# How do you find the asymptotes for f(x) = (x^2) / (x^2 + 1)?

Jan 22, 2016

$\cancel{\exists}$ vertical asymptotes
$y = 1$ horizontal asymptote $x \rightarrow \pm \infty$

#### Explanation:

Given:

$f \left(x\right) = g \frac{h}{h \left(x\right)}$

possible asymptotes are $x = {x}_{i}$ with ${x}_{i}$ values that zero the denominator $h \left(x\right) = 0$.

Indeed, the existence condition of $f \left(x\right) \in \mathbb{R}$ is $h \left(x\right) \ne 0$.

Therefore,

$h \left(x\right) = 0 \implies {x}^{2} + 1 = 0 \implies {x}^{2} = - 1 \implies {x}_{1} , {x}_{2} \in \mathbb{C}$

$h \left(x\right)$ hasn't any zeros $\in \mathbb{R}$

$\therefore f \left(x\right) : \mathbb{R} \rightarrow \mathbb{R} , \forall x \in \mathbb{R}$

We could looking for possible horizontal asymptotes:

$y = L$ is a horizontal asymptotes when:

${\lim}_{x \rightarrow \pm \infty} f \left(x\right) = L$

${\lim}_{x \rightarrow \pm \infty} f \left(x\right) = {\lim}_{x \rightarrow \pm \infty} {x}^{2} / \left({x}^{2} + 1\right) \approx \frac{\cancel{{x}^{2}}}{\cancel{{x}^{2}}} = 1$

$\therefore y = 1$ horizontal asymptote $x \rightarrow \pm \infty$

graph{x^2/(x^2+1) [-6.244, 6.243, -3.12, 3.123]}