# Oblique asymptote coincides with the function

I know when the oblique asymptote exists and I also know how to find it but I've always wondered, if you equalize the remainder you will always get the point at which the function crosses the asymtote. Why does this work?

Example:

$$f(x) = \frac{x^3+x^2-x-1}{x^2+4x+4}$$

$$f(x) = {x-3 + \frac{7x+11}{x^2+4x+4}}$$

Now if we do the following:

$$\frac{7x+11}{x^2+4x+4} = 0$$

we would get the point at which the f(x) coincides with the asymptote of y = x - 3.

Say that you have a function $f(x) = p(x)/q(x)$ with oblique asymptote $\ell(x)$ and remainder $r(x)$ (i.e., $f(x) = \ell(x) + r(x)$, where $\ell(x)$ is linear and $r(x) = \tilde p(x)/q(x)$ tends to $0$ as $x$ tends to $\infty$). If there is a solution to $r(x) = 0$, call it $x_0$, then we have $$f(x_0) = \ell(x_0) + r(x_0) = \ell(x_0).$$ So when there is no remainder, the original function agrees with the oblique asymptote $\ell(x)$ (meaning that the function will touch or cross the asymptote at that point).