Calculating limits at infinity: will taking the highest degree term of the numerator and denominator always work? If I wanted to find the limit of a rational function at positive or negative infinity, will this method always work when possible?
THE METHOD: taking the highest degree of the numerator and denominator. 
I mean, obviously this method will only work if there is a highest degree term. But when this method is applicable to the problem, will it always work?
 A: If your limit is in the form: $$L=\lim_{x \rightarrow ± \infty}{P(x)\over Q(x)}$$
where $P$ and $Q$ are functions of the form $\sum x^k \,\,\,,k\in\mathbb{R}$, then yes, the idea you presented does work, although you perhaps need to be slightly more specific.
Basically - whichever of $P$ and $Q$ has the term with highest degree 'wins'. If $P$ has the term with highest degree, then $L=± \infty$ (note you need to pay attention to the signs), and if $Q$ has the term with highest degree, then $L=0$. If $P$ and $Q$ have the same term with highest degree, then you can just take the coefficients of the largest powers as the limit (unless $P=Q=0$, in which case the limit is undefined).
Essentially what you're doing is dividing top and bottom by the largest power of $x$. You can then take the individual limits of each of the terms, and this will make everything multiplied by a negative power of $x$ become $0$.
It becomes slightly more complicated when other functions are introduced, and you might have terms like $\sin x$, $x!$ or $a^x$ in there, in which case you won't necessarily be able to apply the above rule.
A: Because in a polynomial, the leading term always dominates and the other terms can be discarded.
$$ax_n+a_x^{n-1}+\cdots a_0=a_nx^n\left(1+\frac{a_{n-1}}{a_nx}+\cdots\frac{a_0}{a_nx^n}\right)\sim a_nx^n.$$
