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indeterminate form
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{{Short description|Expression in mathematical analysis}}In calculus, it is usually possible to compute the limit of the sum, difference, product, quotient or power of two functions by taking the corresponding combination of the separate limits of each respective function. For example,begin{align}lim_{x to c} bigl(f(x) + g(x)bigr) &= lim_{x to c} f(x) + lim_{x to c} g(x), [3mu]lim_{x to c} bigl(f(x)g(x)bigr) &= lim_{x to c} f(x) cdot lim_{x to c} g(x),end{align}and likewise for other arithmetic operations; this is sometimes called the algebraic limit theorem. However, certain combinations of particular limiting values cannot be computed in this way, and knowing the limit of each function separately does not suffice to determine the limit of the combination. In these particular situations, the limit is said to take an indeterminate form, described by one of the informal expressionsfrac 00,~ frac{infty}{infty},~ 0timesinfty,~ infty - infty,~ 0^0,~ 1^infty, text{ or } infty^0,where each expression stands for the limit of a function constructed by an arithmetical combination of two functions whose limits respectively tend to {{tmath|0,}} {{tmath|1,}} or {{tmath|infty}} as indicated.{{sfnp|Varberg|Purcell|Rigdon|2007|p=423, 429, 430, 431, 432}}A limit taking one of these indeterminate forms might tend to zero, might tend to any finite value, might tend to infinity, or might diverge, depending on the specific functions involved. A limit which unambiguously tends to infinity, for instance lim_{x to 0} 1/x^2 = infty, is not considered indeterminate.WEB,mathworld.wolfram.com/Indeterminate.html, Indeterminate, Weisstein, Eric W., mathworld.wolfram.com, en, 2019-12-02, 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 0/0. For example, as x approaches 0, the ratios x/x^3, x/x, and x^2/x go to infty, 1, and 0 respectively. In each case, if the limits of the numerator and denominator are substituted, the resulting expression is 0/0, which is indeterminate. In this sense, 0/0 can take on the values 0, 1, or 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, x sin(1/x) / x.So the fact that two functions f(x) and g(x) converge to 0 as x approaches some limit point c is insufficient to determinate the limit{{block indent| 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 0^0. Whether this expression is left undefined, or is defined to equal 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 0^infty and other expressions involving infinity are not indeterminate forms.

Some examples and non-examples

Indeterminate form 0/0

{{Redirect|0/0|the symbol|Percent sign}}File:Indeterminate form - x over x.gif|Fig. 1: {{var|y}} = {{sfrac|{{var|x}}|{{var|x}}}}File:Indeterminate form - x2 over x.gif|Fig. 2: {{var|y}} = {{sfrac|{{var|x}}{{sup|2}}|{{var|x}}}}File:Indeterminate form - sin x over x close.gif|Fig. 3: {{var|y}} = {{sfrac|sin {{var|x}}|{{var|x}}}}File:Indeterminate form - complicated.gif|Fig. 4: {{var|y}} = {{sfrac|x − 49|{{radic|x}} − 7}} (for {{var|x}} = 49)File:Indeterminate form - 2x over x.gif|Fig. 5: {{var|y}} = {{sfrac|{{var|a}}{{var|x}}|{{var|x}}}} where {{var|a}} = 2File:Indeterminate form - x over x3.gif|Fig. 6: {{var|y}} = {{sfrac|{{var|x}}|{{var|x}}{{sup|3}}}}The indeterminate form 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,{{block indent| lim_{x to 0} frac{x}{x} = 1, qquad (see fig. 1)}}while{{block indent| lim_{x to 0} frac{x^{2}}{x} = 0, qquad (see fig. 2)}}This is enough to show that 0/0 is an indeterminate form. Other examples with this indeterminate form include{{block indent| lim_{x to 0} frac{sin(x)}{x} = 1, qquad (see fig. 3)}}and{{block indent| lim_{x to 49} frac{x - 49}{sqrt{x}, - 7} = 14, qquad (see fig. 4)}}Direct substitution of the number that x approaches into any of these expressions shows that these are examples correspond to the indeterminate form 0/0, but these limits can assume many different values. Any desired value a can be obtained for this indeterminate form as follows:{{block indent| lim_{x to 0} frac{ax}{x} = a . qquad (see fig. 5)}}The value infty can also be obtained (in the sense of divergence to infinity):{{block indent| lim_{x to 0} frac{x}{x^3} = infty . qquad (see fig. 6)}}

Indeterminate form 00

{{multiple image
| image1 = Indeterminate form - x0.gif
| caption1 = Graph of {{math|1=y = x{{sup|0}}}}
| image2 = Indeterminate form - 0x.gif
| caption2 = Graph of {{math|1=y = 0{{sup|x}}}}
| total_width = 300
| direction = vertical
}}The following limits illustrate that the expression 0^0 is an indeterminate form:
begin{align}
lim_{x to 0^+} x^0 &= 1,
lim_{x to 0^+} 0^x &= 0.
end{align} Thus, in general, knowing that textstylelim_{x to c} f(x) ;=; 0 and textstylelim_{x to c} g(x) ;=; 0 is not sufficient to evaluate the limit
lim_{x to c} f(x)^{g(x)}.
If the functions f and g are analytic at c, and f is positive for x sufficiently close (but not equal) to c, then the limit of f(x)^{g(x)} will be 1.JOURNAL, 10.2307/2689754, Louis M. Rotando, Henry Korn, The indeterminate form 00, Mathematics Magazine, January 1977, 50, 1, 41–42, 2689754, Otherwise, use the transformation in the table below to evaluate the limit.

