# How do you find the asymptotes for g(x)=x/(root4(x^4+2))?

Feb 15, 2016

Evaluate the limits of $g \left(x\right)$ as $x \to + \infty$ and $x \to - \infty$ to find horizontal asymptotes $y = 1$ and $y = - 1$.

#### Explanation:

$g \left(x\right) = \frac{x}{\sqrt[4]{{x}^{4} + 2}} = \frac{x}{\left\mid x \right\mid} \frac{\left\mid x \right\mid}{\sqrt[4]{{x}^{4} + 2}} = \frac{x}{\left\mid x \right\mid} \frac{1}{\sqrt[4]{1 + \frac{2}{x} ^ 4}}$

So ${\lim}_{x \to + \infty} g \left(x\right) = 1$ and ${\lim}_{x \to - \infty} g \left(x\right) = - 1$

So $g \left(x\right)$ has horizontal asymptotes $y = 1$ and $y = - 1$

${x}^{4} + 2 \ge 2 > 0$ for any Real number $x$

Hence the denominator of $g \left(x\right)$ is always non-zero and $g \left(x\right)$ has no vertical asymptotes.

graph{x/root(4)(x^4+2) [-5.55, 5.55, -2.775, 2.774]}