4.7 Limits at Infinity

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
$$\underset{x\to \mathrm{\infty }}{lim}f(x)=1.$$

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

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
$$\underset{x\to -\mathrm{\infty }}{lim}f(x)=1.$$ [Here is a coincidence that $\underset{x\to +\mathrm{\infty }}{lim}f(x)=\underset{x\to -\mathrm{\infty }}{lim}f(x)=1$].

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

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 $\mathrm{\infty }$ is $L$ and write
$$\underset{x\to +\mathrm{\infty }}{lim}f(x)=L.$$ Instead of $+\mathrm{\infty }$ we may simply write $\mathrm{\infty }$.

• Again $\mathrm{\infty }$ is a symbol and does not represent a number.
• Alternatively when $\underset{x\to \mathrm{\infty }}{lim}f(x)=L$, 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

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The precise definition is as follows:

Definition 1: Let $f$ be a function defined on an infinite interval $(c,+\mathrm{\infty })$ for some $c$. We say $f$ tends to the limit $L$ as $x$ tends to infinity if for every given positive number $ϵ$, there exists a corresponding number $M$, such that for all values of $x>M$, $f(x)$ differs from $L$ by less that $ϵ$ , i.e.
$$|f(x)-L|<ϵ\text{whenever}x>M.$$ When this is the case, we may write
$$\underset{x\to +\mathrm{\infty }}{lim}f(x)=L,$$ or
$$f(x)\to L\text{as }x\to +\mathrm{\infty }.$$

The geometrical interpretation of $\underset{x\to \mathrm{\infty }}{lim}f(x)=L$ is as follows. Consider any horizontal lines $y=L\pm ϵ$ 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 $ϵ>0$, the graph of $y=f(x)$ lies between $y=L-ϵ$ and $y=L+ϵ$ when $x$ is large enough.

• Note that in the above definition, if you change the value of $ϵ$, we may have to change our $M$. Mathematically, it means that $M$ is a function of $ϵ$ and we may write $M(ϵ)$.
•  Similarly, we can define
  $$\underset{x\to -\mathrm{\infty }}{lim}f(x)=L.$$ We say $f(x)$ approaches $L$ as $x$ approaches minus infinity
  if for every given $ϵ>0$, there exists a number $N<0$ such
  that
  $$|f(x)-L|<ϵ\text{whenever}x<N.$$

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

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

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

The next theorem is helpful when evaluating limits at infinity.

• 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 $\underset{x\to -\mathrm{\infty }}{lim}\frac{1}{x^{m/n}}=0$ 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 $-\mathrm{\infty }$.
• The best way to remember the above theorem is to consider the graphs of these functions (see, Figures 4, 5, and 6).

(a) $\underset{x\to \pm \mathrm{\infty }}{lim}{\displaystyle \frac{1}{x^3}}=0$	(b) $\underset{x\to \pm \mathrm{\infty }}{lim}{\displaystyle \frac{1}{x^{1/3}}}=0$

Figure 4

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

Figure 5

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

Figure 6

Show the proof

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Infinite Limits at Infinity

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

Show the precise definition

Hide the precise definition

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

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

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

$$\begin{array}{}\text{(i)}& \underset{x\to +\mathrm{\infty }}{lim}{x}^n=+\mathrm{\infty },n=1,2,3,4,\cdots \end{array}$$

and

$$\begin{array}{}\text{(ii)}& \underset{x\to -\mathrm{\infty }}{lim}{x}^n=\{\begin{array}{ll}+\mathrm{\infty }& n=2,4,6,\cdots \\ -\mathrm{\infty }& n=1,3,5,\cdots \end{array}\end{array}$$

