# I can only find one horizontal asymptote in this function… how to find the negative asymptote?

Function in question is:

$$f(x) = \frac{2x}{\sqrt{x^2 + 4}}$$

I used the technique of dividing both top and bottom by the largest $x$, which is $x$ for the top and $\sqrt{x^2}$ on the bottom, which are MOSTLY equivalent:

$$f(x) = \frac{2}{\sqrt{1 + \frac{4}{\sqrt{x^2}}}}$$

In this form it's easy to see that as $x$ gets infinitely positive, the horizontal asymptote becomes $2$. However, I know that $-2$ is also a horizontal asymptote because $\sqrt{x^2}$ will produce the same numerical result either way and is confirmed on the graph of the function.

I'm not sure if I simplified it incorrectly, but I'm not sure how to prove that $-2$ is also a horizontal asymptote because no matter what I do, dividing by $x$ in the numerator leaves the numerator as a constant, and I keep getting only positive $2$ as an answer.

Thanks!

• Consider both $x \rightarrow - \infty$ and $x \rightarrow \infty$. Keep in mind $\sqrt{x^2}=-x \text{ if } x<0$ and $\sqrt{x^2}=x \text{ if } x>0$ – randomgirl Apr 1 '15 at 23:35
• Also I would write your expression as $\frac{\frac{2x }{\sqrt{x^2}}}{\sqrt{\frac{x^2}{x^2}+\frac{4}{x^2}}}$ and then consider what I have said above. – randomgirl Apr 1 '15 at 23:37

When $x < 0$, $$\sqrt{x^2+4} = -x \sqrt{1+\frac{4}{x^2}}$$ and for for large $|x|$, this goes to $-x$ so the asymptote is $\frac{2x}{-x} = -2$.
It's an odd function, so $y=2$ being a horizontal asymptote to the right implies $y=-2$ is a horizontal asymptote to the left.