In this section, we learn:

## Finite Limits at Infinity

Consider the function $f$ defined by the equation
$f(x)=\frac{x+1}{x+2}.$ Let’s investigate the behavior of $f$ when $x$ is positive and becomes larger and larger. From the following table and the graph of $f$ (Figure 1), we see that $f(x)$ gets closer and closer to 1. In other words, we can make $f(x)$ arbitrarily close to 1 if we choose $x$ sufficiently large. In this case, we say $f$ approaches 1 (or $f$ has limit 1) as $x$ approaches infinity and we write
$\lim_{x\to\infty}f(x)=1.$

\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & 10 & 100 & 1000 & 10,000 & 100,000 & 1,000,000 & …\\
\hline
\hline
f(x) & 0.916667 & 0.990196 & 0.99002 & 0.9999 & 0.99999 & 0.999999 & …\\
\hline
\end{array}

 Figure 1: Graph of $y=\frac{x+1}{x+2}$.

Now let’s investigate the behavior of $f$ when $x$ is negative and its magnitude becomes larger and larger. In this case, we see from the following table and the graph of $f$ that $f(x)$ gets closer and closer to 1 too. In this case, we say $f$ approaches 1 (or $f$ has limit 1) as $x$ approaches minus infinity and write
$\lim_{x\to-\infty}f(x)=1.$ [Here is a coincidence that $\lim_{x\to+\infty}f(x)=\lim_{x\to-\infty}f(x)=1$].

\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & … & –1,000,000 & –100,000 & –10,000 & –1000 & –100 & –10\\
\hline
\hline
f(x) & … & 1.000001 & 1.00001 & 1.00010002 & 1.001002 & 1.010204 & 0.818182\\
\hline
\end{array}

In general, if the graph of $f$ gets closer and closer to the horizontal line $y=L$ as $x$ gets larger and larger, we say the limit of $f$as $x$ approaches $\infty$ is $L$ and write
$\lim_{x\to+\infty}f(x)=L.$ Instead of $+\infty$ we may simply write $\infty$.

• Again $\infty$ is a symbol and does not represent a number.
• Alternatively when $\lim_{x\to\infty}f(x)=L$, we may say:
• $f(x)$ approaches $L$ as $x$ approaches infinity
• the limit of $f(x)$, as $x$ approaches infinity, is $L$
• the limit of $f(x)$, as $x$ increases without bound, is $L$
• the limit of $f(x)$, as $x$ becomes infinite, is $L$
•  the limit of $f$ at (plus) infinity is $L$.

#### The exact definition of limits at infinity

The precise definition is as follows:

Definition 1: Let $f$ be a function defined on an infinite interval $(c,+\infty)$ for some $c$. We say $f$ tends to the limit $L$ as $x$ tends to infinity if for every given positive number $\epsilon$, there exists a corresponding number $M$, such that for all values of $x>M$, $f(x)$ differs from $L$ by less that $\epsilon$ , i.e.
$|f(x)-L|<\epsilon\qquad\text{whenever}\qquad x>M.$ When this is the case, we may write
$\lim_{x\to+\infty}f(x)=L,$ or
$f(x)\to L\ \text{as }\ x\to+\infty.$
The geometrical interpretation of $\lim_{x\to\infty}f(x)=L$ is as follows. Consider any horizontal lines $y=L\pm\epsilon$ as in Figure 2. The above definition tells us that there is some positive number $M$ such that for every $x$ larger than $M$, the point $(x,f(x))$ lies between these horizontal lines.

 Figure 2: For every given $\epsilon>0$, the graph of $y=f(x)$ lies between $y=L-\epsilon$ and $y=L+\epsilon$ when $x$ is large enough.
• Note that in the above definition, if you change the value of $\epsilon$, we may have to change our $M$. Mathematically, it means that $M$ is a function of $\epsilon$ and we may write $M(\epsilon)$.
•  Similarly, we can define
$\lim_{x\to-\infty}f(x)=L.$ We say $f(x)$ approaches $L$ as $x$ approaches minus infinity
if for every given $\epsilon>0$, there exists a number $N<0$ such
that
$|f(x)-L|<\epsilon\quad\text{whenever}\quad x<N.$

The geometrical meaning of the above definition is illustrated in Figure 3.

