Indeterminate form

From Wikipedia, the free encyclopedia

In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression whose limit cannot be evaluated by substituting the limits of the subexpressions.

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

[edit] Discussion

The most common example is 0/0. In the expression x/x, as x approaches 0, the limit is 1; but in the expression x²/x, as x approaches 0, the limit is 0. In each case, however, if the limits of the numerator and denominator are evaluated and plugged into the division operation, the solution is 0/0. So (roughly speaking) 0/0 might mean 1, or it might mean 0 (and in fact, by appropriate examples, it can be made to mean anything); so it is indeterminate.

Note that an indeterminate form is not the same thing as an expression that is undefined. Although most indeterminate forms (such as 0/0) are also undefined, the indeterminate form 0⁰ is usually defined to be 1 (see Empty product). On the other hand, the expression 1/0 is undefined (at least as a real number), but it is not an indeterminate form (because such an expression, such as 1/x as x approaches 0, consistently diverges to infinity in some way).

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

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

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

More generally, an indeterminate form a * b occurs for any algebraic operation *, as long as the facts

$\lim_{x \to c} f(x) = a \mbox{,} \!$
$\lim_{x \to c} g(x) = b \mbox{,} \!$

are not enough information to evaluate

$\lim_{x \to c} f(x) * g(x) \mbox{,} \!$

even assuming that the latter converges.

[edit] Some examples and nonexamples

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):

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

Thus, 0/0 is quite indeterminate; limits of this form can easily be made to take any value.

The indeterminate form 0⁰ is trickier. It's easy to get the value 1:

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

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

$\lim_{x \to 49} (x - 49)^{\sqrt{x}\, - 7} = 1 . \!$

However, other values are not so obvious. Here is one example (from an 1834 debate [1]):

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

(where the notation →⁺ indicates that the domain D consists only of non-negative real numbers). There is also the simple (but valid) example

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

If the -1/x in the above limit is modified slightly, with the x replaced by the absolute value of x, the limit no longer needs to be one-sided; we have convergence to a 0⁰ form in a two-sided or even complex limit. Note, in the context of the paragraph below, that we are able to find such an example because, by using an absolute value, we no longer have a meromorphic function of x in our base.

However, if the functions f and g are meromorphic (on a neighbourhood of the limit point c in the Riemann sphere), then the limit of f^g will always be 1 (as long as f^g is defined on a neighbourhood of c). In this sense, 0⁰ is less indeterminate than the other indeterminate forms, and this is one reason why 0⁰ is usually not left undefined (but instead defined to be 1).

By contrast, 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).

[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, 2/3, 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 appopriate 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.

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 possibilites 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 and the transformations for applying l'Hôpital's rule.

Indeterminate form	Conditions	Transformation to 0/0	Transformation to ∞/∞
0/0	$\lim_{x \to c} f(x) = 0, \lim_{x \to c} g(x) = 0 \!$	$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f(x)}{g(x)} \!$ (no transformation needed)	$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{1/g(x)}{1/f(x)} \!$
∞/∞	$\lim_{x \to c} f(x) = \infty, \lim_{x \to c} g(x) = \infty \!$	$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{1/g(x)}{1/f(x)} \!$	$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f(x)}{g(x)} \!$ (no transformation needed)
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	$\lim_{x \to c} f(x) = \infty, \lim_{x \to c} g(x) = 0 \!$	$\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{g(x)}{1/\ln f(x)} \!$	$\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)} \!$
∞ - ∞	$\lim_{x \to c} f(x) = \infty, \lim_{x \to c} g(x) = \infty \!$	$\lim_{x \to c} (f(x) - g(x)) = \lim_{x \to c} \frac{1/g(x) - 1/f(x)}{1/f(x)g(x)} \!$	$\lim_{x \to c} (f(x) - g(x)) = \ln \lim_{x \to c} \frac{e^{f(x)}}{e^{g(x)}} \!$