Describing asymptotic behaviour of a function 
For question B!
x^2+x+1/x^2
= 1+ [x+1/x^2]
shouldnt the answer be
asymptote at x=0 and y=1 ??
i dont understand the textbook solution


Asymptotic behavior, in general, describes how a function behaves near a limit. For example, if $\lim\limits_{x\to a}f(x)=\infty$ and $\lim\limits_{x\to a}g(x)=\infty$, where $a$ may be $\infty$, then asymptotically,
$$
f(x)\sim g(x)\quad\text{as}\quad x\to a
$$
means that
$$
\lim_{x\to a}\frac{f(x)}{g(x)}=1
$$
The same can be said if $\lim\limits_{x\to a}f(x)=0$ and $\lim\limits_{x\to a}g(x)=0$.
Often, there is a more precise asymptotic expansion that is possible, and it is usually given as
$$
f(x)\sim g_1(x)+g_2(x)+g_3(x)+\dots\quad\text{as}\quad x\to a
$$
where
$$
\lim_{x\to a}\frac{g_{k+1}(x)}{g_k(x)}=0
$$
which means
$$
\lim_{x\to a}\frac{f(x)-(g_1(x)+g_2(x)+g_3(x)+\dots+g_k(x))}{g_k(x)}=0
$$

Applied to $\frac{1+x+x^2}{x^2}$,
As $x\to\pm\infty$, $\frac{1+x+x^2}{x^2}\sim1+\frac1x+\frac1{x^2}\sim1+\frac1x\sim1$
As $x\to0$, $\frac{1+x+x^2}{x^2}\sim\frac1{x^2}+\frac1x+1\sim\frac1{x^2}+\frac1x\sim\frac1{x^2}$

Perhaps the most well-known asymptotic expansion is Stirling's Asymptotic Expansion:
$$
n!\sim\frac{n^n}{e^n}\sqrt{2\pi n}\left(1+\frac{1}{12n}+\frac{1}{288n^2}-\frac{139}{51840n^3}+\dots\right)
$$
as $n\to\infty$.
