Editing L'Hôpital's rule (section)

== Other indeterminate forms ==
Other indeterminate forms, such as {{math|1<sup>∞</sup>}}, {{math|0<sup>0</sup>}}, {{math|∞<sup>0</sup>}}, {{math|0 · ∞}}, and {{math|∞ − ∞}}, can sometimes be evaluated using L'Hôpital's rule. We again indicate applications of L'Hopital's rule by <math>
\ \stackrel{\mathrm{H}}{=}\ 
</math>.

For example, to evaluate a limit involving {{math|∞ − ∞}}, convert the difference of two functions to a quotient:

:<math>
\begin{align}
\lim_{x\to 1}\left(\frac{x}{x-1}-\frac1{\ln x}\right)
& = \lim_{x\to 1}\frac{x\cdot\ln x -x+1}{(x-1)\cdot\ln x} \\[6pt]
& \ \stackrel{\mathrm{H}}{=}\  \lim_{x\to 1}\frac{\ln x}{\frac{x-1}{x}+\ln x} \\[6pt]
& = \lim_{x\to 1}\frac{x\cdot\ln x}{x-1+x\cdot\ln x}   \\[6pt]
& \ \stackrel{\mathrm{H}}{=}\  \lim_{x\to 1}\frac{1+\ln x}{1+1+\ln x}  = \frac{1+0}{1+1+0}.
\end{align}
</math>

L'Hôpital's rule can be used on indeterminate forms involving [[Exponentiation|exponents]] by using [[logarithm]]s to "move the exponent down". Here is an example involving the indeterminate form {{math|0<sup>0</sup>}}:

:<math>
\lim_{x\to 0^+\!}x^x
= \lim_{x\to 0^+\!}e^{\ln (x^x)}
= \lim_{x\to 0^+\!}e^{x\cdot\ln x}
= \lim_{x\to 0^+\!}\exp(x\cdot\ln x)
= \exp({\lim\limits_{x\to 0^+\!\!}\,x\cdot\ln x}).
</math>

It is valid to move the limit inside the [[exponential function]] because this function is [[continuous function|continuous]]. Now the exponent <math>x</math> has been "moved down". The limit <math>\lim_{x\to 0^+}x\cdot\ln x</math> is of the indeterminate form {{math|0 · ∞}} dealt with in an example above: L'Hôpital may be used to determine that

:<math>\lim_{x\to 0^+}x\cdot\ln x = 0.</math>

Thus

:<math>\lim_{x\to 0^+}x^x =\exp(0) = e^0 = 1.</math>

The following table lists the most common indeterminate forms and the transformations which precede applying l'Hôpital's rule:

{| class="wikitable" style="background-color: #ffffff; width: 70%;"
!Indeterminate form with f & g
!Conditions
!Transformation to <math>0/0</math>
|-
|{{sfrac|0|0}}
|<math> \lim_{x \to c} f(x) = 0,\  \lim_{x \to c} g(x) = 0 \! </math>
|{{center|&mdash;}}
|-
|{{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>
|-
|<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>\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>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>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>\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>
|}