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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?

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It's correct. The method will work for any polynomial (factor out the leading term). –  David Mitra Dec 20 '11 at 10:01
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3 Answers 3

up vote 2 down vote accepted

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.

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everything is clear,thanks very much @David Mitra –  dato datuashvili Dec 20 '11 at 10:17
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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 $g(x)=-x$.

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nice example thanks very much –  dato datuashvili Dec 20 '11 at 10:11
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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.

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