Expressions that are not indeterminate forms

The expression 1/0 is not commonly regarded as an indeterminate form, because if the limit of f/g exists then there is no ambiguity as to its value, as it always diverges. Specifically, if f approaches 1 and g approaches 0, then f and g may be chosen so that:
  1. f/g approaches +infty
  2. f/g approaches -infty
  3. The limit fails to exist.
In each case the absolute value |f/g| approaches +infty, and so the quotient 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 infty in all three casesWEB,www.cut-the-knot.org/blue/GhostCity.shtml, Undefined vs Indeterminate in Mathematics, www.cut-the-knot.org, 2019-12-02, ). Similarly, any expression of the form a/0 with ane0 (including a=+infty and a=-infty) is not an indeterminate form, since a quotient giving rise to such an expression will always diverge.The expression 0^infty is not an indeterminate form. The expression 0^{+infty} obtained from considering lim_{x to c} f(x)^{g(x)} gives the limit 0, provided that f(x) remains nonnegative as x approaches c. The expression 0^{-infty} is similarly equivalent to 1/0; if f(x) > 0 as x approaches c, the limit comes out as +infty.To see why, let L = lim_{x to c} f(x)^{g(x)}, where lim_{x to c} {f(x)}=0, and lim_{x to c} {g(x)}=infty. By taking the natural logarithm of both sides and using lim_{x to c} ln{f(x)}=-infty, we get that ln L = lim_{x to c} ({g(x)}timesln{f(x)})=inftytimes{-infty}=-infty, which means that 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 alpha and beta converge to zero at the same limit point and textstyle lim frac{beta}{alpha} = 1, they are called equivalent infinitesimal (equiv. alpha sim beta).Moreover, if variables alpha’ and beta’ are such that alpha sim alpha’ and beta sim beta’, then:{{block indent|lim frac{beta}{alpha} = lim frac{beta’}{alpha’}}}Here is a brief proof:Suppose there are two equivalent infinitesimals alpha sim alpha’ and beta sim beta’.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 0/0, one can make use of the following facts about equivalent infinitesimals (e.g., xsimsin x if x becomes closer to zero):WEB,www.vaxasoftware.com/doc_eduen/mat/infiequi.pdf, Table of equivalent infinitesimals, Vaxa Software, {{block indent|x sim sin x,}}{{block indent|x sim arcsin x,}}{{block indent|x sim sinh x,}}{{block indent|x sim tan x,}}{{block indent|x sim arctan x,}}{{block indent|x sim ln(1 + x),}}{{block indent|1 - cos x sim frac{x^2}{2},}}{{block indent|cosh x - 1 sim frac{x^2}{2},}}{{block indent|a^x - 1 sim x ln a,}}{{block indent|e^x - 1sim x,}}{{block indent|(1 + x)^a - 1 sim ax.}}For example:begin{align}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^{xln{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}+1right) &= lim_{x to 0} frac{cos x -1}{3x^2} &= lim_{x to 0} -frac{x^2}{6x^2} &= -frac{1}{6}end{align}In the 2nd equality, e^y - 1 sim y where y = xln{2+cos x over 3} as y become closer to 0 is used, and y sim ln {(1+y)} where y = {{cos x - 1} over 3} is used in the 4th equality, and 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 0/0 and infty/infty. This rule states that (under appropriate conditions){{block indent| 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 expressions infty/0, 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:{{block indent| 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 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 f and g may (or may not) be as long as f is asymptotically positive. (the domain of logarithms is the set of all positive real numbers.)Although L’Hôpital’s rule applies to both 0/0 and 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 f/g to (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.{| border=1 class=“wikitable” style="background-color: #ffffff; width: 85%;“!Indeterminate form!Conditions!Transformation to 0/0!Transformation to infty/infty
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)} !
infty|infty}}| 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)} ! —}}
|0cdotinfty| 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)} !
|infty - infty| 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)}} !
|0^0| 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)} !
|1^infty| 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)} !
|infty^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)} !

See also

References

Citations

{{reflist}}

Bibliographies

  • BOOK


, Varberg, Dale E.
, Purcell, Edwin J.
, Rigdon, Steven E.
, Calculus
, 2007
, Pearson Prentice Hall
, 9th
, 978-0131469686
, {{Calculus topics}}

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