Indeterminate form: Difference between revisions

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{{Short description|Expression in mathematical analysis}}

In [[calculus]], andit otheris branchesusually ofpossible [[mathematical analysis]],to whencompute the [[limit (mathematics)|limit]] of the sum, difference, product, quotient or power of two functions isby taken,taking itthe maycorresponding oftencombination be possible to simply add, subtract, multiply, divide or exponentiateof the correspondingseparate limits of theseeach tworespective functions respectivelyfunction. ThatFor isexample, in most circumstances

:<math display=block>\begin{align}

\lim_{x \to c} \bigl(f(x) + g(x)\bigr) &= \lim_{x \to c} f(x) + \lim_{x \to c} g(x), \\[3mu]

\lim_{x \to c} \bigl(f(x)g(x)\bigr) &= \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x),

\end{align}</math>

and likewise for other arithmetic operations; this is sometimes called the [[limit of a function#Properties|algebraic limit theorem]]. However, therecertain arecombinations occasionsof whereparticular itlimiting isvalues unclearcannot whatbe thecomputed sum,in difference,this productway, quotient,and orknowing powerthe limit of theseeach twofunction limitsseparately oughtdoes not suffice to bedetermine the limit of the combination. In these particular situations, itthe limit is unclearsaid whatto thetake followingan expressions'''indeterminate oughtform''', todescribed evaluateby to,<refone name=":1"of />the informal expressions

:<math display=block>\frac 00,~ \frac{\infty}{\infty},~ 0\times\infty,~ \infty - \infty,~ 0^0,~ 1^\infty, \text{ andor } \infty^0,</math>

among a wide variety of uncommon others, where each expression stands for the limit of a function constructed by an arithmetical combination of two functions whose limits respectively tend to {{tmath|0,}} {{tmath|1,}} or {{tmath|\infty}} as indicated.{{sfnp|Varberg|Purcell|Rigdon|2007|p=423, 429, 430, 431, 432}}

TheseA sevenlimit expressionstaking areone knownof asthese '''indeterminate forms'''. Moremight specifically, such expressions are obtained by naively applying the [[algebraic limit theorem]]tend to evaluate the limit of the corresponding arithmetic operation of two functionszero, yetmight there are examples of pairs of functions that after being operated on convergetend to 0, converge to anotherany finite value, divergemight tend to infinity, or justmight diverge., Thisdepending inability to decide whaton the limitspecific oughtfunctions to be explains why these forms are regarded as '''indeterminate'''involved. A limit confirmedwhich unambiguously tends to be infinity, isfor notinstance indeterminate<math sincedisplay=inline>\lim_{x it\to has0} been1/x^2 determined= to\infty,</math> haveis anot specificconsidered value (infinity)indeterminate.<ref name=":1">{{Cite web|url=http://mathworld.wolfram.com/Indeterminate.html|title=Indeterminate|last=Weisstein|first=Eric W.|website=mathworld.wolfram.com|language=en|access-date=2019-12-02}}</ref> The term was originally introduced by [[Cauchy]]'s student [[Moigno]] in the middle of the 19th century.

The most common example of an indeterminate form is the quotient of two functions each of which converges to zero. This indeterminate form is denoted by <math>0/0</math>. For example, as <math>x</math> approaches <math>0~,</math>, the ratios <math>x/x^3</math>, <math>x/x</math>, and <math>x^2/x</math> go to <math>\infty</math>, <math>1</math>, and <math>0~</math> respectively. In each case, if the limits of the numerator and denominator are substituted, the resulting expression is <math>0/0</math>, which is indeterminate. In this sense, <math>0/0</math> can take on the values <math>0~</math>, <math>1</math>, or <math>\infty</math>, by appropriate choices of functions to put in the numerator and denominator. A pair of functions for which the limit is any particular given value may in fact be found. Even more surprising, perhaps, the quotient of the two functions may in fact diverge, and not merely diverge to infinity. For example, <math> x \sin(1/x) / x</math>.

