# How do you identify all asymptotes or holes for f(x)=(x^2-2)/x?

Jan 18, 2017

The vertical asymptote is $x = 0$
The slant asymptote is $y = x$
No holes and no horizontal asymptote.

#### Explanation:

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{0\right\}$

As we cannot divide by $0$, $x \ne 0$

The vertical asymptote is $x = 0$

We can rewrite $f \left(x\right)$ as

$f \left(x\right) = \frac{{x}^{2} - 2}{x} = x - \frac{2}{x}$

Therefore,

${\lim}_{x \to - \infty} \left(f \left(x\right) - x\right) = {\lim}_{x \to - \infty} - \frac{2}{x} = {0}^{+}$

${\lim}_{x \to + \infty} \left(f \left(x\right) - x\right) = {\lim}_{x \to + \infty} - \frac{2}{x} = {0}^{-}$

The slant asymptote is $y = x$

graph{(y-(x^2-2)/x)(y-x)=0 [-11.25, 11.25, -5.625, 5.62]}