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# 4.4: Infinite Limits. Operations in E*

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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."