Describing asymptotic behaviour of a function

Question

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

Answer 1

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 $$ \lim_{x\to a}\frac{f(x)-(g_1(x)+g_2(x)+g_3(x)+\dots+g_k(x))}{g_k(x)}=0 $$

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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 $x\to0$, $\frac{1+x+x^2}{x^2}\sim\frac1{x^2}+\frac1x+1\sim\frac1{x^2}+\frac1x\sim\frac1{x^2}$

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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 $n\to\infty$.

Answer 2

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.