For several mathematical models, one generally needs to study the limits at infinity of the solution. It is a kind of stability of systems. We then recall this concept and give some examples.

We already discussed the limits at the finite points of functions. Now we discuss the case of infinite points.

## Definitions and properties of the limits at infinity

Let $a,$ $b$, $L,$ and $K$ be real numbers.

**Limit a plus infinity****$+\infty$:**Informally, we say that the function $f;[a,+\infty)$ has the limit $L$ at $+\infty$ and we write $\lim_{x\to+\infty}f(x)=L$ if when the number $x$ is very large and get closed to $+\infty,$ it image $f(x)$ get very closed to the real number $L$. Formally, this means that for any small real number $\varepsilon>0,$ we can find a sufficiently large real number $A>0,$ such that for any $x>A$ we have $|f(x)-L|<\varepsilon$.-
**Limit at minus infinity $-\infty$:**Similarly, we say that the function $g:(-\infty,b]\to \mathbb{R}$ has $K$ as a limit at $-\infty$ if $g(x)$ approaches $k$ whenever, the real number $x$ is sufficiently small a get closed to $-\infty$. In this case, we write $\lim_{x\to -\infty}g(x)=K$. Formally, it means that for any $\varepsilon>0,$ small enough, there exists a sufficiently large real number $A>0,$ such that for any $x<-A$ we have $|g(x)-K|<\varepsilon$.

### Limits rational functions at infinity

Let us start with the fonction $f(x)=\frac{1}{x}$ with $x\in \mathbb{R}^\ast:=\mathbb{R}\setminus\{0\}$.

We have $$ \lim_{x\to+\infty}\frac{1}{x}=0.$$ In fact, let $\varepsilon>0$ be a very small real number. For $x>0,$ the inequality $\frac{1}<\varepsilon$ is verified whenever $x>\frac{1}{\varepsilon}$. Now if we select $A:=\frac{1}{\varepsilon}$, then for any $x>A$ we have $|\frac{1}{x}-0|=\frac{1}{x}<\frac{1}{A}=\varepsilon$. This ends the proof.

Now if $x\to -\infty,$ then $-x\to +\infty$. So that $$ \lim_{x\to -\infty}\frac{1}{x}=\lim_{x\to -\infty} \frac{-1}{-x}=- \lim_{y\to +\infty} \frac{1}{y}=0.$$

For any $n\in\mathbb{N},$ we have $$ \lim_{n\to \pm\infty} \frac{1}{x^{n}}=0.$$ More generally, let a polynomial $Q(x)=b_n x^n+b_{n-1}x^{n-1}+\cdots+b_1 x+b_0,$ where all the $b_i$ are real numbers with $b_n\neq 0$. Then \begin{align*} \frac{1}{Q(x)}=\frac{1}{x^n}\times \frac{1}{b_n+\frac{b_{n-1}}{x}+\cdots+\frac{a_0}{x^n}}.\end{align*}Thus $$ \lim_{n\to \pm \infty} \frac{1}{Q(x)}=0.$$

Next, we consider rational functions of the form $$ f(x)=\frac{P(x)}{Q(x)}$$ for polynomials $P$ and $Q$ with real coefficients with the degree of $P$ is less than the degree of $Q,$ $\deg P\le \deg Q,$ and $x\in D_f:=\{x\in \mathbb{R}: Q(x)\neq 0\}$. We distinguish two cases:

If $\deg P=\deg Q=n,$, says that $P(x)=a_n x^n+P_1(x)$ and $Q(x)=b_n x^n+Q_1(x)$ with $p=\deg P_1<n$ and $q=\deg Q_1<n$ with coefficient $\alpha_i$ and $\beta_i,$ respectively. Then, factoring by $x^n,$ we obtain \begin{align*} \frac{P(x)}{Q(x)}=\frac{a_n+\alpha_p x^{p-n}+\cdots +\alpha_0 x^{-n}}{b_n+\beta_q x^{q-n}+\cdots+\beta_0 x^{-n}}.\end{align*} As $n>p$ and $n>q,$ then according to the above limits we have $$ \lim_{x\pm \infty}\frac{P(x)}{Q(x)}=\frac{a_n}{b_n}.$$ As an example $$ \lim_{x\to\pm \infty}\frac{2x^6+3x^5+10x^2+4}{3x^6+1}=\frac{2}{3}.$$

