Editing Indeterminate form (section)

== 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 <math>\alpha</math> and <math>\beta</math> converge to zero at the same limit point and <math>\textstyle \lim \frac{\beta}{\alpha} = 1</math>, they are called ''equivalent infinitesimal'' (equiv. <math>\alpha \sim \beta</math>).

Moreover, if variables <math>\alpha'</math> and <math>\beta'</math> are such that <math>\alpha \sim \alpha'</math> and <math>\beta \sim \beta'</math>, then:
{{block indent|<math>\lim \frac{\beta}{\alpha} = \lim \frac{\beta'}{\alpha'}</math>}}

Here is a brief proof:

Suppose there are two equivalent infinitesimals <math>\alpha \sim \alpha'</math> and <math>\beta \sim \beta'</math>.

<math display=block>\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'}</math>

For the evaluation of the indeterminate form <math>0/0</math>, one can make use of the following facts about equivalent [[infinitesimal]]s (e.g., <math>x\sim\sin x</math> if ''x'' becomes closer to zero):<ref>{{Cite web|url=http://www.vaxasoftware.com/doc_eduen/mat/infiequi.pdf|title=Table of equivalent infinitesimals|website=Vaxa Software}}</ref>

{{block indent|<math>x \sim \sin x,</math>}}
{{block indent|<math>x \sim \arcsin x,</math>}}
{{block indent|<math>x \sim \sinh x,</math>}}
{{block indent|<math>x \sim \tan x,</math>}}
{{block indent|<math>x \sim \arctan x,</math>}}
{{block indent|<math>x \sim \ln(1 + x),</math>}}
{{block indent|<math>1 - \cos x \sim \frac{x^2}{2},</math>}}
{{block indent|<math>\cosh x - 1 \sim \frac{x^2}{2},</math>}}
{{block indent|<math>a^x - 1 \sim x \ln a,</math>}}
{{block indent|<math>e^x - 1\sim x,</math>}}
{{block indent|<math>(1 + x)^a - 1 \sim ax.</math>}}

For example:

<math display=block>\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^{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{align}</math>

In the 2nd equality, <math>e^y - 1 \sim y</math> where <math>y = x\ln{2+\cos x \over 3}</math> as ''y'' become closer to 0 is used, and <math>y \sim \ln {(1+y)}</math> where <math>y = {{\cos x - 1} \over 3}</math> is used in the 4th equality, and <math>1-\cos x \sim {x^2 \over 2}</math> is used in the 5th equality.

===L'Hôpital's rule===
{{main|L'Hôpital's rule}}
L'Hôpital's rule is a general method for evaluating the indeterminate forms <math>0/0</math> and <math>\infty/\infty</math>. This rule states that (under appropriate conditions)
{{block indent|<math> \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} ,</math>}}
where <math>f'</math> and <math>g'</math> are the [[derivative (calculus)|derivative]]s of <math>f</math> and <math>g</math>. (Note that this rule does ''not'' apply to expressions <math>\infty/0</math>, <math>1/0</math>, 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 0<sup>0</sup>:
{{block indent|<math> \ln \lim_{x \to c} f(x)^{g(x)} = \lim_{x \to c} \frac{\ln f(x)}{1/g(x)} .</math>}}
The right-hand side is of the form <math>\infty/\infty</math>, 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 <math>f</math> and <math>g</math> may (or may not) be as long as <math>f</math> is asymptotically positive. (the domain of logarithms is the set of all positive real numbers.)

Although L'Hôpital's rule applies to both <math>0/0</math> and <math>\infty/\infty</math>, 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 <math>f/g</math> to <math>(1/g)/(1/f)</math>.

== 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="width: 85%;"
!Indeterminate form
!Conditions
!Transformation to <math>0/0</math>
!Transformation to <math>\infty/\infty</math>
|-
|{{sfrac|<math>0</math>|<math>0</math>}}
|<math> \lim_{x \to c} f(x) = 0,\  \lim_{x \to c} g(x) = 0 \! </math>
|{{center|&mdash;}}
|<math> \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{1/g(x)}{1/f(x)} \! </math>
|-
|{{sfrac|<math>\infty</math>|<math>\infty</math>}}
|<math> \lim_{x \to c} f(x) = \infty,\  \lim_{x \to c} g(x) = \infty \! </math>
|<math> \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{1/g(x)}{1/f(x)} \! </math>
|{{center|&mdash;}}
|-
|<math>0\cdot\infty</math>
|<math> \lim_{x \to c} f(x) = 0,\  \lim_{x \to c} g(x) = \infty \! </math>
|<math> \lim_{x \to c} f(x)g(x) = \lim_{x \to c} \frac{f(x)}{1/g(x)} \! </math>
|<math> \lim_{x \to c} f(x)g(x) = \lim_{x \to c} \frac{g(x)}{1/f(x)} \! </math>
|-
|<math>\infty - \infty</math>
|<math> \lim_{x \to c} f(x) = \infty,\  \lim_{x \to c} g(x) = \infty \! </math>
|<math> \lim_{x \to c} (f(x) - g(x)) = \lim_{x \to c} \frac{1/g(x) - 1/f(x)}{1/(f(x)g(x))} \! </math>
|<math> \lim_{x \to c} (f(x) - g(x)) = \ln \lim_{x \to c} \frac{e^{f(x)}}{e^{g(x)}} \! </math>
|-
|<math>0^0</math>
|<math> \lim_{x \to c} f(x) = 0^+, \lim_{x \to c} g(x) = 0 \! </math>
|<math> \lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{g(x)}{1/\ln f(x)} \! </math>
|<math> \lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)} \! </math>
|-
|<math>1^\infty</math>
|<math> \lim_{x \to c} f(x) = 1,\  \lim_{x \to c} g(x) = \infty \! </math>
|<math> \lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)} \! </math>
|<math> \lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{g(x)}{1/\ln f(x)} \! </math>
|-
|<math>\infty^0</math>
|<math> \lim_{x \to c} f(x) = \infty,\  \lim_{x \to c} g(x) = 0 \! </math>
|<math> \lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{g(x)}{1/\ln f(x)} \! </math>
|<math> \lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)} \! </math>
|}