By Rationalising the denominator lim_x → 0x/√(1+x) √(1 x)×(√(1+x)+√(1 x))/(√(1+x)+√(1 x)) =lim_x → 0x(√(1+x)+√(1 x))/(√(1+x))^2 (√(1 x))^2 =lim_x → 0x(√(1+x)+√( ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

lim x → 0x/√1...

Question

${lim}_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}$

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Solution

By Rationalising the denominator

${lim}_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}} \times \frac{(\sqrt{1 + x} + \sqrt{1 - x})}{(\sqrt{1 + x} + \sqrt{1 - x})}$

$= {lim}_{x \to 0} \frac{x (\sqrt{1 + x} + \sqrt{1 - x})}{(\sqrt{1 + x})^{2} - (\sqrt{1 - x})^{2}}$

$= {lim}_{x \to 0} \frac{x (\sqrt{1 + x} + \sqrt{1 - x})}{1 + x - 1 + x}$

$= {lim}_{x \to 0} \frac{x (\sqrt{1 + x} + \sqrt{1 - x})}{2 x}$

$= \frac{1}{2} {lim}_{x \to 0} (\frac{\sqrt{1 + x} + \sqrt{1 - x}}{x}) x$

$= \frac{1}{2} {lim}_{x \to 0} (\sqrt{1 + x} + \sqrt{1 - x})$

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

$= \frac{2}{2} = 1$

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limx→0x√1+x−√1−x

By Rationalising the denominator limx→0x√1+x−√1−x×(√1+x+√1−x)(√1+x+√1−x) =limx→0x(√1+x+√1−x)(√1+x)2−(√1−x)2 =limx→0x(√1+x+√1−x)1+x−1+x =limx→0x(√1+x+√1−x)2x =12limx→0(√1+x+√1−xx)x =12limx→0(√1+x+√1−x) =12(√1+√1) =22=1

${lim}_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}$