 Figure 3: $\lim_{x\to-\infty}f(x)=L$ means that for every given $\epsilon>0$, the graph of $y=f(x)$ lies between $y=L-\epsilon$ and $y=L+\epsilon$ when $x$ is sufficiently large negative.

We can show that for a constant number $c$
$\lim_{x\to\infty}c=c\qquad\lim_{x\to-\infty}c=c$ Also the limit laws (Theorems 2 and 5 in Section 4.4) carry over without changes to limits at $+\infty$ or $-\infty$, namely:

If $f(x)\to L$ and $g(x)\to M$ as $x\to\infty$ or $x\to-\infty$ then

• $cf(x)\to cL$
• $f(x)\pm g(x)\to L\pm M$
• $f(x)g(x)\to LM$
• $\dfrac{f(x)}{g(x)}\to\dfrac{L}{M}\qquad$ provided $M\neq0$
• $(f(x))^{n}\to L^{n}\qquad$ where $n$ is a positive integer
• $\sqrt[n]{f(x)}\to\sqrt[n]{L}\qquad$ where $n$ is a positive integer and when $n$ is even, we assume $L>0$.

The next theorem is helpful when evaluating limits at infinity.

Theorem 1 (Important Limits at Infinity): (1) If $r>0$ is a rational number, then
$\lim_{x\to+\infty}\frac{1}{x^{r}}=0,\qquad\lim_{x\to-\infty}\frac{1}{x^{r}}=0$ The second limit is valid only if $x^{r}$ is defined when $x<0$.

(2)
$\lim_{x\to+\infty}e^{-x}=0,\qquad\lim_{x\to-\infty}e^{x}=0.$

(3)
$\lim_{x\to+\infty}\arctan x=\frac{\pi}{2},\qquad\lim_{x\to-\infty}\arctan x=-\frac{\pi}{2}.$ [Another notation for the inverse of tangent is $\tan^{-1} x$.]

• Note that when $r=m/n$ where $m$ and $n$ are two integers, $x^{r}$ is defined for $x<0$ only when $n$ is an odd integer. So we can say $\lim_{x\to-\infty}\frac{1}{x^{m/n}}=0$ only when $n$ is odd (see Section 3.1). For example, if $r=\frac{1}{2}$, $\frac{1}{x^{1/2}}=\frac{1}{\sqrt{x}}$ is not defined for $x<0$, so it is meaningless to talk about its limit as $x$ approaches $-\infty$.
• The best way to remember the above theorem is to consider the graphs of these functions (see, Figures 4, 5, and 6).

 (a) $\lim_{x\to\pm\infty}\dfrac{1}{x^{3}}=0$ (b) $\lim_{x\to\pm\infty}\dfrac{1}{x^{1/3}}=0$

Figure 4

 (a) $\lim_{x\to-\infty}e^{x}=0$ (b) $\lim_{x\to+\infty}e^{-x}=0$

Figure 5

 Figure 6: Graph of $y=\arctan x$ (or $y=\tan^{-1}x$). $\lim_{x\to-\infty}\arctan x=-\frac{\pi}{2}$ and $\lim_{x\to\infty}\arctan x=\frac{\pi}{2}$

Figure 6

#### Show the proof

We just prove the first part. The second and third parts are clear from their graphs (see Figures 5(a,b), and 6)

