In calculus and other branches of mathematical analysis, when the limit of the sum, difference, product, quotient or power of two functions is taken, it may often be possible to simply add, subtract, multiply, divide or exponentiate the corresponding limits of these two functions respectively. However, there are occasions where it is unclear what the sum, difference, product, quotient, or power of these two limits ought to be. For example, it is unclear what the following expressions ought to evaluate to:^[1]

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

These seven expressions are known as indeterminate forms. More specifically, such expressions are obtained by naively applying the algebraic limit theorem to evaluate the limit of the corresponding arithmetic operation of two functions, yet there are examples of pairs of functions that after being operated on converge to 0, converge to another finite value, diverge to infinity or just diverge. This inability to decide what the limit ought to be explains why these forms are regarded as indeterminate. A limit confirmed to be infinity is not indeterminate since it has been determined to have a specific value (infinity).^[1] 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 ${\displaystyle 0/0}$. 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 indeterminate. In this sense, ${\displaystyle 0/0}$ can take on the values ${\displaystyle 0~}$, ${\displaystyle 1}$, or ${\displaystyle \infty }$, 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, ${\displaystyle x\sin(1/x)/x}$.

So the fact that two functions ${\displaystyle f(x)}$ and ${\displaystyle g(x)}$ converge to ${\displaystyle 0~}$ as ${\displaystyle x}$ approaches some limit point ${\displaystyle c}$ is insufficient to determinate the limit

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

An 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. An 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

• 
  Fig. 1: y = x/x
• 
  Fig. 2: y = x²/x
• 
  Fig. 3: y = sin x/x
• 
  Fig. 4: y = x − 49/√x − 7 (for x = 49)
• 
  Fig. 5: y = ax/x where a = 2
• 
  Fig. 6: y = x/x³

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

• 
  Fig. 7: y = x⁰
• 
  Fig. 8: y = 0^x

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 ${\displaystyle 1}$.^[2] 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 if the limit of ${\displaystyle f/g}$ exists then there is no ambiguity as to its value, as it always diverges. 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.

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):^[4]

${\displaystyle x\sim \sin x,}$

${\displaystyle x\sim \arcsin x,}$

${\displaystyle x\sim \sinh x,}$

${\displaystyle x\sim \tan x,}$

${\displaystyle x\sim \arctan x,}$

${\displaystyle x\sim \ln(1+x),}$

${\displaystyle 1-\cos x\sim {\frac {x^{2)){2)),}$

${\displaystyle \cosh x-1\sim {\frac {x^{2)){2)),}$

${\displaystyle a^{x}-1\sim x\ln a,}$

${\displaystyle e^{x}-1\sim x,}$

${\displaystyle (1+x)^{a}-1\sim ax.}$

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