Indefinite expression (mathematics)

In mathematics, an indefinite expression is a term whose occurrence plays a special role in the investigation of limit values. The term must be distinguished from the undefined expression .

Problem description

Since division by zero is not defined, the term 1: 0 does not represent a number. If you compare with 1:  x , where x should be a very small (but positive) number, the result is a very large value. If x is negative , however, there is a corresponding negative value of large magnitude. It therefore makes sense to introduce the symbol ∞ so that one can at least make the statement of the amount . The calculation with real numbers extended by infinite elements is possible with minor restrictions ( see detailed extended real number ). On the other hand, some terms like 0: 0 cannot be assigned a number or the symbol ∞ even in such an extension. ${\displaystyle |{\tfrac {1}{0}}|=\infty }$

If one compares the term 0: 0 with x  :  y , where both x and y are small numbers, their quotient can have a very large amount, as above, but just as easily any other value. Even with the help of ∞ there is no suitable value for 0: 0, it is therefore an indefinite expression.

definition

Usually the term "indefinite expression" is used for:

${\displaystyle {\frac {0}{0}},\quad 0\cdot \infty ,\quad \infty -\infty ,\quad {\frac {\infty }{\infty }},\quad 0^{0},\quad \infty ^{0},\quad 1^{\infty }}$

These are precisely those expressions in which limit value statements about the expression do not result solely from the limit values ​​of the operands and different finite limit values ​​are possible even in the case of convergence.

Demarcation

Indefinite term does not mean the same as

undefined expression

Numerous other expressions are not defined - even in the area of affine extended real numbers - such as 1: 0 or . Conversely, it is quite common to define. ${\displaystyle (-1)^{\infty }}$ ${\displaystyle 0^{0}=1}$

Point of discontinuity or definition gap of the arithmetic operation that cannot be removed

Otherwise 1: 0 would also have to be counted among the indefinite expressions.

No indefinite expressions are (independent of existence or finiteness) limit values of concrete functions, like

${\displaystyle \lim _{x\to 0}{\frac {x^{2}}{x}}}$or . ${\displaystyle \lim _{x\to 0}x\cdot \cot x}$

The indefinite expression 0: 0 or 0 · ∞ results here by naive substitution. The limit value can be determined through more detailed examination with suitable methods such as the de l'Hospital rule . It applies

${\displaystyle \lim _{x\to 0}{\frac {x^{2}}{x}}=0}$ such as ${\displaystyle \lim _{x\to 0}x\cdot \cot x=1}$

and not about

${\displaystyle \lim _{x\to 0}{\frac {x^{2}}{x}}={\frac {0}{0}}}$or . ${\displaystyle \lim _{x\to 0}x\cdot \cot x=0\cdot \infty }$

Occurrence at impact limits

Are and two sequences of real numbers, so you can see the consequences , , and - if - define; as far as applies, for example , also . If the output sequences converge in the affinely extended real numbers, for example and , then for the linked sequences it usually also applies , where one of the basic arithmetic operations denotes the exponentiation . However, if there is any of the indefinite expressions listed above, then the limit behavior of is indefinite. In fact, one can (well sometimes) any sequence are given and then with , , be constructed as the following list shows. ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ ${\displaystyle (a_{n}+b_{n})_{n\in \mathbb {N} }}$ ${\displaystyle (a_{n}-b_{n})_{n\in \mathbb {N} }}$ ${\displaystyle (a_{n}\cdot b_{n})_{n\in \mathbb {N} }}$ ${\displaystyle b_{n}\neq 0}$ ${\displaystyle \left({\tfrac {a_{n}}{b_{n}}}\right)_{n\in \mathbb {N} }}$ ${\displaystyle a_{n}>0}$ ${\displaystyle \left(a_{n}^{b_{n}}\right)_{n\in \mathbb {N} }}$ ${\displaystyle a_{n}\to a}$ ${\displaystyle b_{n}\to b}$ ${\displaystyle a_{n}\circ b_{n}\to a\circ b}$ ${\displaystyle \circ }$ ${\displaystyle a\circ b}$ ${\displaystyle (a_{n}\circ b_{n})}$ ${\displaystyle (c_{n})_{n\in \mathbb {N} }}$ ${\displaystyle a_{n},b_{n}}$ ${\displaystyle a_{n}\circ b_{n}=c_{n}}$ ${\displaystyle a_{n}\to a}$ ${\displaystyle b_{n}\to b}$

