqrt{x}}{x^{3} - 1}$