## Section1.8(Optional) — Making Infinite Limits a Little More Formal

For those of you who made it through the formal $\epsilon-\delta$ definition of limits we give the formal definition of limits involving infinity:

###### Definition1.8.1Limits involving infinity — formal
1. Let $f$ be a function defined on the whole real line. We say that

the limit as $x$ approaches $\infty$ of $f(x)$ is $L$

or equivalently

$f(x)$ converges to $L$ as $x$ goes to $\infty$

and write

\begin{align*} \lim_{x \to \infty} f(x) &= L \end{align*}

if and only if for every $\epsilon \gt 0$ there exists $M \in \mathbb{R}$ so that $|f(x)-L| \lt \epsilon$ whenever $x \gt M\text{.}$

Similarly we write

\begin{align*} \lim_{x \to -\infty} f(x) &= K \end{align*}

if and only if for every $\epsilon \gt 0$ there exists $N \in \mathbb{R}$ so that $|f(x)-K| \lt \epsilon$ whenever $x \lt N\text{.}$

2. Let $a$ be a real number and $f(x)$ be a function defined for all $x\ne a\text{.}$ We write
\begin{equation*} \lim_{x \to a} f(x) = \infty \end{equation*}
if and only if for every $P \gt 0$ there exists $\delta\gt 0$ so that $f(x) \gt P$ whenever $0\lt |x-a|\lt \delta\text{.}$
3. Let $f$ be a function defined on the whole real line. We write
\begin{equation*} \lim_{x \to \infty} f(x) = \infty \end{equation*}
if and only if for every $P \gt 0$ there exists $M\gt 0$ so that $f(x) \gt P$ whenever $x \gt M\text{.}$

Note that we can loosen the above requirements on the domain of definition of $f$ — for example, in part (a) all we actually require is that $f(x)$ be defined for all $x$ larger than some value. It would be sufficient to require “there is some $x_0 \in \mathbb{R}$ so that $f$ is defined for all $x \gt x_0$”. Also note that there are obvious variations of parts (b) and (c) with $\infty$ replaced by $-\infty\text{.}$

For completeness let's prove Theorem 1.5.3 using this form definition. The layout of the proof will be very similar to our proof of Theorem 1.4.1.

Proof of Theorem 1.5.3.

There are four limits to prove in total and we do each in turn. Let $c \in \mathbb{R}\text{.}$

• Let $\epsilon \gt 0$ and set $f(x)=c\text{.}$ Choose $M=0\text{,}$ then for any $x$ satisfying $x \gt M$ (or indeed any real number $x$ at all) we have $|f(x)-c| = 0 \lt \epsilon\text{.}$ Hence $\ds \lim_{x \to \infty} c = c$ as required.
• The proof that $\ds \lim_{x \to -\infty} c = c$ is nearly identical. Again, let $\epsilon \gt 0$ and set $f(x)=c\text{.}$ Choose $N=0\text{,}$ then for any $x$ satisfying $x \lt N$ we have $|f(x)-c| = 0 \lt \epsilon\text{.}$ Hence $\ds \lim_{x \to -\infty} c = c$ as required.
• Let $\epsilon \gt 0$ and set $f(x)=x\text{.}$ Choose $M = \frac{1}{\epsilon}\text{.}$ Then when $x \gt M$ we have
\begin{align*} 0 \lt M & \lt x & \text{divide through by $xM$ to get}\\ 0 \lt \frac{1}{x} & \lt \frac{1}{M} = \epsilon \end{align*}
Since $x \gt 0\text{,}$ $\frac{1}{x} = |\frac{1}{x}| = |\frac{1}{x} - 0| \lt \epsilon$ as required.
• Again, the proof in the limit to $-\infty$ is similar but we have to be careful of signs. Let $\epsilon \gt 0$ and set $f(x)=x\text{.}$ Choose $N = -\frac{1}{\epsilon}\text{.}$ Then when $x \lt N$ we have
\begin{align*} 0 \gt N & \gt x &\text{ divide through by $xN$ to get}\\ 0 \gt \frac{1}{x} & \gt \frac{1}{N} = -\epsilon \end{align*}
Notice that by assumption both $x,N \lt 0\text{,}$ so $xN \gt 0\text{.}$ Now since $x \lt 0\text{,}$ $\frac{1}{x} = -|\frac{1}{x}| = |\frac{1}{x} - 0| \lt \epsilon$ as required.

This completes the proof.