# How do you find the horizontal asymptote for (x^2+1)/(x^2-1)?

Jan 7, 2016

See explanation...

#### Explanation:

One way is to divide both numerator and denominator by ${x}^{2}$ to find:

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

Then note that $\frac{1}{x} ^ 2 \to 0$ as $x \to \pm \infty$

So

$\frac{{x}^{2} + 1}{{x}^{2} - 1} \to \frac{1 + 0}{1 - 0} = 1$ as $x \to \pm \infty$

So the horizontal asymptote is $y = 1$

Alternatively, separate out the "polynomial part" $1$ from the rational expression as follows:

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

Then note that $\frac{2}{{x}^{2} - 1} \to 0$ as $x \to \pm \infty$

So again we see that the horizontal asymptote is $y = 1$