(Optional) — Making Infinite Limits a Little More Formal

Section 1.8 (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:

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Definition 1.8.1. Limits involving infinity — formal.

Let $f$ be a function defined on the whole real line. We say that
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the limit as $x$ approaches $\infty$ of $f(x)$ is $L$
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or equivalently
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$f(x)$ converges to $L$ as $x$ goes to $\infty$
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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 $x \gt M\text{.}$

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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 $x \lt N\text{.}$

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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 $0\lt |x-a|\lt \delta\text{.}$

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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 $x \gt M\text{.}$

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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 $-\infty\text{.}$

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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.

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Proof.

Proof of Theorem 1.5.3.

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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.

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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.

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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.

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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.

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This completes the proof.

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