1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ Consider both $x \rightarrow - \infty$ and $x \rightarrow \infty$. Keep in mind $\sqrt{x^2}=-x \text{ if } x<0$ and $\sqrt{x^2}=x \text{ if } x>0$ $\endgroup$ – randomgirl Apr 1 '15 at 23:35
  • 1
    $\begingroup$ 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. $\endgroup$ – randomgirl Apr 1 '15 at 23:37
3
$\begingroup$

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 $\frac{2x}{-x} = -2$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.