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

Figure 7

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

• In a more general case: (a)
  $$\begin{array}{}\text{(iii)}& \underset{x\to +\mathrm{\infty }}{lim}{x}^{m/n}=+\mathrm{\infty }\end{array}$$ and (b) if $n$ is odd
  $$\begin{array}{}\text{(iv)}& \underset{x\to -\mathrm{\infty }}{lim}{x}^{m/n}=\{\begin{array}{ll}+\mathrm{\infty }& \text{if }m\text{ is even}\\ -\mathrm{\infty }& \text{if }m\text{ is odd}\end{array}\end{array}$$

• Show the formal proof

  Hide 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 $M\ge K^{n/m}$ then
  $$x>M⟹x^{m/n}>K.$$ For the formal proof of (b), we first proof that
  $$\underset{x\to -\mathrm{\infty }}{lim}{x}^{1/n}=-\mathrm{\infty }.$$ 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 $x<C^n$
  $$x<N=C^n⟹\sqrt[n]{x}<C.$$ [For example, if $n=3$ and $C=-10$, then
  $$x<-1000=(-10{)}^3⟹\sqrt[3]{x}<-10]$$ Because $x^{m/n}=(\sqrt[n]{x}{)}^m$, by part (d) of Theorem 2 in Section 4.6, we have
  $$\underset{x\to -\mathrm{\infty }}{lim}{\left(\sqrt[n]{x}\right)}^m=\{\begin{array}{ll}+\mathrm{\infty }& \text{if }m\text{ is even}\\ -\mathrm{\infty }& \text{if }m\text{ is odd}\end{array}$$

• Show the informal proof

  Hide the informal proof

   For the informal proof, we note that $\sqrt[n]{+\mathrm{\infty }}=+\mathrm{\infty }$ because
  $$\underset{n\text{ times}}{\underbrace{\mathrm{\infty }\cdot \mathrm{\infty }\cdots \mathrm{\infty }}}=\mathrm{\infty }$$ and if $n$ is odd $\sqrt[n]{-\mathrm{\infty }}=-\mathrm{\infty }$ because
  $$\underset{n\text{ times }(n\text{ is odd})}{\underbrace{(-\mathrm{\infty })(-\mathrm{\infty })\cdots (-\mathrm{\infty })}}=-\mathrm{\infty }$$ Therefore,
  $$\begin{array}{rl}\underset{x\to +\mathrm{\infty }}{lim}{x}^{m/n}& =\underset{x\to +\mathrm{\infty }}{lim}(\sqrt[n]{x}{)}^m\\ & ={\left(\sqrt[n]{\underset{x\to +\mathrm{\infty }}{lim}x}\right)}^m\\ & ={\left(\sqrt[n]{+\mathrm{\infty }}\right)}^m\\ & ={(+\mathrm{\infty })}^m\\ & =+\mathrm{\infty }\end{array}$$
  and similarly
  $$\begin{array}{rl}\underset{x\to -\mathrm{\infty }}{lim}{x}^{m/n}& ={\left(\underset{x\to -\mathrm{\infty }}{lim}\sqrt[n]{x}\right)}^m\\ & ={\left(\sqrt[n]{\underset{x\to -\mathrm{\infty }}{lim}x}\right)}^m\\ & =(-\mathrm{\infty }{)}^m\\ & =\underset{m\text{ times}}{\underbrace{(-\mathrm{\infty })\cdots (-\mathrm{\infty })}}\\ & =\{\begin{array}{ll}+\mathrm{\infty }& \text{if }m\text{ is even}\\ -\mathrm{\infty }& \text{if }n\text{ is even}\end{array}\end{array}$$
  

• If $f(x)\to L(\ne 0)$ as $x\to +\mathrm{\infty }$ or $x\to -\mathrm{\infty }$, 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 $L<0$.
  $$\underset{x\to \pm \mathrm{\infty }}{lim}(f(x)x^n)=\{\begin{array}{ll}\underset{x\to \pm \mathrm{\infty }}{lim}{x}^n& \text{if }L>0\\ \text{indeterminate}& \text{if }L=0\\ -\underset{x\to \pm \mathrm{\infty }}{lim}{x}^n& \text{if }L<0\end{array}.$$

