Why 1^∞ is an indeterminate form?

For the 1^∞ form:

Assume lim f(x)＝1 and lim g(x)＝∞
　　　x→p　　　　　x→p

For f(x)＜1, we have lim f(x)^g(x)＝0
　　　　　　　　　x→p

For f(x)＞1, we have lim f(x)^g(x)＝∞
　　　　　　　　　x→p

Therefore, we cannot determine the limit, it could be any real number between 0 and ∞.

Therefore, 1^∞ is an indeterminate form.


For the 0^∞ form:

Assume lim f(x)＝0 and lim g(x)＝∞
　　　x→p　　　　　x→p

As lim f(x)＝0, there exists a real δ1＞0 such that 0＜|x－p|＜δ1 implies |f(x)|＜0.5
　x→p

As lim g(x)＝∞, there exists a real δ2＞0 such that 0＜|x－p|＜δ2 implies |g(x)|＞1
　x→p

Now take 0＜|x－p|＜δ1＋δ2, we have

　　　0＜|f(x)|^g(x)＜0.5^g(x)

＝＞　0≤ lim |f(x)|^g(x)≤ lim 0.5^g(x)＝0
　　　　x→p　　　　x→p

＝＞　lim |f(x)|^g(x)＝0
　　　x→p

Hence,

lim f(x)^g(x)＝0　　[0^∞ form]
x→p

Therefore, we can determine that the limit is 0 and 0^∞ is not an indeterminate form.


Definition of “indeterminate form” from Wikipedia:

In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form.

2008-07-17 22:40:36 補充：
Correction:

WRONG:

Now take 0＜|x－p|＜δ1＋δ2, we have

RIGHT:

Now take 0＜|x－p|＜min(δ1,δ2), we have