4.4: Infinite Limits. Operations in E*

As we have noted, Theorem 1 of §3 does not apply to infinite limits, even if the function values $f(x), g(x), h(x)$ remain finite (i.e., $i n E^{1} ) .$ Only in certain cases (stated below) can we prove some analogues.

There are quite a few such separate cases. Thus, for brevity, we shall adopt a kind of mathematical shorthand. The letter $q$ will not necessarily denote a constant; it will stand for

$$\text{"a function } f : A \rightarrow E^{1}, A \subseteq(S, \rho),\text{ such that } f(x) \rightarrow q \in E^{1}\text{ as } x \rightarrow p.\text{"}$$

Similarly, "0" and "$\pm\infty$" will stand for analogous expressions, with $q$ replaced by 0 and $\pm \infty,$ respectively.

For example, the "shorthand formula" $(+\infty)+(+\infty)=+\infty$ means

$$\text{"The sum of two real functions, with limit}+\infty\text{ at } p\text{ } (p \in S),\text{ is itself a function with limit}+\infty\text{ at } p.\text{"}$$

The point $p$ is fixed, possibly $\pm \infty\left(\text {if } A \subseteq E^{*}\right).$ With this notation, we have the following theorems.

Theorems

1. $(\pm \infty)+(\pm \infty)=\pm \infty$.

2. $(\pm \infty)+q=q+(\pm \infty)=\pm \infty$.

3. $(\pm \infty) \cdot(\pm \infty)=+\infty$.

4. $(\pm \infty) \cdot(\mp \infty)=-\infty$.

5. $|\pm \infty|=+\infty$.

6. $(\pm \infty) \cdot q=q \cdot(\pm \infty)=\pm \infty$ if $q>0$.

7. $(\pm \infty) \cdot q=q \cdot(\pm \infty)=\mp \infty$ if $q<0$.

8. $-(\pm \infty)=\mp \infty$.

9. $\frac{(\pm \infty)}{q}=(\pm \infty) \cdot \frac{1}{q}$ if $q \neq 0$.

10. $\frac{q}{(\pm \infty)}=0$.

11. $(+\infty)^{+\infty}=+\infty$.

12. $(+\infty)^{-\infty}=0$.

13. $(+\infty)^{q}=+\infty$ if $q>0$.

14. $(+\infty)^{q}=0$ if $q<0$.

15. If $q>1,$ then $q^{+\infty}=+\infty$ and $q^{-\infty}=0$.

16. If $0<q<1,$ then $q^{+\infty}=0$ and $q^{-\infty}=+\infty$.

Proof

We prove Theorems 1 and 2, leaving the rest as problems. (Theorems 11-16 are best postponed until the theory of logarithms is developed.)

1. Let $f(x)$ and $g(x) \rightarrow+\infty$ as $x \rightarrow p.$ We have to show that

$$f(x)+g(x) \rightarrow+\infty,$$

i.e., that

$$\left(\forall b \in E^{1}\right)(\exists \delta>0)\left(\forall x \in A \cap G_{\neg p}(\delta)\right) \quad f(x)+g(x)>b$$

(we may assume $b>0 ).$ Thus fix $b>0.$ As $f(x)$ and $g(x) \rightarrow+\infty,$ there are $\delta^{\prime}, \delta^{\prime \prime}>0$ such that

$$\left(\forall x \in A \cap G_{\neg p}\left(\delta^{\prime}\right)\right) f(x)>b\text{ and } \left(\forall x \in A \cap G_{\neg p}\left(\delta^{\prime \prime}\right)\right) g(x)>b.$$

Let $\delta=\min \left(\delta^{\prime}, \delta^{\prime \prime}\right).$ Then

$$\left(\forall x \in A \cap G_{\neg p}(\delta)\right) \quad f(x)+g(x)>b+b>b,$$

as required; similarly for the case of $-\infty$.

2. Let $f(x) \rightarrow+\infty$ and $g(x) \rightarrow q \in E^{1}.$ Then there is $\delta^{\prime}>0$ such that for $x$ in $A \cap G_{\neg p}\left(\delta^{\prime}\right),|q-g(x)|<1,$ so that $g(x)>q-1$.

Also, given any $b \in E^{1},$ there is $\delta^{\prime \prime}$ such that

$$\left(\forall x \in A \cap G_{-p}\left(\delta^{\prime \prime}\right)\right) \quad f(x)>b-q+1.$$

Let $\delta=\min \left(\delta^{\prime}, \delta^{\prime \prime}\right).$ Then

$$\left(\forall x \in A \cap G_{\neg p}(\delta)\right) \quad f(x)+g(x)>(b-q+1)+(q-1)=b,$$

as required; similarly for the case of $f(x) \rightarrow-\infty$.

