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

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

#### Solution

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

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.6 | Q 13 | Page 39

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