Find all vertical and horizontal asymptotes. Find all vertical and horizontal asymptotes:
$$f(x)= \frac{4x}{x^2+16}$$

My Work: (Vertical)
1) $x^2 + 16 = 0$
2) $x^2=-16$
3) not possible so there is no vertical asymptote 



My Work: (Horizontal)
1) $4x/x^2$
2) $4/x$
3) What do I do from here?
 A: Vertical asymptote: the only undefined output of $f(x)$ is at $x = 0$. Consider $x\to 0$:
$$\lim_{x\to0} \left(\frac{4x}{x^2}+16\right) = \lim_{x\to0} \left(\frac{4}{x}+16\right) = \infty$$
and so $x=0$ is an asymptote.

Horizontal asymptote:
$$\lim_{x\to\infty} \left(\frac{4x}{x^2}+16\right) = 
\lim_{x\to\infty} \frac{4}{x} +16= 16$$
and so $y=16$ is another asymptote.

Horizontal asymptote:
$$\lim_{x\to\infty} \frac{4x}{x^2+16} = 
\lim_{x\to\infty} \frac{4/x}{1+16/x^2}= 0$$
and so $y=0$ is another asymptote.
A: You don't have to worry about the limit here, that's usually taught in calculus.
You are right about the vertical asymptote as there is no real number that makes the denominator equal to  $0$.
Now the horizontal asymptotes are NOT calculated by finding the roots of the numerator in
$f(x)= \frac{4x}{x^2+16}$. The rules for finding the horizontal asymptote are on this fun little site here:
 http://www.coolmath.com/precalculus-review-calculus-intro/precalculus-algebra/18-rational-functions-finding-horizontal-slant-asymptotes-01
In this case since the degree of the numerator is less than the degree of the denominator the (the degree of numerator-1 < 2- degree of denominator) we automatically know that the x-axis is the the horizontal asymptote. 
Good Luck!
