Indeterminate form
Expressions that are not indeterminate forms

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

In each case the absolute value | f / g | {\displaystyle |f/g|} approaches + ∞ {\displaystyle +\infty }, and so the quotient f / g {\displaystyle 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 ∞ {\displaystyle \infty } in all three cases). Similarly, any expression of the form a / 0 {\displaystyle a/0} with a ≠ 0 {\displaystyle a\neq 0} (including a = + ∞ {\displaystyle a=+\infty } and a = − ∞ {\displaystyle a=-\infty }) is not an indeterminate form, since a quotient giving rise to such an expression will always diverge.

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

To see why, let L = lim x → c f (x) g (x), {\displaystyle L=\lim _{x\to c}f(x)^{g(x)},} where lim x → c f (x) = 0, {\displaystyle \lim _{x\to c}{f(x)}=0,} and lim x → c g (x) = ∞. {\displaystyle \lim _{x\to c}{g(x)}=\infty.} By taking the natural logarithm of both sides and using lim x → c ln f (x) = − ∞, {\displaystyle \lim _{x\to c}\ln {f(x)}=-\infty,} we get that ln L = lim x → c (g (x) × ln f (x)) = ∞ × − ∞ = − ∞, {\displaystyle \ln L=\lim _{x\to c}({g(x)}\times \ln {f(x)})=\infty \times {-\infty }=-\infty,} which means that L = e − ∞ = 0. {\displaystyle L={e}^{-\infty }=0.}
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