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


You are right : there is a vertical asymptote for $x=0$ and an horizontal asymptote at $y=1$. 
But the answers given in the textbook correpond to the analysis of the behavior of $y$ when $x$ goes to $0^+$ or $0^-$ as well as when $x$ goes to $+ \infty$ or to $- \infty$. $$y=\frac{x^2+x+1}{x^2}=1+\frac{1}{x}+\frac{1}{x^2}$$ When $x$ goes to infinite values, we can ignore $\frac{1}{x^2}$ which is very small compared to $\frac{1}{x}$; so, for this case, $y \simeq 1+\frac{1}{x}$. If $x$ goes to $+ \infty$, $\frac{1}{x}$ is a small positive number which is added to $1$; then $y$ goes to $1^+$. If $x$ goes to $-
 \infty$, $\frac{1}{x}$ is a small negative number which is added to $1$; then $y$ goes to $1^-$
Now, when $x$ goes to zero, the system is dominated by $\frac{1}{x^2}$ which the largest; since it is a square, it is positive. So, if $x$ goes to $0$ by positive or negative values, it does not matter and $y$ goes to $+ \infty$.  
Is this making things clearer to you ? If not, just post. 
