Indeterminate form

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In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form. The indeterminate forms include 0⁰, 0/0, 1^∞, ∞ - ∞, ∞/∞, 0×∞, and ∞⁰.

1 Discussion
2 Some examples and nonexamples
3 Evaluating indeterminate forms
4 List of indeterminate forms
5 See also
6 References

[edit] Discussion

The most common example of an indeterminate form is 0/0. As x approaches 0, the ratios x²/x, x/x, and x/x³ go to 0, 1, and $\scriptstyle\infty$ correspondingly. In each case, however, if the limits of the numerator and denominator are evaluated and plugged into the division operation, the resulting expression is 0/0. So (roughly speaking) 0/0 can be 0 or it can be $\scriptstyle\infty$ and, in fact, it is possible to construct similar examples converging to any particular value. That is why the expression 0/0 is indeterminate.

More formally, the fact that the functions f and g both approach 0 as x approaches some limit point c is not enough information to evaluate the limit

$\lim_{x \to c} \frac{f(x)}{g(x)}. \!$

That limit could converge to any number, or diverge to infinity, or might not exist, depending on what the functions f and g are.

Not every undefined algebraic expression is an indeterminate form. For example, the expression 1/0 is undefined as a real number but is not indeterminate. This is because any limit that gives rise to this form will diverge to infinity.

An expression representing an indeterminate form may sometimes be given a numerical value in settings other than the computation of limits. The expression 0⁰ is defined as 1 when it represents an empty product. In the theory of power series, it is also often treated as 1 by convention, to make certain formulas more concise. (See the section "Zero to the zero power" in the article on exponentiation.) In the context of measure theory, it is necessary to take $\scriptstyle 0\cdot\infty$ to be 0.

[edit] Some examples and nonexamples

[edit] The form 0/0

The indeterminate form 0/0 is particularly common in calculus because it often arises in the evaluation of derivatives using their limit definition.

As mentioned above,

$\lim_{x \to 0} \frac{x}{x} = 1, \!$

while

$\lim_{x \to 0} \frac{x^{2}}{x} = 0 . \!$

This is enough to show that 0/0 is an indeterminate form. Other examples with this indeterminate form include

$\lim_{x \to 0} \frac{\sin(x)}{x} = 1 \!$

and

$\lim_{x \to 49} \frac{x - 49}{\sqrt{x}\, - 7} = 14 . \!$

Direct substitution of the number that x approaches into any of these expressions leads to the indeterminate form 0/0, but the limits take many different values. In fact, any desired value A can be obtained for this indeterminate form as follows:

$\lim_{x \to 0} \frac{Ax}{x} = A . \!$

Furthermore, the value infinity can also be obtained (in the sense of divergence to infinity):

$\lim_{x \to 0} \frac{x}{x^3} = \infty . \!$

[edit] The form 0⁰

The indeterminate form 0⁰ has been discussed since at least 1834.^[1] The following examples illustrate that the form is indeterminate:

$\lim_{x \to^{+} 0} (2^{-1/x})^{x} = 1/2, \!$

$\lim_{x \to^{+} 0} x^{x} = 1 , \!$

$\lim_{x \to^{+} 0} 0^{x} = 0. \!$

Thus, in general, knowing that $\scriptstyle\lim_{x \to c} f(x) \;=\; 0^+\!$ and $\scriptstyle\lim_{x \to c} g(x) \;=\; 0$ is not sufficient to calculate the limit

$\lim_{x \to c} f(x)^{g(x)}.\!$

However, if the functions f and g are additionally assumed to be meromorphic on a neighbourhood of 0 in the complex plane then the limit of f^g as z approaches 0 will always be 1.

There are settings where 0⁰ is taken to be defined even though it is an indeterminate form, as discussed in the article on exponentiation.

