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  • 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:

enter image description here

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  • $\begingroup$ The polynomial part of the result of the polynomial division is the asymptote. $\endgroup$ – Peter Jan 27 '15 at 18:21
  • $\begingroup$ 2nd question, $\lim f(x)/x = \infty$, so any "asymptote" would be vertical. A good reason to conclude there is no oblique asymptote. $\endgroup$ – GEdgar Jan 27 '15 at 18:21
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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 $d(x)$.

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  • $\begingroup$ By "principle branch of the arctangent" you mean $f(x)=\arctan x$? $\endgroup$ – shinzou Jan 27 '15 at 18:59
  • $\begingroup$ @kuhaku Yes - the solution $y$ of the equation $x=\tan y$ with $-\frac\pi2\lt y\lt\frac\pi2$. $\endgroup$ – Steven Stadnicki Jan 27 '15 at 19:01

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