Need to find the horizontal asymptotes of a given expression.

Question

I need to determine the horizontal asymptotes of the following expression:$$\frac{2x^{1/3}}{(1x^2+4)^{1/6}}$$ I’m not even sure where to begin. I understand I need to divide all terms by the highest degree in the denominator (which normally I’d assume was $x^2$), but I’m unsure of how to interpret it since it’s contained within the 6th root radicand. The third root in the numerator throws me off as well. Any help would be appreciated. Thank you.

Answer 1

I shall assume here that $x \in \mathbb R ^ +$, because defining fractional powers of negative numbers is tedious in $\mathbb R$ (you could define $x^{\frac 1 p}$ for $x < 0$ when $p$ is odd by $x^{\frac 1 p} = -\left((-x)^{\frac 1 p}\right)$, but then you would get $\left((-1)^2\right)^\frac 1 6 = 1$ but $(-1)^\frac 2 6 = (-1)^\frac 1 3 = -1$ and $\left((-1)^\frac 1 6\right)^2$ undefined...)

Then the equality $(x^a)^b = x^{ab}$ is valid for any $x \gt 0$ and $(a,b)\in\mathbb R^2$. Hence:

$$\frac{2x^\frac 1 3}{(1x^2+4)^\frac 1 6} = \frac{2x^\frac 1 3}{\left(x^2\times(1+\frac 4{x^2})\right)^\frac 1 6} = \frac{2x^\frac 1 3}{x^\frac 1 3(1+\frac 4 {x^2})^\frac 1 6} = \frac 2 {(1+\frac 4 {x^2})^\frac 1 6}$$

And you can conclude:

$$\lim_{x\to\infty}{\frac 1 {x^2}} = 0 \implies\lim_{x\to\infty}{\frac 1{1+\frac 4 {x^2}}} = 1 \implies\lim_{x\to\infty}{\frac 2{(1+\frac 4 {x^2})^\frac 1 6}}= 2 \implies\lim_{x\to\infty}{\frac{2x^\frac 1 3}{(1x^2+4)^\frac 1 6}}=2$$