• 0-0

  Put and . Then and , because of or . ${\displaystyle b_{n}={\tfrac {1}{n\cdot \max(1,|c_{n}|)}}}$ ${\displaystyle a_{n}=b_{n}\cdot c_{n}}$ ${\displaystyle {\tfrac {a_{n}}{b_{n}}}=c_{n}}$ ${\displaystyle a_{n}\to 0}$ ${\displaystyle b_{n}\to 0}$ ${\displaystyle |a_{n}|\leq {\tfrac {1}{n}}}$ ${\displaystyle |b_{n}|\leq {\tfrac {1}{n}}}$

• 0 · ∞

  Put and . Then and , because of or . ${\displaystyle b_{n}=n\cdot \max(1,|c_{n}|)}$ ${\displaystyle a_{n}={\tfrac {c_{n}}{b_{n}}}}$ ${\displaystyle {a_{n}}\cdot {b_{n}}=c_{n}}$ ${\displaystyle a_{n}\to 0}$ ${\displaystyle b_{n}\to +\infty }$ ${\displaystyle |a_{n}|\leq {\tfrac {1}{n}}}$ ${\displaystyle b_{n}\geq n}$

• ∞ - ∞

  Put and . Then and it is because of , because of , if , and if . ${\displaystyle b_{n}=\max(n,-2c_{n})}$ ${\displaystyle a_{n}=b_{n}+c_{n}}$ ${\displaystyle a_{n}-b_{n}=c_{n}}$ ${\displaystyle b_{n}\to +\infty }$ ${\displaystyle b_{n}\geq n}$ ${\displaystyle a_{n}\to +\infty }$ ${\displaystyle a_{n}\geq {\tfrac {n}{2}}}$ ${\displaystyle c_{n}\geq -{\tfrac {n}{2}}}$ ${\displaystyle a_{n}=-c_{n}>{\tfrac {n}{2}}}$ ${\displaystyle c_{n}<-{\tfrac {n}{2}}}$

• ∞: ∞

  It is assumed. Put and . Then , so , of course . ${\displaystyle c_{n}>0}$ ${\displaystyle a_{n}=n\cdot \max(1,c_{n})}$ ${\displaystyle b_{n}={\tfrac {a_{n}}{c_{n}}}}$ ${\displaystyle |a_{n}|\geq n}$ ${\displaystyle b_{n}\geq n}$ ${\displaystyle a_{n}\to +\infty }$ ${\displaystyle b_{n}\to \infty }$ ${\displaystyle c_{n}={\tfrac {a_{n}}{b_{n}}}}$

• 0 ⁰ , ∞ ⁰ , 1 ^∞

  It is assumed. Set and determine episodes , with , and as above . ${\displaystyle c_{n}>0}$ ${\displaystyle \gamma _{n}=\ln c_{n}}$ ${\displaystyle (\alpha _{n})}$ ${\displaystyle (\beta _{n})}$ ${\displaystyle \alpha _{n}\to 0}$ ${\displaystyle \beta _{n}\to +\infty }$ ${\displaystyle \alpha _{n}\beta _{n}=\gamma _{n}}$

  • With and one settles the case 0 ⁰ , ${\displaystyle a_{n}=e^{-\beta _{n}}}$ ${\displaystyle b_{n}=-\alpha _{n}}$
  • with and the case ∞ ⁰ , ${\displaystyle a_{n}=e^{\beta _{n}}}$ ${\displaystyle b_{n}=\alpha _{n}}$
  • with and the case 1 ^∞ ${\displaystyle a_{n}=e^{\alpha _{n}}}$ ${\displaystyle b_{n}=\alpha _{n}}$

