Do all polynomials with order $> 1$ go to $\pm$ infinity? Background
As background, I have found that taylor expansion provides poor estimates of a function at extreme parameter values. Indeed, the approximation at extreme values can get worse (more rapid exponential increase) as the order of the taylor series increases.
This seems intuitive, but I don't know that it is a rule... or if there is a proof. 
Questions


*

*Do all polynomials with order $> 1$ go to $\pm$ infinity?

*Is there a good reference where I can find answers to a qustion such as this?

 A: By scaling by $\rm\:x^k\:,\:$ a nonzero rational function (polynomial fraction) $\rm\:f(x)\:$ can be written uniquely in the form $\rm\ c\ x^n\: \bar f(x)\ $ where $\rm\ \lim_{x\to\infty}\:\bar f(x) = 1\:.\:$ The degree $\rm\:n\:$ is known as the order of $\rm\:f\:$ at $\rm\infty\:.\:$ These integer exponents serve as an (archimedean) scale to compare the asymptotic growth of rational functions in a neighborhood of $\rm\:\infty\:.\:$ More generally one can extend to wider classes of asymptotically well-behaved functions such as $\rm\ log(x),\ exp(x)\ $ etc to obtain more general (non-archimedean) scales that measure so-called "orders of infinity". For example see Hardy's classic textbook by that name, google "transseries", and see my post on boundaries of convergence.
A: Even linear polynomials get larger (in absolute value) than any fixed number at the variable goes to $\pm \infty$  Please see the discussion cited by lhf in the comment.
A: Yes all non-constant polynomials are unbounded.  You could see Liouville's Theorem, or just notice that the leading coefficient of the polynomial will dominate.  That is if I look at $$p(x)=a_n x^n +a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ then as $x\rightarrow\infty$ we have $p(x)\sim a_nx^n$ which goes to infinity.
Edit:  I will elaborate, but it may not be needed, and may not clarify things further.  Factoring out $x^n$, we see $$p(x)=x^n\cdot \left( a_n+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_0}{x^n} \right)$$ taking $x\rightarrow \infty$ we find $$ \left( a_n+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_0}{x^n} \right)\rightarrow a_n$$ so that $p(x)\sim a_nx^n$.  As $x^n\rightarrow \infty$ when $x\rightarrow \infty$ we see that for nonzero $n$ the polynomial will go to infinity.
Remark:  The symbol $g(x)\sim f(x)$ is short hand for $\lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=1$, and reads "f(x) is asymptotic to g(x)."
