# Evalute lim_(x->oo) [sqrt(x^2+x+2) - sqrt(x^2-3x-5)]?

May 10, 2017

${\lim}_{x \to \infty} \sqrt{{x}^{2} + x + 2} - \sqrt{{x}^{2} - 3 x - 5}$

$= {\lim}_{x \to \infty} x \left(\sqrt{1 + \frac{1}{x} + \frac{2}{x} ^ 2} - \sqrt{1 - \frac{3}{x} - \frac{5}{x} ^ 2}\right)$

This next bit is an algebraic shortcut as the $\frac{1}{x} ^ 2$ terms seem to disappear, but those terms are going to be order of magnitudes smaller. We could do a full expansion but that would look quite messy. So, from there:

$= {\lim}_{x \to \infty} x \left(\sqrt{1 + \frac{1}{x}} - \sqrt{1 - \frac{3}{x}}\right)$

By the binomial series expansion, ${\left(1 + z\right)}^{\alpha} = 1 + \alpha z + O \left({z}^{2}\right)$:

= lim_(x->oo) x ( (1+1/(2x) + O(1/x^2)) - (1-3/(2x) + O(1/x^2) )

$= {\lim}_{x \to \infty} x \cdot \left(\frac{1}{2 x} + \frac{3}{2 x}\right) = 2$

May 10, 2017

2

#### Explanation:

${\lim}_{x \to \infty} \left[\sqrt{{x}^{2} + x + 2} - \sqrt{{x}^{2} - 3 x - 5}\right]$
${\lim}_{x \to \infty} \left[\frac{\left(\sqrt{{x}^{2} + x + 2} - \sqrt{{x}^{2} - 3 x - 5}\right) \left(\sqrt{{x}^{2} + x + 2} + \sqrt{{x}^{2} - 3 x - 5}\right)}{\sqrt{{x}^{2} + x + 2} + \sqrt{{x}^{2} - 3 x - 5}}\right]$
${\lim}_{x \to \infty} \frac{{\left(\sqrt{{x}^{2} + x + 2}\right)}^{2} - {\left(\sqrt{{x}^{2} - 3 x - 5}\right)}^{2}}{\sqrt{{x}^{2} + x + 2} + \sqrt{{x}^{2} - 3 x - 5}}$
${\lim}_{x \to \infty} \left[\frac{\left({x}^{2} + x + 2 - {x}^{2} + 3 x + 5\right)}{\sqrt{{x}^{2} + x + 2} + \sqrt{{x}^{2} - 3 x - 5}}\right]$
${\lim}_{x \to \infty} \left[\frac{4 x + 7}{\sqrt{{x}^{2} + x + 2} + \sqrt{{x}^{2} - 3 x - 5}}\right]$
${\lim}_{x \to \infty} \left[\frac{\frac{4 x + 7}{x}}{\frac{\sqrt{{x}^{2} + x + 2} + \sqrt{{x}^{2} - 3 x - 5}}{x}}\right]$
${\lim}_{x \to \infty} \left[\frac{4 + \left(\frac{7}{x}\right)}{\left(\sqrt{1 + \frac{1}{x} + \frac{2}{x} ^ 2}\right) + \left(\sqrt{1 - \frac{3}{x} - \frac{5}{x} ^ 2}\right)}\right]$

As we know:
$\left({\lim}_{x \to \infty} \frac{1}{x} = 0\right) \mathmr{and} \left({\lim}_{x \to \infty} \frac{1}{x} ^ 2 = 0\right)$

Therefore:
$= \left[\frac{4 + 0}{\sqrt{1 + 0 + 0} + \sqrt{1 - 0 - 0}}\right]$
$= \left[\frac{4}{2}\right]$
$= 2$