Occurrence at functional limit values

The methods used above for sequences can easily be generalized to functions. In this way, for every real number (or also or ), every indefinite expression , every real function (possibly with the restriction ), two real functions are found and with for all as well as and . So here can take any finite or infinite value (if only non-negative) or do not exist. In other words: From the knowledge of and no inference can be drawn about if is an indefinite expression. On the other hand, for the basic arithmetic operations and exponentiation , it is absolutely true if it is a question of a defined and not indefinite expression (and is defined in a dotted environment of at all); If necessary, the calculation rules for how they apply to the extended real numbers must be observed. ${\displaystyle x_{0}}$ ${\displaystyle x_{0}=\infty }$ ${\displaystyle x_{0}=-\infty }$ ${\displaystyle a\circ b}$ ${\displaystyle h(x)}$ ${\displaystyle h(x)>0}$ ${\displaystyle f(x)}$ ${\displaystyle g(x)}$ ${\displaystyle h(x)=f(x)\circ g(x)}$ ${\displaystyle x\neq x_{0}}$ ${\displaystyle \lim _{x\to x_{0}}f(x)=a}$ ${\displaystyle \lim _{x\to x_{0}}g(x)=b}$ ${\displaystyle \lim _{x\to x_{0}}h(x)}$ ${\displaystyle \lim _{x\to x_{0}}f(x)=a}$ ${\displaystyle \lim _{x\to x_{0}}g(x)=b}$ ${\displaystyle \lim _{x\to x_{0}}f(x)\circ g(x)}$ ${\displaystyle a\circ b}$ ${\displaystyle \lim _{x\to x_{0}}f(x)\circ g(x)=a\circ b}$ ${\displaystyle f(x)\circ g(x)}$ ${\displaystyle x_{0}}$ ${\displaystyle \infty }$

If the functions and the more stringent requirements of de l'Hospital's rule, in particular with regard to differentiability, can be used to make a statement about the limit value sought . ${\displaystyle f(x)}$ ${\displaystyle g(x)}$ ${\displaystyle \lim _{x\to x_{0}}f(x)\circ g(x)}$

Overview

Be and be real functions and be a real number or one of the two symbolic values or . It is assumed that the limit values and either exist or that certain divergence is present, which is symbolically expressed or expressed as a limit value . In most cases, the following limit values ​​also exist with the specified values ​​(or certain divergence exists if it results on the right ): ${\displaystyle f}$ ${\displaystyle g}$ ${\displaystyle x_{0}}$ ${\displaystyle +\infty }$ ${\displaystyle -\infty }$ ${\displaystyle a:=\lim _{x\to x_{0}}{f(x)}}$ ${\displaystyle b:=\lim _{x\to x_{0}}{g(x)}}$ ${\displaystyle +\infty }$ ${\displaystyle -\infty }$ ${\displaystyle \pm \infty }$

• ${\displaystyle \lim _{x\to x_{0}}{(f(x)\pm g(x))}=a\pm b}$,
• ${\displaystyle \lim _{x\to x_{0}}{(f(x)\cdot g(x))}=a\cdot b}$,
• ${\displaystyle \lim _{x\to x_{0}}{\frac {f(x)}{g(x)}}={\frac {a}{b}}}$,
• ${\displaystyle \lim _{x\to x_{0}}{f(x)^{g(x)}}=a^{b}}$.

Here, the calculation rules for , for , for , for , for , for , for , for and corresponding sign variants are agreed. ${\displaystyle \infty +b=\infty }$ ${\displaystyle b\neq -\infty }$ ${\displaystyle \infty \cdot b=\infty }$ ${\displaystyle b>0}$ ${\displaystyle {\tfrac {\infty }{b}}=\infty }$ ${\displaystyle 0<b<\infty }$ ${\displaystyle {\tfrac {a}{\infty }}=0}$ ${\displaystyle a\neq \pm \infty }$ ${\displaystyle \infty ^{b}=\infty }$ ${\displaystyle b>0}$ ${\displaystyle \infty ^{b}=0}$ ${\displaystyle b<0}$ ${\displaystyle a^{\infty }=\infty }$ ${\displaystyle a>1}$ ${\displaystyle a^{\infty }=0}$ ${\displaystyle 0\leq a<1}$

