# How do you find the asymptotes of y=sqrt(x^2+x+1) - sqrt(x^2-x)?

Jul 8, 2017

There are 2 horizontal asymptotes. On the right $y = 1$ is an asymptote and on the left $y = - 1$ is an asymptote.

#### Explanation:

$y$ never becomes infinite, so there is no vertical asymptote.

$\frac{\left(\sqrt{{x}^{2} + x + 1} - \sqrt{{x}^{2} - x}\right)}{1} \cdot \frac{\left(\sqrt{{x}^{2} + x + 1} + \sqrt{{x}^{2} - x}\right)}{\left(\sqrt{{x}^{2} + x + 1} + \sqrt{{x}^{2} - x}\right)} = \frac{\left({x}^{2} + x + 1\right) - \left({x}^{2} - x\right)}{\sqrt{{x}^{2} + x + 1} + \sqrt{{x}^{2} - x}}$

$= \frac{2 x + 1}{\sqrt{{x}^{2} + x + 1} + \sqrt{{x}^{2} - x}}$

 = (x(2+1/x))/(sqrt(x^2)(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))

Recall that $\sqrt{{x}^{2}} = \left\mid x \right\mid$.

As $x \rightarrow \infty$, $x$ is positive so we have

lim_(xrarroo)(x(2+1/x))/(sqrt(x^2)(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))) = lim_(xrarroo)(x(2+1/x))/(x(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))

$= {\lim}_{x \rightarrow \infty} \frac{2 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x} ^ 2} + \sqrt{1 - \frac{1}{x}}}$

$= \frac{2 + 0}{\sqrt{1 + 0 + 0} + \sqrt{x - 0}} = \frac{2}{2} = 1$

As $x \rightarrow - \infty$, $x$ is negative so we have

lim_(xrarr-oo)(x(2+1/x))/(sqrt(x^2)(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))) = lim_(xrarr-oo)(x(2+1/x))/((-x)(sqrt(1+1/x+1/x^2)+sqrt(1-1/x))

$= {\lim}_{x \rightarrow \infty} - \frac{2 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x} ^ 2} + \sqrt{1 - \frac{1}{x}}}$

$= - \frac{2 + 0}{\sqrt{1 + 0 + 0} + \sqrt{x - 0}} = - \frac{2}{2} = - 1$