# What are the asymptote(s) and hole(s), if any, of  f(x) =1/(x^2+2) ?

Dec 4, 2015

$f \left(x\right)$ has a horizontal asymptote $y = 0$ and no holes

#### Explanation:

${x}^{2} \ge 0$ for all $x \in \mathbb{R}$

So ${x}^{2} + 2 \ge 2 > 0$ for all $x \in \mathbb{R}$

That is, the denominator is never zero and $f \left(x\right)$ is well defined for all $x \in \mathbb{R}$, but as $x \to \pm \infty$, $f \left(x\right) \to 0$. Hence $f \left(x\right)$ has a horizontal asymptote $y = 0$.

graph{1/(x^2+2) [-2.5, 2.5, -1.25, 1.25]}