# Indeterminate form

In calculus and other branches of mathematical analysis, limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits; if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. More specifically, an indeterminate form is a mathematical expression involving ${\displaystyle 0}$, ${\displaystyle 1}$ and ${\displaystyle \infty }$, obtained by applying the algebraic limit theorem in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity (if a limit is confirmed as infinity, then it is not indeterminate since the limit is determined as infinity) and thus does not yet determine the limit being sought.[1][2] The term was originally introduced by Cauchy's student Moigno in the middle of the 19th century.

There are seven indeterminate forms which are typically considered in the literature:[2]

${\displaystyle {\frac {0}{0}},~{\frac {\infty }{\infty }},~0\times \infty ,~\infty -\infty ,~0^{0},~1^{\infty },{\text{ and }}\infty ^{0}.}$

The most common example of an indeterminate form occurs when determining the limit of the ratio of two functions, in which both of these functions tend to zero in the limit, and is referred to as "the indeterminate form ${\displaystyle 0/0}$". For example, as ${\displaystyle x}$ approaches ${\displaystyle 0}$, the ratios ${\displaystyle x/x^{3}}$, ${\displaystyle x/x}$, and ${\displaystyle x^{2}/x}$ go to ${\displaystyle \infty }$, ${\displaystyle 1}$, and ${\displaystyle 0}$ respectively. In each case, if the limits of the numerator and denominator are substituted, the resulting expression is ${\displaystyle 0/0}$, which is undefined. In a loose manner of speaking, ${\displaystyle 0/0}$ can take on the values ${\displaystyle 0}$, ${\displaystyle 1}$, or ${\displaystyle \infty }$, and it is easy to construct similar examples for which the limit is any particular value.

So, given that two functions ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$ both approaching ${\displaystyle 0}$ as ${\displaystyle x}$ approaches some limit point ${\displaystyle c}$, that fact alone does not give enough information for evaluating the limit

${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}.}$

Not every undefined algebraic expression corresponds to an indeterminate form. For example, the expression ${\displaystyle 1/0}$ is undefined as a real number but does not correspond to an indeterminate form, because any limit that gives rise to this form will diverge to infinity if the denominator gets closer to 0 but never be 0.[3]

A expression that arises by ways other than applying the algebraic limit theorem may have the same form of an indeterminate form. However it is not appropriate to call an expression "indeterminate form" if the expression is made outside the context of determining limits. For example, ${\displaystyle 0/0}$ which arise from substituting ${\displaystyle 0}$ for ${\displaystyle x}$ in the equation ${\displaystyle f(x)=|x|/(|x-1|-1)}$ is not an indeterminate form since this expression is not made in the determination of a limit (it is in fact undefined as division by zero). Another example is the expression ${\displaystyle 0^{0}}$. Whether this expression is left undefined, or is defined to equal ${\displaystyle 1}$, depends on the field of application and may vary between authors. For more, see the article Zero to the power of zero. Note that ${\displaystyle 0^{\infty }}$ and other expressions involving infinity are not indeterminate forms.

## Some examples and non-examples

### Indeterminate form 0/0

The indeterminate form ${\displaystyle 0/0}$ is particularly common in calculus, because it often arises in the evaluation of derivatives using their definition in terms of limit.

As mentioned above,

${\displaystyle \lim _{x\to 0}{\frac {x}{x}}=1,\qquad }$ (see fig. 1)

while

${\displaystyle \lim _{x\to 0}{\frac {x^{2}}{x}}=0,\qquad }$ (see fig. 2)

This is enough to show that ${\displaystyle 0/0}$ is an indeterminate form. Other examples with this indeterminate form include

${\displaystyle \lim _{x\to 0}{\frac {\sin(x)}{x}}=1,\qquad }$ (see fig. 3)

and

${\displaystyle \lim _{x\to 49}{\frac {x-49}{{\sqrt {x}}\,-7}}=14,\qquad }$ (see fig. 4)

Direct substitution of the number that ${\displaystyle x}$ approaches into any of these expressions shows that these are examples correspond to the indeterminate form ${\displaystyle 0/0}$, but these limits can assume many different values. Any desired value ${\displaystyle a}$ can be obtained for this indeterminate form as follows:

${\displaystyle \lim _{x\to 0}{\frac {ax}{x}}=a.\qquad }$ (see fig. 5)

The value ${\displaystyle \infty }$ can also be obtained (in the sense of divergence to infinity):

${\displaystyle \lim _{x\to 0}{\frac {x}{x^{3}}}=\infty .\qquad }$ (see fig. 6)

