Editing Indeterminate form (section)

=== Expressions that are not indeterminate forms ===

The expression <math>1/0</math> is not commonly regarded as an indeterminate form, because if the limit of <math>f/g</math> exists then there is no ambiguity as to its value, as it always diverges. Specifically, if <math>f</math> approaches <math>1</math> and <math>g</math> approaches <math>0,</math> then <math>f</math> and <math>g</math> may be chosen so that:

# <math>f/g</math> approaches <math>+\infty</math>
# <math>f/g</math> approaches <math>-\infty</math>
# The limit fails to exist.

In each case the absolute value <math>|f/g|</math> approaches <math>+\infty</math>, and so the quotient <math>f/g</math> must diverge, in the sense of the [[extended real number]]s (in the framework of the [[projectively extended real line]], the limit is the [[Point at infinity|unsigned infinity]] <math>\infty</math> in all three cases<ref name=":3">{{Cite web|url=https://www.cut-the-knot.org/blue/GhostCity.shtml|title=Undefined vs Indeterminate in Mathematics|website=www.cut-the-knot.org|access-date=2019-12-02}}</ref>). Similarly, any expression of the form <math>a/0</math> with <math>a\ne0</math> (including <math>a=+\infty</math> and <math>a=-\infty</math>) is not an indeterminate form, since a quotient giving rise to such an expression will always diverge.

The expression <math>0^\infty</math> is not an indeterminate form. The expression <math>0^{+\infty}</math> obtained from considering <math>\lim_{x \to c} f(x)^{g(x)}</math> gives the limit <math>0,</math> provided that <math>f(x)</math> remains nonnegative as <math>x</math> approaches <math>c</math>. The expression <math>0^{-\infty}</math> is similarly equivalent to <math>1/0</math>; if <math>f(x) > 0</math> as <math>x</math> approaches <math>c</math>, the limit comes out as <math>+\infty</math>.

To see why, let <math>L = \lim_{x \to c} f(x)^{g(x)},</math> where <math> \lim_{x \to c} {f(x)}=0,</math> and <math> \lim_{x \to c} {g(x)}=\infty.</math> By taking the natural logarithm of both sides and using <math> \lim_{x \to c} \ln{f(x)}=-\infty,</math> we get that <math>\ln L = \lim_{x \to c} ({g(x)}\times\ln{f(x)})=\infty\times{-\infty}=-\infty,</math> which means that <math>L = {e}^{-\infty}=0.</math>