What is the connection between slant/oblique asymptote calculation to the polynomial part of the function and polynomial division?
To find the slant asymptote $y=mx+n$ we can can calculate it in two methods:
$m=\displaystyle\lim_{x\to\pm\infty}\frac {f(x)} x,\\n=\displaystyle\lim_{x\to\pm\infty}{f(x)} -mx $
Or if we have $f(x)=\frac {p(x)}{q(x)}$ where $p,q$ are polynomials, then by doing a polynomial long division on them we can get the slant asymptote.
What is the relation between the two methods? How is the limit calculation actually does a polynomial division?
Another related question is how is it that some functions don't have infinite slant asymptotes, like $f(x)=x^2$ for example, it looks like it has infinite asymptotes below the graph:
It's not that the limit calculation 'does' a long division behind the scenes; the limit calculation is viable even for functions (say, the principle branch of the arctangent) that aren't polynomial at all. Instead, polynomial long division gives you a different form of the function, one that makes it easier to compute the asymptote. If $p(x)=d(x)\cdot q(x)+r(x)$ with the degree of $r$ less than that of $q$, then we have (except at the roots of $q$!) $f(x) = \frac{p(x)}{q(x)} = d(x)+\frac{r(x)}{q(x)}$; but since $\lim_{x\to\infty}\frac{r(x)}{q(x)}=0$ (convince yourself of this, or prove it!) then the limiting behavior of $f(x)$ as $x$ goes to infinity is the same as that of $d(x)$.