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

Question

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

Solution

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

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Concept of Limits - Algebra of Limits

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Chapter 29: Limits - Exercise 29.6 [Page 38]

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Chapter 29 Limits
Exercise 29.6 | Q 7 | Page 38

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

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