Vertical/horizontal asymptotes We know  that for rational functions
$$f(x)/g(x)$$
vertical asymptotes are defined as the lines  $x=x_0$ where $g(x_0)=0$, and for horizontal asymptotes it is the limit as $x$ approaches $\infty$ or $x$ approaches $-\infty$.  My question is the following: what is the relationship between vertical and horizontal asymptotes?  For instance, if we have
$$y=f(x)$$
and we  express $x$ as function of $y$, or
$$x=f^{-1}(y),$$
is the horizontal asymptote of inverse function the same as the vertical one of the original?
Thanks in advance.
 A: Good question, and the answer is not a simple "yes" or "no".
First of all, it's not quite true what you said about vertical asymptotes. If we're considering $y=\frac{f(x)}{g(x)}$, points where $g(x)=0$ could give rise to either vertical asymptotes or to removable discontinuities. For example, $y=\frac{\sin x}{x}$ does not have a vertical asymptote at $x=0$
Now, when inverting a function $y=f(x)$, we need to be sure that the function $f$ is one-to-one. If it's not, we may be able to obtain various invertible functions by restricting the domain of $f$ in different ways. For example: There is no inverse for the function $f(x)=x^2$ defined on all reals, but if we restrict its domain to $[0,\infty)$, we get the inverse $f^{-1}(y)=\sqrt{y}$, while restricting the domain instead to $(-\infty,0]$ affords the inverse $f^{-1}(y)=-\sqrt{y}$.
If $f$ is a function with a vertical asymptote at $x=a$, and we've got some interval with $a$ on its boundary and on which $f$ is one-to-one, then yes. That vertical asymptote will turn into a horizontal asymptote for $f^{-1}$.
I'm assuming here that we're talking about simple types of asymptotes. Check out the function $f(x)=\frac{\sin\frac1x}{x}$ for an example of how weird things might get. You're going to have a hard time finding an interval on which to invert that function and see a horizontal asymptote.
A: 
is the horizontal asymptote of inverse function the same as the vertical one of the original?

For functions of the form $f(x)=\dfrac{ax+b}{cx+d}, c\neq 0$, this statement is true. The inverse is $f^{-1}(x) = \dfrac{dx-b}{-cx+a}$.