So the fact that two [[function (mathematics)|functions]] <math>f(x)</math> and <math>g(x)</math> converge to <math>0~</math> as <math>x</math> approaches some [[limit point]] <math>c</math> is insufficient to determinate the [[limit of a function|limit]]

{{block indent|<math> \lim_{x \to c} \frac{f(x)}{g(x)} .</math>}}

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== Some examples and non-examples ==

=== Indeterminate form 0/0 ===

{{Redirect|0/0|the symbol|Percent sign|0 divided by 0|Division by zero}}

<gallery>

File:Indeterminate form - x over x.gif|Fig. 1: {{var|y}} = {{sfrac|{{var|x}}|{{var|x}}}}

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===Indeterminate form 0⁰ ===

{{main|Zero to the power of zero}}

{{multiple image

| image1 = Indeterminate form - x0.gif

<gallery>

File:Indeterminate | formcaption1 -= x0.gif|Fig.Graph 7:of {{varmath|1=''y}}'' = {{var|''x}}''{{sup|0}}}}

File: | image2 = Indeterminate form - 0x.gif|Fig. 8: {{var|y}} = 0{{sup|{{var|x}}}}

| caption2 = Graph of {{math|1=''y'' = 0{{sup|''x''}}}}

</gallery>

| total_width = 300

| direction = vertical

}}

The following limits illustrate that the expression <math>0^0</math> is an indeterminate form:

<math display="block"> \begin{align}

{{block indent|<math> \lim_{x \to 0^+} x^0 &= 1 , \qquad </math> (see fig. 7)}}\

{{block indent|<math> \lim_{x \to 0^+} 0^x &= 0 . \qquad </math> (see fig. 8)}}

\end{align} </math>

Thus, in general, knowing that <math>\textstyle\lim_{x \to c} f(x) \;=\; 0</math> and <math>\textstyle\lim_{x \to c} g(x) \;=\; 0</math> is not sufficient to evaluate the limit

{{block indent|<math display="block"> \lim_{x \to c} f(x)^{g(x)}. .</math>}}

{{block indent|<math>\lim_{x \to c} f(x)^{g(x)} .</math>}}

If the functions <math>f</math> and <math>g</math> are [[Analytic function|analytic]] at <math>c</math>, and <math>f</math> is positive for <math>x</math> sufficiently close (but not equal) to <math>c</math>, then the limit of <math>f(x)^{g(x)}</math> will be <math>1</math>.<ref>{{cite journal |doi=10.2307/2689754 |author1=Louis M. Rotando |author2=Henry Korn |title=The indeterminate form 0⁰ |journal=Mathematics Magazine |date=January 1977 |volume=50 |issue=1 |pages=41&ndash;42|jstor=2689754 }}</ref> Otherwise, use the transformation in the [[#List of indeterminate forms|table]] below to evaluate the limit.

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=== Expressions that are not indeterminate forms ===

The expression <math>1/0</math> is not commonly regarded as an indeterminate form, because if the limit of <math>f/g</math> exists then there is no ambiguity as to its value, as it always diverges. Specifically, if <math>f</math> approaches <math>1</math> and <math>g</math> approaches <math>0~,</math>, then <math>f</math> and <math>g</math> may be chosen so that:

# <math>f/g</math> approaches <math>+\infty</math>

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In each case the absolute value <math>|f/g|</math> approaches <math>+\infty</math>, and so the quotient <math>f/g</math> must diverge, in the sense of the [[extended real number]]s (in the framework of the [[projectively extended real line]], the limit is the [[Point at infinity|unsigned infinity]] <math>\infty</math> in all three cases<ref name=":3">{{Cite web|url=https://www.cut-the-knot.org/blue/GhostCity.shtml|title=Undefined vs Indeterminate in Mathematics|website=www.cut-the-knot.org|access-date=2019-12-02}}</ref>). Similarly, any expression of the form <math>a/0</math> with <math>a\ne0</math> (including <math>a=+\infty</math> and <math>a=-\infty</math>) is not an indeterminate form, since a quotient giving rise to such an expression will always diverge.

The expression <math>0^\infty</math> is not an indeterminate form. The expression <math>0^{+\infty}</math> obtained from considering <math>\lim_{x \to c} f(x)^{g(x)}</math> gives the limit <math>0~,</math>, provided that <math>f(x)</math> remains nonnegative as <math>x</math> approaches <math>c</math>. The expression <math>0^{-\infty}</math> is similarly equivalent to <math>1/0</math>; if <math>f(x) > 0</math> as <math>x</math> approaches <math>c</math>, the limit comes out as <math>+\infty</math>.