However, the existence of the limit value on the left, let alone its value, does not result in this simple way from the limit values ​​of the operands if one of the indefinite expressions given above were to result on the right. In the following, example functions with corresponding limit values are listed, for which various limit values or divergence result: ${\displaystyle f(x),g(x)}$ ${\displaystyle a,b}$ ${\displaystyle c\in \mathbb {R} }$

• 0-0

  ${\displaystyle \lim _{x\to 0}{\frac {f(x)}{g(x)}}=c}$with , . ${\displaystyle f(x)=c\cdot x}$ ${\displaystyle g(x)=x}$

  ${\displaystyle \lim _{x\to 0}{\frac {f(x)}{g(x)}}=\pm \infty }$with , . ${\displaystyle f(x)=\pm x}$ ${\displaystyle g(x)=x^{3}}$

• ∞: ∞

  ${\displaystyle \lim _{x\to +\infty }{\frac {f(x)}{g(x)}}=c}$with , . ${\displaystyle f(x)=c\cdot x}$ ${\displaystyle g(x)=x}$

  ${\displaystyle \lim _{x\to +\infty }{\frac {f(x)}{g(x)}}=\pm \infty }$with , . ${\displaystyle f(x)=\pm x^{2}}$ ${\displaystyle g(x)=x}$

• 0 · ∞

  ${\displaystyle \lim _{x\to 0}{\bigl (}f(x)\cdot g(x){\bigr )}=c}$with , . ${\displaystyle f(x)=c\cdot x}$ ${\displaystyle g(x)={\tfrac {1}{x}}}$

  ${\displaystyle \lim _{x\to 0}{\bigl (}f(x)\cdot g(x){\bigr )}=\pm \infty }$with , . ${\displaystyle f(x)=\pm x}$ ${\displaystyle g(x)={\tfrac {1}{x^{2}}}}$

• ∞ - ∞

  ${\displaystyle \lim _{x\to +\infty }{\bigl (}f(x)-g(x){\bigr )}=c}$with , . ${\displaystyle f(x)=x+c}$ ${\displaystyle g(x)=x}$

  ${\displaystyle \lim _{x\to +\infty }{\bigl (}f(x)-g(x){\bigr )}=\pm \infty }$with , . ${\displaystyle f(x)=(3\pm 1)x}$ ${\displaystyle g(x)=2x}$

• 1  ^∞

  ${\displaystyle \lim _{x\to +\infty }{f(x)}^{g(x)}=c}$with , unless . ${\displaystyle f(x)=c^{\tfrac {1}{x}}}$ ${\displaystyle g(x)=x}$ ${\displaystyle c>0}$

  ${\displaystyle \lim _{x\to +\infty }{f(x)}^{g(x)}=+\infty }$with , . ${\displaystyle f(x)=x^{\tfrac {1}{x}}}$ ${\displaystyle g(x)=x}$

• 0  ⁰

  ${\displaystyle \lim _{x\to +\infty }{f(x)}^{g(x)}=c}$with , unless . ${\displaystyle f(x)=c^{x}}$ ${\displaystyle g(x)={\tfrac {1}{x}}}$ ${\displaystyle 0<c<1}$

• ∞  ⁰

  ${\displaystyle \lim _{x\to +\infty }{f(x)}^{g(x)}=c}$with , unless . ${\displaystyle f(x)=c^{x}}$ ${\displaystyle g(x)={\tfrac {1}{x}}}$ ${\displaystyle c>1}$

  ${\displaystyle \lim _{x\to +\infty }{f(x)}^{g(x)}=+\infty }$with , unless . ${\displaystyle f(x)=x^{x}}$ ${\displaystyle g(x)={\tfrac {1}{x}}}$ ${\displaystyle c>1}$

The various types of indefinite expressions can be reduced to type 1 through mathematical transformations. In the case of an indefinite expression of type 2, for example, the transformation results in an expression of type 1. ${\displaystyle {\frac {f(x)}{g(x)}}={\frac {\frac {1}{g(x)}}{\frac {1}{f(x)}}}}$

Expressions of types 5 to 7 can be reduced to type 1 by taking the logarithm .