### Indeterminate form 00

The following limits illustrate that the expression ${\displaystyle 0^{0}}$ is an indeterminate form:

${\displaystyle \lim _{x\to 0^{+}}x^{0}=1,\qquad }$ (see fig. 7)
${\displaystyle \lim _{x\to 0^{+}}0^{x}=0.\qquad }$ (see fig. 8)

Thus, in general, knowing that ${\displaystyle \textstyle \lim _{x\to c}f(x)\;=\;0\!}$ and ${\displaystyle \textstyle \lim _{x\to c}g(x)\;=\;0}$ is not sufficient to evaluate the limit

${\displaystyle \lim _{x\to c}f(x)^{g(x)}.}$

If the functions ${\displaystyle f}$ and ${\displaystyle g}$ are analytic at ${\displaystyle c}$, and ${\displaystyle f}$ is positive for ${\displaystyle x}$ sufficiently close (but not equal) to ${\displaystyle c}$, then the limit of ${\displaystyle f(x)^{g(x)}}$ will be ${\displaystyle 1}$.[4] Otherwise, use the transformation in the table below to evaluate the limit.

### Expressions that are not indeterminate forms

The expression ${\displaystyle 1/0}$ is not commonly regarded as an indeterminate form, because there is not an infinite range of values that ${\displaystyle f/g}$ could approach. Specifically, if ${\displaystyle f}$ approaches ${\displaystyle 1}$ and ${\displaystyle g}$ approaches ${\displaystyle 0}$, then ${\displaystyle f}$ and ${\displaystyle g}$ may be chosen so that:

1. ${\displaystyle f/g}$ approaches ${\displaystyle +\infty }$
2. ${\displaystyle f/g}$ approaches ${\displaystyle -\infty }$
3. The limit fails to exist.

In each case the absolute value ${\displaystyle |f/g|}$ approaches ${\displaystyle +\infty }$, and so the quotient ${\displaystyle f/g}$ must diverge, in the sense of the extended real numbers (in the framework of the projectively extended real line, the limit is the unsigned infinity ${\displaystyle \infty }$ in all three cases[3]). Similarly, any expression of the form ${\displaystyle a/0}$ with ${\displaystyle a\neq 0}$ (including ${\displaystyle a=+\infty }$ and ${\displaystyle a=-\infty }$) is not an indeterminate form, since a quotient giving rise to such an expression will always diverge.

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

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

## Evaluating indeterminate forms

The adjective indeterminate does not imply that the limit does not exist, as many of the examples above show. In many cases, algebraic elimination, L'Hôpital's rule, or other methods can be used to manipulate the expression so that the limit can be evaluated.[1]

### Equivalent infinitesimal

When two variables ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ converge to zero at the same limit point and ${\displaystyle \textstyle \lim {\frac {\beta }{\alpha }}=1}$, they are called equivalent infinitesimal (equiv. ${\displaystyle \alpha \sim \beta }$).

Moreover, if variables ${\displaystyle \alpha '}$ and ${\displaystyle \beta '}$ are such that ${\displaystyle \alpha \sim \alpha '}$ and ${\displaystyle \beta \sim \beta '}$, then:

${\displaystyle \lim {\frac {\beta }{\alpha }}=\lim {\frac {\beta '}{\alpha '}}}$

Here is a brief proof:

Suppose there are two equivalent infinitesimals ${\displaystyle \alpha \sim \alpha '}$ and ${\displaystyle \beta \sim \beta '}$.

${\displaystyle \lim {\frac {\beta }{\alpha }}=\lim {\frac {\beta \beta '\alpha '}{\beta '\alpha '\alpha }}=\lim {\frac {\beta }{\beta '}}\lim {\frac {\alpha '}{\alpha }}\lim {\frac {\beta '}{\alpha '}}=\lim {\frac {\beta '}{\alpha '}}}$

For the evaluation of the indeterminate form ${\displaystyle 0/0}$, one can make use of the following facts about equivalent infinitesimals (e.g., ${\displaystyle x\sim \sin x}$ if x becomes closer to zero):[5]

${\displaystyle x\sim \sin x,}$
${\displaystyle x\sim \arcsin x,}$
${\displaystyle x\sim \sinh x,}$
${\displaystyle x\sim \tan x,}$
${\displaystyle x\sim \arctan x,}$
${\displaystyle x\sim \ln(1+x),}$
${\displaystyle 1-\cos x\sim {\frac {x^{2}}{2}},}$
${\displaystyle \cosh x-1\sim {\frac {x^{2}}{2}},}$
${\displaystyle a^{x}-1\sim x\ln a,}$
${\displaystyle e^{x}-1\sim x,}$
${\displaystyle (1+x)^{a}-1\sim ax.}$

