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.
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.
(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 ."