Now assume that $p=\deg P<q=\deg Q$ and assume that $a_i$ are the coefficients of $P$ and $b_i$ are the coefficients of $Q$. Then $$ \frac{P(x)}{Q(x)}=\frac{1}{x^p} \frac{a_p x^{p-q}+\cdots+a_0 x^{-q}}{b_q+b_{q-1}x^{-1}+\cdots+b_0 x^{-q}}.$$ The fact that $q>q$ implies that \begin{align*} \lim_{x\to \pm \infty} \frac{P(x)}{Q(x)}= 0\times \frac{0}{b_q}=0.\end{align*} As an example $$ \lim_{x\to\pm\infty} \frac{x^3+x+1}{x^8+1}=0.$$

## Limits at infinity problems and solutions

**Exercise 1:** Determine the limits at infinity of the following functions \begin{align*} f(x)=\frac{\sin(x)}{x^2},\quad g(x)=2^{\frac{1}{x}}. \end{align*} **Solution:** We know that $|\sin(x)|\le 1$ for any $x\in \mathbb{R}$. Thus for $x\in\mathbb{R}\setminus\{0\},$ we have $|f(x)|\le \frac{1}{x^2}$. This means that \begin{align*} -\frac{1}{x^2}\le f(x)\le \frac{1}{x^2}.\end{align*} According to the squeeze theorem for the limits, we deduce that $\lim_{x\to\pm\infty}f(x)=0$. On the other hand, we have $$ g(x)={\rm exp }(\ln(2^{\frac{1}{x}}))=e^{\frac{\ln(2)}{x}}.$$ We now that $\frac{\ln(2)}{x}\to 0$ as $x\to \pm\infty$ and the exponential function is continuous then $g(x)\to e^0=1$ as $x\to\pm\infty$.

**Exercise 2:** Determine the limit at infinity of functions $$ f(x)={x+\sin(x)}{x+\arctan(x)},\quad g(x)=x^2\sin\left( \frac{1}{x^4}\right).$$ **Solution:** Factoring by $x,$ we have $$ f(x)=\frac{1+\frac{\sin(x)}{x}}{1+\frac{\arctan(x)}{x}}.$$ As in Exercise 1, we have $\lim_{x\to\pm \infty}\frac{\sin(x)}{x}=0$. Similarly, as $|\arctan(x)|\le \frac{\pi}{2},$ we also have $\lim_{x\to\pm \infty}\frac{\arctan(x)}{x}=0$. Thus $f(x)\to 1$ as $x\to\pm\infty$. For the function $g,$ we estimate $$ |g(x)|\le x^2 \frac{1}{x^4}=\frac{1}{x^2}.$$ According to the squeeze theorem, we have $g(x)\to 0$ as $\to\pm \infty$.

**Exercise 3:** Determine the limit $$ \lim_{x\to+\infty}\left(1+\frac{1}{x}\right)^x.$$ **Proof:** We ewrite \begin{align*} \left(1+\frac{1}{x}\right)^x=e^{x\ln\left(1+\frac{1}{x}\right)}.\end{align*} On the other hand, we know that when $x\to\infty,$ we have \begin{align*} \ln\left(1+\frac{1}{x}\right)= \frac{1}{x}-\frac{1}{2}\frac{1}{x^2}+o(\frac{1}{x^2}).\end{align*} Then \begin{align*} \left(1+\frac{1}{x}\right)^x= e^{1-\frac{1}{2x}+o(\frac{1}{x})}.\end{align*} Thus $$ \lim_{x\to+\infty} \left(1+\frac{1}{x}\right)^x=e.$$