For example:

{\displaystyle {\begin{aligned}\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{aligned}}}

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

### L'Hôpital's rule

L'Hôpital's rule is a general method for evaluating the indeterminate forms ${\displaystyle 0/0}$ and ${\displaystyle \infty /\infty }$. This rule states that (under appropriate conditions)

${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}},}$

where ${\displaystyle f'}$ and ${\displaystyle g'}$ are the derivatives of ${\displaystyle f}$ and ${\displaystyle g}$. (Note that this rule does not apply to expressions ${\displaystyle \infty /0}$, ${\displaystyle 1/0}$, and so on, as these expressions are not indeterminate forms.) These derivatives will allow one to perform algebraic simplification and eventually evaluate the limit.

L'Hôpital's rule can also be applied to other indeterminate forms, using first an appropriate algebraic transformation. For example, to evaluate the form 00:

${\displaystyle \ln \lim _{x\to c}f(x)^{g(x)}=\lim _{x\to c}{\frac {\ln f(x)}{1/g(x)}}.}$

The right-hand side is of the form ${\displaystyle \infty /\infty }$, so L'Hôpital's rule applies to it. Note that this equation is valid (as long as the right-hand side is defined) because the natural logarithm (ln) is a continuous function; it is irrelevant how well-behaved ${\displaystyle f}$ and ${\displaystyle g}$ may (or may not) be as long as ${\displaystyle f}$ is asymptotically positive. (the domain of logarithms is the set of all positive real numbers.)

Although L'Hôpital's rule applies to both ${\displaystyle 0/0}$ and ${\displaystyle \infty /\infty }$, one of these forms may be more useful than the other in a particular case (because of the possibility of algebraic simplification afterwards). One can change between these forms, if necessary, by transforming ${\displaystyle f/g}$ to ${\displaystyle (1/g)/(1/f)}$.

## List of indeterminate forms

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

Indeterminate form Conditions Transformation to ${\displaystyle 0/0}$ Transformation to ${\displaystyle \infty /\infty }$
0/0 ${\displaystyle \lim _{x\to c}f(x)=0,\ \lim _{x\to c}g(x)=0\!}$
${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {1/g(x)}{1/f(x)}}\!}$
${\displaystyle \infty }$/${\displaystyle \infty }$ ${\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=\infty \!}$ ${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {1/g(x)}{1/f(x)}}\!}$
${\displaystyle 0\cdot \infty }$ ${\displaystyle \lim _{x\to c}f(x)=0,\ \lim _{x\to c}g(x)=\infty \!}$ ${\displaystyle \lim _{x\to c}f(x)g(x)=\lim _{x\to c}{\frac {f(x)}{1/g(x)}}\!}$ ${\displaystyle \lim _{x\to c}f(x)g(x)=\lim _{x\to c}{\frac {g(x)}{1/f(x)}}\!}$
${\displaystyle \infty -\infty }$ ${\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=\infty \!}$ ${\displaystyle \lim _{x\to c}(f(x)-g(x))=\lim _{x\to c}{\frac {1/g(x)-1/f(x)}{1/(f(x)g(x))}}\!}$ ${\displaystyle \lim _{x\to c}(f(x)-g(x))=\ln \lim _{x\to c}{\frac {e^{f(x)}}{e^{g(x)}}}\!}$
${\displaystyle 0^{0}}$ ${\displaystyle \lim _{x\to c}f(x)=0^{+},\lim _{x\to c}g(x)=0\!}$ ${\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {g(x)}{1/\ln f(x)}}\!}$ ${\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {\ln f(x)}{1/g(x)}}\!}$
${\displaystyle 1^{\infty }}$ ${\displaystyle \lim _{x\to c}f(x)=1,\ \lim _{x\to c}g(x)=\infty \!}$ ${\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {\ln f(x)}{1/g(x)}}\!}$ ${\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {g(x)}{1/\ln f(x)}}\!}$
${\displaystyle \infty ^{0}}$ ${\displaystyle \lim _{x\to c}f(x)=\infty ,\ \lim _{x\to c}g(x)=0\!}$ ${\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {g(x)}{1/\ln f(x)}}\!}$ ${\displaystyle \lim _{x\to c}f(x)^{g(x)}=\exp \lim _{x\to c}{\frac {\ln f(x)}{1/g(x)}}\!}$