\[\lim_{x \to \infty} \left[ x\left\{ \sqrt{x^2 + 1} - \sqrt{x^2 - 1} \right\} \right]\]

\[\lim_{x \to \infty} \left[ x\left\{ \sqrt{x^2 + 1} - \sqrt{x^2 - 1} \right\} \right]\]\[\text{ Rationalising the numerator }: \]\[ \lim_{x \to \infty} \left[ x \left\{ \frac{\left( \sqrt{x^2 + 1} - \sqrt{x^2 - 1} \right) \left( \sqrt{x^2 + 1} + \sqrt{x^2 - 1} \right)}{\left( \sqrt{x^2 + 1} + \sqrt{x^2 - 1} \right)} \right\} \right]\]\[ = \lim_{x \to \infty} \left[ x \left\{ \frac{\left( x^2 + 1 \right) - \left( x^2 - 1 \right)}{\left( \sqrt{x^2 + 1} + \sqrt{x^2 - 1} \right)} \right\} \right]\]\[ = \lim_{x \to \infty} \left[ \frac{x \times 2}{\left( \sqrt{x^2 + 1} + \sqrt{x^2 - 1} \right)} \right]\]\[\text{ Dividing the numerator and the denominator by } x: \]\[ \lim_{x \to \infty} \left[ \frac{2}{\sqrt{\frac{x^2 + 1}{x^2}} + \sqrt{\frac{x^2 - 1}{x^2}}} \right]\]\[ = \lim_{x \to \infty} \left[ \frac{2}{\sqrt{\frac{x^2 + 1}{x^2}} + \sqrt{\frac{x^2 - 1}{x^2}}} \right]\]\[ = \lim_{x \to \infty} \left[ \frac{2}{\sqrt{1 + \frac{1}{x^2}} + \sqrt{1 - \frac{1}{x^2}}} \right]\]\[ = \frac{2}{\sqrt{1} + \sqrt{1}}\]\[ = 2\]

Lim X → ∞ [ X { √ X 2 + 1 − √ X 2 − 1 } ] - Mathematics

Question

$lim_{x \to \infty} [x {\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}}]$

Solution

$lim_{x \to \infty} [x {\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}}]$
$Rationalising the numerator :$
$lim_{x \to \infty} [x {\frac{(\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}) (\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}{(\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}}]$
$= lim_{x \to \infty} [x {\frac{(x^{2} + 1) - (x^{2} - 1)}{(\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}}]$
$= lim_{x \to \infty} [\frac{x \times 2}{(\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}]$
$Dividing the numerator and the denominator by x :$
$lim_{x \to \infty} [\frac{2}{\sqrt{\frac{x^{2} + 1}{x^{2}}} + \sqrt{\frac{x^{2} - 1}{x^{2}}}}]$
$= lim_{x \to \infty} [\frac{2}{\sqrt{\frac{x^{2} + 1}{x^{2}}} + \sqrt{\frac{x^{2} - 1}{x^{2}}}}]$
$= lim_{x \to \infty} [\frac{2}{\sqrt{1 + \frac{1}{x^{2}}} + \sqrt{1 - \frac{1}{x^{2}}}}]$
$= \frac{2}{\sqrt{1} + \sqrt{1}}$
$= 2$

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Concept of Limits - Algebra of Limits

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Q 12 Q 11 Q 13

Chapter 29: Limits - Exercise 29.6 [Page 39]

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RD Sharma Mathematics [English] Class 11

Chapter 29 Limits
Exercise 29.6 | Q 12 | Page 39

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Lim X → ∞ [ X { √ X 2 + 1 − √ X 2 − 1 } ] - Mathematics

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