First slide

Introduction to limits

Question

The value of $lim_{x \to - \infty} (\sqrt{x^{2} - x + 1} + x)$ , is

Moderate

A

$- \frac{1}{2}$
B

$\frac{1}{2}$
C

$1$
D

$- 1$

Solution

We have,
$lim_{x \to - \infty} (\sqrt{x^{2} - x + 1} + x)$
$= lim_{y \to \infty} (\sqrt{y^{2} + y + 1} - y)$ , where $y = - x$
$= lim_{y \to \infty} \frac{\{\sqrt{y^{2} + y + 1} - y\} \{\sqrt{y^{2} + y + 1} + y\}}{\{\sqrt{y^{2} + y + 1 + y}\}}$
$\begin{array}{l} = lim_{y \to \infty} \frac{y^{2} + y + 1 - y^{2}}{\sqrt{y^{2} + y + 1 + y}} \\ = lim_{y \to \infty} \frac{y + 1}{\sqrt{y^{2} + y + 1 + y}} = lim_{y \to \infty} \frac{1 + \frac{1}{y}}{\sqrt{1 + \frac{1}{y} + \frac{1}{y^{2}}} + 1} = \frac{1}{2} \end{array}$
ALITER $lim_{x \to - \infty} (\sqrt{x^{2} - x + 1} + x)$
$\begin{array}{l} = lim_{x \to - \infty} \frac{\{\sqrt{x^{2} - x + 1} + x\} \{\sqrt{x^{2} - x + 1} - x\}}{\{\sqrt{x^{2} - x} + 1 + x\}} \\ = lim_{x \to - \infty} \frac{x^{2} - x + 1 - x^{2}}{\{\sqrt{x^{2} - x + 1} - x\}} \\ = lim_{x \to - \infty} \frac{- x + 1}{\sqrt{x^{2} - x + 1} - x} \end{array}$
$= lim_{x \to - \infty} \frac{- \frac{x}{| x |} + \frac{1}{| x |}}{\frac{\sqrt{x^{2} - x + 1}}{| x |} - \frac{x}{| x |}}$ [ Dividing $N r$ and $O'$ by $|x|$ ]
$= lim_{x \to - \infty} \frac{{\frac{- x}{- x}}^{} - \frac{1}{x}}{\sqrt{\frac{x^{2}}{x^{2}} - x^{2} + \frac{1}{x^{2}} + \frac{x}{x}}} [∵ \sqrt{x^{2}} = | x | = - x as x < 0]$
$= lim_{x \to - \infty} \frac{1 - \frac{1}{x}}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^{2}} + 1}} = \frac{1}{2}$

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