Find limit of polynomials Suppose we  want to  find limit of the following polynomial
$$\lim_{x\to-\infty}(x^4+x^5).$$
If we directly put here $-\infty$, we   get  "$-\infty +\infty$" which is definitely undefined form, but otherwise  if factor out $x^5$, our polynomial will be of the form $x^5(1/x+1)$.
$\lim_{x\to-\infty}\frac 1x=0$, so our result will be $-\infty*(0+1)$,which equal to $-\infty$. I have exam in a 3 days and interested if  my last procedure is correct? Directly putting $x$ values  gives me undefined form, but factorization  on the other hand, negative infinity, which one is correct?
 A: Your factoring method is fine.
In general
given a polynomial, $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1 x+a_0,\quad a_n\ne0,$$ 
you can factor out the leading term when $x\ne0$:
$$
P(x)= x^n\Bigl(\,a_n+{ a_{n-1}\over x}+ \cdots +{a_1\over x^{n-1}}  +{a_0\over x^n} \,\Bigr),\quad x\ne0.
$$
When taking the limit as $x$ tends to an infinity, the parenthetical term above will tend towards $a_n$.
From this,
$$
\lim_{x\rightarrow-\infty} P(x) = \cases{\phantom{-}{\rm sign}(a_n) \infty, & n\text{ even}\cr  -{\rm sign}(a_n) \infty, & n\text{  odd}}
$$
and
$$
\lim_{x\rightarrow \infty} P(x) =  {\rm sign}(a_n)  \infty.
$$
Informally, for large $x$, a polynomial behaves as its leading term. So, to compute a limit "at infinity",  you could just drop all but the leading term of the polynomial and take the limit of just the leading term.
A: The second method is correct, since it shows that when $x$ becomes "really negative", $x^4$ will weight less than $x^5$. 
The first method is not correct, since "$-\infty+\infty$" can give anything: for example with $f(x)=x+a$ and $g(x)=-x$, where $a$ is a real number, we have $\lim_{x\to+\infty}f(x)=+\infty$, $\lim_{x\to+\infty}g(x)=-\infty$ but $\lim_{x\to+\infty}f(x)+g(x)=a$, so we can get any real number, $+\infty$ or $-\infty$ as your example shows. It's possible that $f(x)+g(x)$ doesn't converge, for example with $f(x)=x+\cos x$ and $g(x)=-x$.
A: Yes. generally if $\rm\:L\:$ is a field map preserving sums and products and $\rm\ L(g/f) = 0\ $ then
$$\rm L(f+g)\ =\ L(f\ (1+g/f))\ =\ L(f)\ \left(L(1) + L(g/f)\right)\ =\ L(f) $$
Induction yields the dominant term lemma: the L-value of a sum is that of the dominant term(s).
When $\rm\:L\:$ is a limit the map need not be total (everywhere defined). This concept is abstracted algebraically in the important notion of a valuation or place.