Caution: No theorems of this kind exist for the following cases (which therefore are called indeterminate expressions):

$$(+\infty)+(-\infty), \quad( \pm \infty) \cdot 0, \quad \frac{ \pm \infty}{ \pm \infty}, \quad \frac{0}{0}, \quad( \pm \infty)^{0}, \quad 0^{0}, \quad 1^{ \pm \infty}.$$

In these cases, it does not suffice to know only the limits of $f$ and $g.$ It is necessary to investigate the functions themselves to give a definite answer, since in each case the answer may be different, depending on the properties of $f$ and $g.$ The expressions (1*) remain indeterminate even if we consider the simplest kind of functions, namely sequences, as we show next.

Examples

(a) Let

$$u_{m}=2 m\text{ and } v_{m}=-m.$$

(This corresponds to $f(x)=2 x$ and $g(x)=-x.)$ Then, as is readily seen,

$$u_{m} \rightarrow+\infty, v_{m} \rightarrow-\infty,\text{ and } u_{m}+v_{m}=2 m-m=m \rightarrow+\infty.$$

If, however, we take $x_{m}=2 m$ and $y_{m}=-2 m,$ then

$$x_{m}+y_{m}=2 m-2 m=0;$$

thus $x_{m}+y_{m}$ is constant, with limit 0 (for the limit of a constant function equals its value; see §1, Example (a)).

Next, let

$$u_{m}=2 m\text{ and } z_{m}=-2 m+(-1)^{m}.$$

Then again

$$u_{m} \rightarrow+\infty\text{ and } z_{m} \rightarrow-\infty,\text{ but } u_{m}+z_{m}=(-1)^{m};$$

$u_{m}+z_{m}$ "oscillates" from $-1$ to 1 as $m \rightarrow+\infty,$ so it has no limit at all.

These examples show that $(+\infty)+(-\infty)$ is indeed an indeterminate expression since the answer depends on the nature of the functions involved. No general answer is possible.

(b) We now show that $1^{+\infty}$ is indeterminate.

Take first a constant $\left\{x_{m}\right\}, x_{m}=1,$ and let $y_{m}=m.$ Then

$$x_{m} \rightarrow 1, y_{m} \rightarrow+\infty,\text{ and } x_{m}^{y_{m}}=1^{m}=1=x_{m} \rightarrow 1.$$

If, however, $x_{m}=1+\frac{1}{m}$ and $y_{m}=m,$ then again $y_{m} \rightarrow+\infty$ and $x_{m} \rightarrow 1$ (by Theorem 10 above and Theorem 1 of Chapter 3, §15), but

$$x_{m}^{y_{m}}=\left(1+\frac{1}{m}\right)^{m}$$

does not tend to $1 ;$ it tends to $e>2,$ as shown in Chapter 3, §15. Thus again the result depends on $\left\{x_{m}\right\}$ and $\left\{y_{m}\right\}.$

In a similar manner, one shows that the other cases (1*) are indeterminate.

Note 1. It is often useful to introduce additional "shorthand" conventions. Thus the symbol $\infty$ (unsigned infinity) might denote a function$f$ such that

$$|f(x)| \rightarrow+\infty\text{ as } x \rightarrow p;$$

we then also write $f(x) \rightarrow \infty.$ The symbol $0^{+}$ (respectively, $0^{-})$ denotes a function $f$ such that

$$f(x) \rightarrow 0\text{ as } x \rightarrow p$$

and, moreover

$$f(x)>0\text{ } (f(x)<0,\text { respectively})\text{ on some } G_{\neg p}(\delta).$$

We then have the following additional formulas:

(i) $\frac{( \pm \infty)}{0^{+}}=\pm \infty, \frac{( \pm \infty)}{0^{-}}=\mp \infty$.

(ii) If $q>0,$ then $\frac{q}{0^{+}}=+\infty$ and $\frac{q}{0^{-}}=-\infty$.

(iii) $\frac{\infty}{0}=\infty$.

(iv) $\frac{q}{\infty}=0$.

The proof is left to the reader.

Note 2. All these formulas and theorems hold for relative limits, too.

So far, we have defined no arithmetic operations in $E^{*}.$ To fill this gap (at least partially), we shall henceforth treat Theorems 1-16 above not only as certain limit statements (in "shorthand") but also as definitions of certain operations in $E^{*}.$ For example, the formula $(+\infty)+(+\infty)=+\infty$ shall be treated as the definition of the actual sum of $+\infty$ and $+\infty$ in $E^{*},$ with $+\infty$ regarded this time as an element of $E^{*}$ (not as a function). This convention defines the arithmetic operations for certain cases only; the indeterminate expressions (1*) remain undefined, unless we decide to assign them some meaning.

In higher analysis, it indeed proves convenient to assign a meaning to at least some of them. We shall adopt these (admittedly arbitrary) conventions:

$\left\{\begin{array}{l}{( \pm \infty)+(\mp \infty)=( \pm \infty)-( \pm \infty)=+\infty ; 0^{0}=1;} \\ {0 \cdot( \pm \infty)=( \pm \infty) \cdot 0=0 \text { (even if } 0 \text { stands for the zero-vector } ).}\end{array}\right.$

Caution: These formulas must not be treated as limit theorems (in "short-hand"). Sums and products of the form (2*) will be called "unorthodox."