Solution

$$\underset{x\to -\mathrm{\infty }}{lim}(x\arctan x)=\left(\underset{x\to -\mathrm{\infty }}{lim}x\right)(\underset{x\to -\mathrm{\infty }}{lim}\arctan x)$$ By Part (3) of Theorem 1, $\underset{x\to -\mathrm{\infty }}{lim}\arctan x=-\pi /2$. Therefore
$$\begin{array}{rl}\underset{x\to -\mathrm{\infty }}{lim}(x\arctan x)& =\left(\underset{x\to -\mathrm{\infty }}{lim}x\right)(\underset{x\to -\mathrm{\infty }}{lim}\arctan x)\\ & =(-\mathrm{\infty })(-\frac{\pi }{2})\\ & =+\mathrm{\infty }.\end{array}$$
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 $+\mathrm{\infty }$ (or $-\mathrm{\infty }$) does not exist and it does not approaches $+\mathrm{\infty }$ or $-\mathrm{\infty }$, 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 $+\mathrm{\infty }$ and $-\mathrm{\infty }$) as $x$ approaches$\pm \mathrm{\infty }$. The sine and cosine functions osciallate finitely (between $-1$ and $1$) and the tangent and cotangent osciallate infinitely.

$$\underset{x\to \pm \mathrm{\infty }}{lim}\sin x=\text{does not exist}\underset{x\to \pm \mathrm{\infty }}{lim}\cos x=\text{does not exist}$$

$$\underset{x\to \pm \mathrm{\infty }}{lim}\tan x=\text{does not exist}\underset{x\to \pm \mathrm{\infty }}{lim}\cot x=\text{does not exist}$$

​

(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 $x\to +\mathrm{\infty }$ or $x\to -\mathrm{\infty }$

Reducing Limits at Infinity to Limits at Zero

Suppose $\underset{x\to +\mathrm{\infty }}{lim}f(x)=L$. Let $u=1/x$. As $x\to +\mathrm{\infty }$, $u$ approaches zero through positive values. Therefore, we can say
$$\underset{x\to +\mathrm{\infty }}{lim}f(x)=L⟹\underset{u\to 0^{+}}{lim}f\left(\frac{1}{u}\right)=L.$$ Conversely we can show if $\underset{u\to 0^{+}}{lim}f(1/u)=L$, then $\underset{x\to +\mathrm{\infty }}{lim}f(x)=L$.
$$\underset{x\to +\mathrm{\infty }}{lim}f(x)=L⟺\underset{u\to 0^{+}}{lim}f\left(\frac{1}{u}\right)=L$$

Similarly if $u=1/x$, as $x\to -\mathrm{\infty }$, $u$ approaches zero through negative values, $u\to 0^{-}$, and we can show
$$\underset{x\to -\mathrm{\infty }}{lim}f(x)=L⟺\underset{u\to 0^{-}}{lim}f\left(\frac{1}{u}\right)=L$$ 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.

Solution

Let $u=1/x$. Thus,
$$\underset{x\to +\mathrm{\infty }}{lim}x\sin \frac{1}{x}=\underset{u\to 0^{+}}{lim}\frac{1}{u}\sin u,$$ and
$$\underset{x\to -\mathrm{\infty }}{lim}x\sin \frac{1}{x}=\underset{u\to 0^{-}}{lim}\frac{1}{u}\sin u.$$ Recall that (See Section $\text{???}$)
$$\underset{u\to 0}{lim}\frac{\sin u}{u}=1.$$ Therefore,
$$\underset{x\to +\mathrm{\infty }}{lim}x\sin \frac{1}{x}=\underset{x\to -\mathrm{\infty }}{lim}x\sin \frac{1}{x}=1.$$ Also, recall that by the Sandwich Theorem we showed
$$\underset{x\to 0}{lim}x\sin \frac{1}{x}=0.$$ The graph of $y=x\sin \frac{1}{x}$ is shown in the following figure.

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