First, let us show
$\lim_{x\to+\infty}\frac{1}{x}=0.$ For a given $\epsilon>0$, we need to find $M>0$ such that
$\left|\frac{1}{x}-0\right|<\epsilon\ \text{ whenever } x>M$ Because $x>0$
$\left|\frac{1}{x}-0\right|=\frac{1}{x}<\epsilon,$ and we can rewrite the above inequality as [Recall if $a<b$ and $a$ and $b$ are both positive or both negative, then $1/a>1/b$. See property 7 in the Section on Inequalities] $x>\frac{1}{\epsilon}.$ So if we choose $M>1/\epsilon$ then $x>M$ implies
$\left|\frac{1}{x}-0\right|<\epsilon.$ Similarly, we can show
$\lim_{x\to-\infty}\frac{1}{x}=0$ For a given $\epsilon>0$, we need to find $N<0$ such that
$\left|\frac{1}{x}-0\right|<\epsilon\ \text{ whenever \ }x<N$ Because $x<0$,
$\left|\frac{1}{x}-0\right|=\frac{-1}{x}<\epsilon$ which can be rewritten as [both $-1/x$ and $\epsilon$ are positive] $-x>\frac{1}{\epsilon}$ or [multiplying each side of an inequality by a negative number reverses the direction of the inequality. Recall property 6 in the Section on Inequalities] $x<-\frac{1}{\epsilon}$ If we choose $N<-1/\epsilon$, then
$x<N\implies\left|\frac{1}{x}-0\right|<\epsilon.$ Now we consider the general case when $r=m/n$ and show
$\lim_{x\to\pm\infty}\frac{1}{x^{m/n}}=0$ using the limit laws.
\begin{align*}
\lim_{x\to\pm\infty}\frac{1}{x^{m/n}} & =\lim_{x\to\pm\infty}\frac{1}{\sqrt[n]{x^{m}}}\\
& =\sqrt[n]{\lim_{x\to\pm\infty}\frac{1}{x^{m}}}\\
& =\sqrt[n]{\left(\lim_{x\to\pm\infty}\frac{1}{x}\right)^{m}}\\
& =\sqrt[n]{0^{m}}\\
& =0.
\end{align*}

Example 1
Find the following limits
(a) ${\displaystyle \lim_{x\to+\infty}\left(3-\frac{5}{x^{3}}+\frac{1}{\sqrt{x}}\right)}$
(b) ${\displaystyle \lim_{x\to+\infty}\frac{4}{e^{x}}}$

Solution
(a)
\begin{align*}
\lim_{x\to+\infty}\left(3-\frac{5}{x^{3}}+\frac{1}{\sqrt{x}}\right) & =\lim_{x\to+\infty}3-5\lim_{x\to+\infty}\frac{1}{x^{3}}+\lim_{x\to+\infty}\frac{1}{\sqrt{x}}\\
& =3-5(0)+0\\
& =3
\end{align*}
(b)
\begin{align*}
\lim_{x\to+\infty}\frac{4}{e^{x}} & =\lim_{x\to+\infty}4e^{-x}\\
& =4\lim_{x\to+\infty}e^{-x}\\
& =4(0)\\
& =0.
\end{align*}

## Infinite Limits at Infinity

If $f(x)$gets larger and larger as $x$ gets larger and larger, we
write
$\lim_{x\to+\infty}f(x)=+\infty$ [We may also write $\infty$ instead of $+\infty$.]

#### Show the precise definition

The precise definition is as follows.

Definition 2: Let the function $f$ be defined on some open interval $(c,+\infty)$. The function $f(x)$ is said to tend to $+\infty$ with $x$ if you
give me any number $K$, however large, I can choose a number $M$ such that
$f(x)>K\qquad\text{whenever}\qquad x>M.$ When this condition is met, we may write
$\lim_{x\to+\infty}f(x)=+\infty.$
•  Again, in this definition, I choose my number $M$ depending on what your number $K$ is.

• Similarly, we can define
$\lim_{x\to+\infty}f(x)=-\infty,$ or
$\lim_{x\to-\infty}f(x)=+\infty\quad\text{or}\quad\lim_{x\to-\infty}f(x)=-\infty.$

It can be shown that the theorems in the Section on Infinite Limits  carry over without change if $x\to a$ is replaced by $x\to+\infty$ or $x\to-\infty$, namely

• number + infinity = infinity
• number ($\neq0$)$\times$ infinity = $+\infty$ or $-\infty$ (depending on the sign of the number and sign of infinity)
• number / infinity = 0

where infinity can be $+\infty$ or $-\infty$

### Limits of $x^r$ as x →±∞

The end behavior of the power functions $y=x^{n}$ ($n$ is an integer) for some special values of $n$ is illustrated in Figure 7. The general results are

