Indeterminate form: Difference between revisions

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In [[calculus]], it is usually possible to compute the [[limit (mathematics)|limit]] of the sum, difference, product, quotient or power of two functions by taking the corresponding combination of the separate limits of each respective function. For example,

:<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),

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and likewise for other arithmetic operations; this is sometimes called the [[limit of a function#Properties|algebraic limit theorem]]. However, certain combinations of particular limiting values cannot be computed in this way, and knowing the limit of each function separately does not suffice to determine the limit of the combination. In these particular situations, the limit is said to take an '''indeterminate form''', described by one of the informal expressions

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

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|Rigdon|Purcell|2007|p=423, 429, 430, 431, 432}}

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|Rigdon|Purcell|Rigdon|2007|p=423, 429, 430, 431, 432}}

A limit taking one of these indeterminate forms might tend to zero, might tend to any finite value, might tend to infinity, or might diverge, depending on the specific functions involved. A limit which unambiguously tends to infinity, for instance <math display=inline>\lim_{x \to 0} 1/x^2 = \infty,</math> is not considered 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.

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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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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