Indeterminate form

• 
  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)

• 
  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 ${\displaystyle 1}$.[4] Otherwise, use the transformation in the table below to evaluate the limit.

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 ${\displaystyle L={e}^{-\infty }=0.}$

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 2^nd 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 4^th equality, and ${\displaystyle 1-\cos x\sim {x^{2} \over 2}}$ is used in the 5^th equality.

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 ${\displaystyle (1/g)/(1/f)}$.