$\bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to+\infty}x^{n}=+\infty,\quad n=1,2,3,4,\cdots}\tag{i}$

and

$\bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to-\infty}x^{n}=\begin{cases} +\infty & n=2,4,6,\cdots\\ -\infty & n=1,3,5,\cdots \end{cases}}\tag{ii}$

 (a) As we can see here $\lim_{x\to\pm}x^{4}=\infty$. Similarly $\lim_{x\to\pm\infty}x^{n}=\infty$ if $n$ is even (b) As we can see here $\lim_{x\to\infty}x^{3}=\infty$ and $\lim_{x\to-\infty}x^{3}=-\infty$. Similarly $\lim_{x\to\infty}x^{n}=\infty$ and $\lim_{x\to-\infty}x^{n}=\infty$ if $n$ is odd.

Figure 7

As explained above, a nonzero number multiplied by infinity is $+\infty$ or $-\infty$. Therefore, the limit of $ax^{n}$ does not affect the limit if $a>0$ and reverses the sign if $a<0$.

• In a more general case: (a)
$\bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to+\infty}x^{m/n}=+\infty}\tag{iii}$ and (b) if $n$ is odd
$\bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to-\infty}x^{m/n}=\begin{cases} +\infty & \text{if }m\text{ is even}\\ -\infty & \text{if }m\text{ is odd} \end{cases}}\tag{iv}$

• #### Show the formal proof

Formal proof of part (a). Let $K>0$ be given. To make $x^{m/n}$ exceed $K$, we just need to choose $x>K^{n/m}$. Therefore if $M\geq K^{n/m}$ then
$x>M\implies x^{m/n}>K.$ For the formal proof of (b), we first proof that
$\lim_{x\to-\infty}x^{1/n}=-\infty.$ Because $n$ is odd, $\sqrt[n]{x}$ is defined for negative $x$. Let $C<0$ be given. To make $x^{1/n}$ less than $C$, we just need to choose $x<C^{n}$
$x<N=C^{n}\implies\sqrt[n]{x}<C.$ [For example, if $n=3$ and $C=-10$, then
$x<-1000=(-10)^{3}\implies\sqrt[3]{x}<-10]$ Because $x^{m/n}=(\sqrt[n]{x})^{m}$, by part (d) of Theorem 2 in Section 4.6, we have
$\lim_{x\to-\infty}\left(\sqrt[n]{x}\right)^{m}=\begin{cases} +\infty & \text{if }m\text{ is even}\\ -\infty & \text{if }m\text{ is odd} \end{cases}$
• #### Show the informal proof