[edit] Undefined forms that are not indeterminate

The expression 1/0 is not an indeterminate form because there is no range of distinct values that f/g could approach. Specifically, if f approaches 1 and g approaches 0, then |f/g| must diverge to infinity. Notice that although f and g may be chosen (on an appropriate domain) so that f/g approaches either positive or negative infinity (in the sense of the extended real numbers), this variation does not create an indeterminate form (from one point of view, because they both diverge; from another point of view, because all infinities are equivalent in the real projective line).

Similarly, the expressions $\scriptstyle 0/\infty$ and $\scriptstyle\infty/0$ are not indeterminate because any limit that gives rise to one of these forms will converge to 0 or diverge to infinity, respectively.

[edit] Evaluating indeterminate forms

The indeterminate nature of a limit's form 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.

For example, the expression x²/x can be simplified to x at any point other than x = 0. Thus, the limit of this expression as x approaches 0 (which depends only on points near 0, not at x = 0 itself) is the limit of x, which is 0. Most of the other examples above can also be evaluated using algebraic simplification.

L'Hôpital's rule is a general method for evaluating the indeterminate forms 0/0 and ∞/∞. This rule states that (under appropriate conditions)

$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} , \!$

where f' and g' are the derivatives of f and g. (Note that this rule does not apply to forms like 0/∞, 1/0, and so on; but these forms are not indeterminate either.) With luck, 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:

$\ln \lim_{x \to c} f(x)^{g(x)} = \lim_{x \to c} \frac{\ln f(x)}{1/g(x)} . \!$

The right-hand side is of the form ∞/∞, so L'Hôpital's rule applies to it. Notice that this equation is valid (as long as the right-hand side is defined) because the natural logarithm (ln) is a continuous function; it's irrelevant how well-behaved f and g may (or may not) be as long as f is asymptotically positive.

Although L'Hôpital's rule applies both to 0/0 and to ∞/∞, one of these may be better than the other in a particular case (because of the possibilities for algebraic simplification afterwards). You can change between these forms, if necessary, by transforming f/g to (1/g)/(1/f).

[edit] List of indeterminate forms

The following table lists the indeterminate forms for the standard arithmetic operations and the transformations for applying l'Hôpital's rule.

Indeterminate form	Conditions	Transformation to 0/0	Transformation to ∞/∞
0/0	$\lim_{x \to c} f(x) = 0, \lim_{x \to c} g(x) = 0 \!$	—	$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{1/g(x)}{1/f(x)} \!$
∞/∞	$\lim_{x \to c} f(x) = \infty, \lim_{x \to c} g(x) = \infty \!$	$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{1/g(x)}{1/f(x)} \!$	—
0 × ∞	$\lim_{x \to c} f(x) = 0, \lim_{x \to c} g(x) = \infty \!$	$\lim_{x \to c} f(x)g(x) = \lim_{x \to c} \frac{f(x)}{1/g(x)} \!$	$\lim_{x \to c} f(x)g(x) = \lim_{x \to c} \frac{g(x)}{1/f(x)} \!$
1^∞	$\lim_{x \to c} f(x) = 1, \lim_{x \to c} g(x) = \infty \!$	$\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)} \!$	$\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{g(x)}{1/\ln f(x)} \!$
0⁰	$\lim_{x \to c} f(x) = 0^+, \lim_{x \to c} g(x) = 0 \!$	$\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{g(x)}{1/\ln f(x)} \!$	$\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)} \!$
∞⁰	$\lim_{x \to c} f(x) = \infty, \lim_{x \to c} g(x) = 0 \!$	$\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{g(x)}{1/\ln f(x)} \!$	$\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)} \!$
∞ - ∞	$\lim_{x \to c} f(x) = \infty, \lim_{x \to c} g(x) = \infty \!$	$\lim_{x \to c} (f(x) - g(x)) = \lim_{x \to c} \frac{1/g(x) - 1/f(x)}{1/f(x)g(x)} \!$	$\lim_{x \to c} (f(x) - g(x)) = \ln \lim_{x \to c} \frac{e^{f(x)}}{e^{g(x)}} \!$