In principle, the expression also does not allow a complete statement about the limit behavior, but in this case at least unlike in the cases listed above, no finite limit value can result, but at most a certain divergence to or . As an example, consider with for and optionally ${\displaystyle \infty :0}$ ${\displaystyle +\infty }$ ${\displaystyle -\infty }$ ${\displaystyle \lim _{x\to 0}{\frac {f(x)}{g(x)}}}$ ${\displaystyle f(x)={\tfrac {1}{x}}}$ ${\displaystyle x\neq 0}$

• ${\displaystyle g(x)=x}$: certain divergence after , ${\displaystyle +\infty }$
• ${\displaystyle g(x)=-x}$: certain divergence after , ${\displaystyle -\infty }$
• ${\displaystyle g(x)=|x|}$: different specific divergence on the left and on the right, i.e. overall indefinite divergence,
• ${\displaystyle g(x)=x\cdot \sin(x)}$: even one-sided there is indefinite divergence.

The expression 0 ⁰

→ Main article : Zero to zero

The expression , which in itself is well defined, namely as . Note that the exponentiation, i.e. the calculation of the expression , is initially only defined as repeated multiplication, which consequently has to be a non-negative integer. Then the empty product is, which - independently of - is defined as 1: It should apply, which results at least for mandatory . The empty product has no factors, and to that extent it does not matter what value the factor that does not appear at all has, so that it also results. The definition also makes sense for other reasons. For example, if both are nonnegative integers, there are always exactly mappings from a -element set to a -element set. This also applies in the case only with the definition . ${\displaystyle 0^{0}}$ ${\displaystyle 0^{0}=1}$ ${\displaystyle a^{b}}$ ${\displaystyle b}$ ${\displaystyle a^{0}}$ ${\displaystyle a}$ ${\displaystyle a^{1}=a\cdot a^{0}}$ ${\displaystyle a\neq 0}$ ${\displaystyle a^{0}=1}$ ${\displaystyle a}$ ${\displaystyle 0^{0}=1}$ ${\displaystyle 0^{0}=1}$ ${\displaystyle a,b}$ ${\displaystyle a^{b}}$ ${\displaystyle b}$ ${\displaystyle a}$ ${\displaystyle 0^{0}=1}$ ${\displaystyle a=b=0}$

The operation of exponentiation, defined in this way as a mapping of after , can also be continued in the real by to the case , and for nonnegative by extracting the roots initially to nonnegative rational exponents and then by considering the limit value . The latter is by definition continuous in , but the exponentiation as a mapping from to is not continuous at that point : For example , but . The above-mentioned indeterminacy in connection with limit values ​​results from this discontinuity. ${\displaystyle \mathbb {R} \times \mathbb {N} _{0}}$ ${\displaystyle \mathbb {R} }$ ${\displaystyle \textstyle a^{-b}:={\frac {1}{a^{b}}}}$ ${\displaystyle a\neq 0}$ ${\displaystyle b\in \mathbb {Z} }$ ${\displaystyle a}$ ${\displaystyle b\in [0,\infty )}$ ${\displaystyle b}$ ${\displaystyle \textstyle \left((\mathbb {R} \setminus \{0\})\times \mathbb {Z} \right)\cup \left([0,\infty )\times [0,\infty )\right)}$ ${\displaystyle \mathbb {R} }$ ${\displaystyle (0,0)}$ ${\displaystyle \textstyle \lim _{x\to 0^{+}}0^{x}=\lim _{x\to 0^{+}}0=0}$ ${\displaystyle \textstyle \lim _{x\to 0^{+}}x^{0}=\lim _{x\to 0^{+}}1=1}$

Web links

• Eric Weisstein: Indeterminates . In: MathWorld (English).

Individual evidence

1. ↑ Augustin-Louis Cauchy , Cours d'Analyse de l'École Royale Polytechnique (1821). Oeuvres Complètes, part 2, volume 3, page 70.
2. ↑ Eric Weisstein: Indeterminate . In: MathWorld (English).