For the informal proof, we note that $\sqrt[n]{+\infty}=+\infty$ because
$\underbrace{\infty\cdot\infty\cdots\infty}_{n\text{ times}}=\infty$ and if $n$ is odd $\sqrt[n]{-\infty}=-\infty$ because
$\underbrace{(-\infty)(-\infty)\cdots(-\infty)}_{n\text{ times }(n\text{ is odd})}=-\infty$ Therefore,
\begin{align*}
\lim_{x\to+\infty}x^{m/n} & =\lim_{x\to+\infty}(\sqrt[n]{x})^{m}\\
& =\left(\sqrt[n]{\lim_{x\to+\infty}x}\right)^{m}\\
& =\left(\sqrt[n]{+\infty}\right)^{m}\\
& =\left(+\infty\right)^{m}\\
& =+\infty
\end{align*}
and similarly
\begin{align*}
\lim_{x\to-\infty}x^{m/n} & =\left(\lim_{x\to-\infty}\sqrt[n]{x}\right)^{m}\\
& =\left(\sqrt[n]{\lim_{x\to-\infty}x}\right)^{m}\\
& =(-\infty)^{m}\\
& =\underbrace{(-\infty)\cdots(-\infty)}_{m\text{ times}}\\
& =\begin{cases}
+\infty & \text{if }m\text{ is even}\\
-\infty & \text{if }n\text{ is even}
\end{cases}
\end{align*}
Example 2
Find
(a) $\lim_{x\to\infty}4x^{7}$
(b) $\lim_{x\to-\infty}4x^{7}$
(c) $\lim_{x\to\infty}-3x^{8}$
(d) $\lim_{x\to-\infty}-3x^{8}$
Solution
(a)
$\lim_{x\to\infty}4x^{7}=4\lim_{x\to\infty}x^{7}=4(+\infty)=\infty$ (b)
$\lim_{x\to-\infty}4x^{7}=4\lim_{x\to-\infty}x^{7}=4(-\infty)=-\infty$ (c)
$\lim_{x\to\infty}-3x^{8}=-3\lim_{x\to\infty}x^{8}=-3(+\infty)=-\infty$ (d)
$\lim_{x\to-\infty}-3x^{8}=-3\lim_{x\to-\infty}x^{8}=-3(+\infty)=-\infty$
Example 3
Find
(a) ${\displaystyle \lim_{x\to\infty}x^{3/4}}$
(b) ${\displaystyle \lim_{x\to-\infty}x^{3/5}}$
(c) ${\displaystyle \lim_{x\to-\infty}x^{4/5}}$
(d) ${\displaystyle \lim_{x\to-\infty}x^{3/4}}$
Solution
(a) Because $x$ approaches plus infinity
$\lim_{x\to+\infty}x^{3/4}=+\infty$ (b) Here $m=3$ is odd and thus
$\lim_{x\to-\infty}x^{3/5}=-\infty$ (c) Here $m=4$ is even and thus
$\lim_{x\to-\infty}x^{4/5}=+\infty$ (d) Here $n=4$ is even. Thus, $x^{3/4}$ is not defined for negative $x$ and $\lim_{x\to-\infty}x^{3/4}$ is meaningless.
• If $f(x)\to L(\neq0)$ as $x\to+\infty$ or $x\to-\infty$, then the limit of $f(x)x^{n}$ is the same as the limit of $x^{n}$ if $L>0$ and the opposite of the limit of $x^{n}$ if $L<0$.
$\bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to\pm\infty}\left(f(x)x^{n}\right)=\begin{cases} \lim_{x\to\pm\infty}x^{n} & \text{if }L>0\\ \text{indeterminate} & \text{if }L=0\\ -\lim_{x\to\pm\infty}x^{n} & \text{if }L<0 \end{cases}.}$

Example 4
Find $\lim_{x\to-\infty}\left(x\arctan x\right)$.
Solution
$\lim_{x\to-\infty}\left(x\arctan x\right)=\left(\lim_{x\to-\infty}x\right)\left(\lim_{x\to-\infty}\arctan x\right)$ By Part (3) of Theorem 1, $\lim_{x\to-\infty}\arctan x=-\pi/2$. Therefore
\begin{align*}
\lim_{x\to-\infty}\left(x\arctan x\right) & =\left(\lim_{x\to-\infty}x\right)\left(\lim_{x\to-\infty}\arctan x\right)\\
& =(-\infty)\left(-\frac{\pi}{2}\right)\\
& =+\infty.
\end{align*}
The graph of $y=x\arctan x$ shown in the following figure.

 Figure 8: Graph of $y=x\arctan x$ (or $y=x\tan^{-1}x$)

If the limit of $f(x)$ as $x$ approaches $+\infty$ (or $-\infty$) does not exist and it does not approaches $+\infty$ or $-\infty$, then $f(x)$is said to oscillate as $x$ approaches infinity. In this case, if $f(x)$ is bounded, we say $f(x)$ oscillates finitely, and otherwise infinitely.

### End Behavior of Trigonometric Functions

The trigonometric functions are periodic and they do not approach any values (inlcuding $+\infty$ and $-\infty$) as $x$ approaches$\pm\infty$. The sine and cosine functions osciallate finitely (between $-1$ and $1$) and the tangent and cotangent osciallate infinitely.

