# Show that Lim X → ∞ ( √ X 2 + X + 1 − X ) ≠ Lim X → ∞ ( √ X 2 + 1 − X ) - Mathematics

Show that $\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)$

#### Solution

$\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)$
$\text{ LHS }:$
$\lim_{x \to \infty} \left( \left( \sqrt{x^2 + x + 1} - x \right) \right)$
$\text{ Rationalising the numerator }:$
$\lim_{x \to \infty} \left[ \frac{\left( \sqrt{x^2 + x + 1} - x \right) \left( \sqrt{x^2 + x + 1} + x \right)}{\left( \sqrt{x^2 + x + 1} + x \right)} \right]$
$= \lim_{x \to \infty} \left[ \frac{\left( x^2 + x + 1 \right) - x^2}{\left( \sqrt{x^2 + x + 1} + x \right)} \right]$
$= \lim_{x \to \infty} \left[ \frac{x + 1}{\left( \sqrt{x^2 + x + 1} + x \right)} \right]$
$\text{ Dividing the numerator and the denominator by x }:$
$\lim_{x \to \infty} \left[ \frac{1 + \frac{1}{x}}{\frac{\sqrt{x^2 + x + 1}}{x} + 1} \right]$
$= \lim_{x \to \infty} \left[ \frac{1 + \frac{1}{x}}{\sqrt{\frac{x^2 + x + 1}{x^2}} + 1} \right]$
$= \lim_{x \to \infty} \left[ \frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + 1} \right]$
$\text{ When x } \to \infty , \text{ then } \frac{1}{x} \to 0 .$
$\frac{1}{\sqrt{1} + 1}$
$= \frac{1}{2}$
$RHS:$
$\lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right) \left[ \text{ from } \infty - \infty \right]$

Rationalising the numerator:

$\lim_{x \to \infty} \left[ \frac{\left( \sqrt{x^2 + 1} - x \right) \left( \sqrt{x^2 + 1} + x \right)}{\left( \sqrt{x^2 + 1} + x \right)} \right]$
$= \lim_{x \to \infty} \left[ \frac{x^2 + 1 - x^2}{\left( \sqrt{x^2 + 1} + x \right)} \right]$
$= \frac{1}{\infty}$
$= 0$
$\therefore \lim_{x \to \infty} \left[ \sqrt{x^2 + x + 1} - x \right] \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.6 | Q 22 | Page 39