# lim x → ∞ x √ 4 x 2 + 1 − 1 - Mathematics

$\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}$

#### Solution

$\lim_{x \to \infty} \left[ \frac{x}{\sqrt{4 x^2 + 1} - 1} \right]$
$\text{ Rationalising the denominator }:$
$\lim_{x \to \infty} \left[ \frac{x}{\left( \sqrt{4 x^2 + 1} - 1 \right)} \frac{\left( \sqrt{4 x^2 + 1} + 1 \right)}{\left( \sqrt{4 x^2 + 1} + 1 \right)} \right]$
$= \lim_{x \to \infty} \left[ \frac{x\left( \sqrt{4 x^2 + 1} + 1 \right)}{4 x^2 + 1 - 1} \right]$
$= \lim_{x \to \infty} \left[ \frac{\sqrt{4 x^2 + 1} + 1}{4x} \right]$
$\text{ Dividing the numerator and the denominator by } x:$
$\lim_{x \to \infty} \left[ \frac{\frac{\sqrt{4 x^2 + 1}}{x} + \frac{1}{x}}{4} \right]$
$= \lim_{x \to \infty} \left[ \frac{\sqrt{\frac{4 x^2 + 1}{x^2}} + \frac{1}{x}}{4} \right]$
$= \lim_{x \to \infty} \left[ \frac{\sqrt{4 + \frac{1}{x^2}} + \frac{1}{x}}{4} \right]$
$x \to \infty$
$\therefore \frac{1}{x}, \frac{1}{x^2} \to 0$
$= \frac{\sqrt{4}}{4}$
$= \frac{2}{4}$
$= \frac{1}{2}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.6 | Q 7 | Page 38