To see why, let <math>L = \lim_{x \to c} f(x)^{g(x)},</math> where <math> \lim_{x \to c} {f(x)}=0,</math> and <math> \lim_{x \to c} {g(x)}=\infty.</math> By taking the natural logarithm of both sides and using <math> \lim_{x \to c} \ln{f(x)}=-\infty,</math> we get that <math>\ln L = \lim_{x \to c} ({g(x)}\times\ln{f(x)})=\infty\times{-\infty}=-\infty,</math> which means that <math>L = {e}^{-\infty}=0.</math>

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Suppose there are two equivalent infinitesimals <math>\alpha \sim \alpha'</math> and <math>\beta \sim \beta'</math>.

:<math display=block>\lim \frac{\beta}{\alpha} = \lim \frac{\beta \beta' \alpha'}{\beta' \alpha' \alpha} = \lim \frac{\beta}{\beta'} \lim \frac{\alpha'}{\alpha} \lim \frac{\beta'}{\alpha'} = \lim \frac{\beta'}{\alpha'}</math>

For the evaluation of the indeterminate form <math>0/0</math>, one can make use of the following facts about equivalent [[infinitesimal]]s (e.g., <math>x\sim\sin x</math> if ''x'' becomes closer to zero):<ref>{{Cite web|url=http://www.vaxasoftware.com/doc_eduen/mat/infiequi.pdf|title=Table of equivalent infinitesimals|website=Vaxa Software}}</ref>

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{{block indent|<math>e^x - 1\sim x,</math>}}

{{block indent|<math>(1 + x)^a - 1 \sim ax.</math>}}

For example:

:<math>\begin{align}\lim_{x \to 0} \frac{1}{x^3} \left[\left(\frac{2+\cos x}{3}\right)^x - 1 \right] & = \lim_{x \to 0} \frac{e^{x\ln{\frac{2 + \cos x}{3}}}-1}{x^3} \\ & = \lim_{x \to 0} \frac{1}{x^2} \ln \frac{2+ \cos x}{3} \\ & = \lim_{x \to 0} \frac{1}{x^2} \ln \left(\frac{\cos x -1}{3}+1\right) \\ &= \lim_{x \to 0} \frac{\cos x -1}{3x^2} \\ &= \lim_{x \to 0} -\frac{x^2}{6x^2} \\ & = -\frac{1}{6}\end{align}</math>

<math display=block>\begin{align}

\lim_{x \to 0} \frac{1}{x^3} \left[\left(\frac{2+\cos x}{3}\right)^x - 1 \right]

&= \lim_{x \to 0} \frac{e^{x\ln{\frac{2 + \cos x}{3}}}-1}{x^3} \\

&= \lim_{x \to 0} \frac{1}{x^2} \ln \frac{2+ \cos x}{3} \\

&= \lim_{x \to 0} \frac{1}{x^2} \ln \left(\frac{\cos x -1}{3}+1\right) \\

&= \lim_{x \to 0} \frac{\cos x -1}{3x^2} \\

&= \lim_{x \to 0} -\frac{x^2}{6x^2} \\

&= -\frac{1}{6}

\end{align}</math>

In the 2nd equality, <math>e^y - 1 \sim y</math> where <math>y = x\ln{2+\cos x \over 3}</math> as ''y'' become closer to 0 is used, and <math>y \sim \ln {(1+y)}</math> where <math>y = {{\cos x - 1} \over 3}</math> is used in the 4th equality, and <math>1-\cos x \sim {x^2 \over 2}</math> is used in the 5th equality.

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The following table lists the most common indeterminate forms and the transformations for applying l'Hôpital's rule.

{| border=1 class="wikitable" style="background-color: #ffffff; width: 85%;"

!Indeterminate form

!Conditions

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== References ==

=== Citations ===

{{reflist}}

=== Bibliographies ===

* {{cite book

| last1 = Varberg | first1 = Dale E.

| last2 = Purcell | first2 = Edwin J.

| last3 = Rigdon | first3 = Steven E.

| title = Calculus

| year = 2007

| publisher = [[Pearson Prentice Hall]]

| edition = 9th

| isbn = 978-0131469686

}}

{{Calculus topics}}