$\bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to\pm\infty}\sin x=\text{does not exist}\qquad \lim_{x\to \pm\infty}\cos x=\text{does not exist}}$

$\bbox[#F2F2F2,5px,border:2px solid black]{ \lim_{x\to\pm\infty}\tan x=\text{does not exist}\qquad \lim_{x\to \pm\infty}\cot x=\text{does not exist}}$

 (a) Graphs of $y=\sin x$ and $y=\cos x$ (b) Graphs of $y=\tan x$ and $y=\cot x$

Figure 9: Trignonmetric functions fail to have limits as $x\to+\infty$ or $x\to-\infty$

### Reducing Limits at Infinity to Limits at Zero

Suppose $\lim_{x\to+\infty}f(x)=L$. Let $u=1/x$. As $x\to+\infty$, $u$ approaches zero through positive values. Therefore, we can say
$\lim_{x\to+\infty}f(x)=L\implies\lim_{u\to0^{+}}f\left(\frac{1}{u}\right)=L.$ Conversely we can show if $\lim_{u\to0^{+}}f(1/u)=L$, then $\lim_{x\to+\infty}f(x)=L$.
$\bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to+\infty}f(x)=L\quad\Longleftrightarrow\quad\lim_{u\to0^{+}}f\left(\frac{1}{u}\right)=L}$

Similarly if $u=1/x$, as $x\to-\infty$, $u$ approaches zero through negative values, $u\to0^{-}$, and we can show
$\bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to-\infty}f(x)=L\quad\Longleftrightarrow\quad\lim_{u\to0^{-}}f\left(\frac{1}{u}\right)=L}$ So we have shown that any problem that involves limits at infinity can be reduced to a problem involving limits at zero, and vice versa.

Example 5
Find
(a) ${\displaystyle \lim_{x\to+\infty}x\sin\frac{1}{x}}$
(b) ${\displaystyle \lim_{x\to-\infty}x\sin\frac{1}{x}}$
Solution
Let $u=1/x$. Thus,
$\lim_{x\to+\infty}x\sin\frac{1}{x}=\lim_{u\to0^{+}}\frac{1}{u}\sin u,$ and
$\lim_{x\to-\infty}x\sin\frac{1}{x}=\lim_{u\to0^{-}}\frac{1}{u}\sin u.$ Recall that (see here)
$\lim_{u\to0}\frac{\sin u}{u}=1.$ Therefore,
$\lim_{x\to+\infty}x\sin\frac{1}{x}=\lim_{x\to-\infty}x\sin\frac{1}{x}=1.$ Also, recall that by the Sandwich Theorem we showed
$\lim_{x\to0}x\sin\frac{1}{x}=0.$ The graph of $y=x\sin\frac{1}{x}$ is shown in the following figure.

 Figure 10: From this graph we can see that $\lim_{x\to0}x\sin\frac{1}{x}=0$ and $\lim_{x\to\pm\infty}x\sin\frac{1}{x}=1$.

Example 6
Find $\lim_{x\to\infty}(x^{2}-x)$.
Solution
Because $\lim_{x\to+\infty}x^{2}=\lim_{x\to+\infty}x=+\infty$, we obtain the indeteminate form $\infty-\infty$. Letting $u=1/x$, we have
\begin{align*}
\lim_{x\to\infty}(x^{2}-x) & =\lim_{u\to0^{+}}\left(\frac{1}{u^{2}}-\frac{1}{u}\right)\\
& =\lim_{u\to0^{+}}\frac{1-u}{u^{2}}.
\end{align*}
Because the limit of the numerator is one$>0$ ($\lim_{u\to0^{+}}(1-u)=1-0=1$), the limit of the denominator is zero $\lim_{u\to0^{+}}u^{2}=0$, and the denominator approaches zero through positive numbers $u^{2}\geq 0$, by Theorem 1 in Section 4.6 we obtain
$\lim_{u\to0^{+}}\frac{1-u}{u^{2}}=+\infty.$ In Section 4.10, we will learn a fast way to find the limits of polynomials at infinity.