What Happens When You Calculate One to the Infinity Power?

[arpitm08]

one raised to the infinity power help please!

Let a and b be positive real numbers. For real number p define, f(p) = ((a^p + b^p)/2)^(1/p). Evaluate the limit of f(p) as p approaches 0.

By directly plugging in zero, you would get (1)^inf. Wouldn't that equal 1 or would it be something else? When I put it into my 89 i got 1 as my answer, but for some reason I don't think that it is right. Please help!

 

[00PS]

why wouldn't it? 1x1x1x1... is still 1 isn't it?

you can also check the graph at 0 to see if it converges to 1 when you plug in positive real numbers for a and b. don't the 89s take limits?

 

[sutupidmath]

why wouldn't it? 1x1x1x1... is still 1 isn't it?

?

$$1^{\infty}$$ is undefined!

 

[arildno]

If a=b, note that the limit is "a".

Let then "a" be greater than "b", and rewrite:
$$(\frac{(a^{p}+b^{p}}{2})^{\frac{1}{p}}=\frac{a}{2^{\frac{1}{p}}}({1+\frac{1}{N})^{\frac{1}{p}}, \frac{1}{N}=(\frac{b}{a})^{p}$$
Then,
$$\frac{1}{p}=N\frac{\ln(\frac{a}{b})}{N\ln(N)}$$


Therefore, you may rewrite this as:
$$((1+\frac{1}{N})^{N})^{\frac{\ln(\frac{a}{b})}{N\ln(N}}}$$
which remains nasty..

 

[lurflurf]

"If a=b, note that the limit is "a"."
Indeed.
"Wouldn't that equal 1"
No.
That is what is called an indeterminite form.
further analysis is needed.
f(p)=((a^p + b^p)/2)^(1/p)
Let us consider an approximation that is exact in the limit.
a^p~1+p log(a)
b^p~1+p log(b)
if p~0
thus
(a^p + b^p)/2~1+p log sqrt(ab)
if p~0
(1+x)^(1/P)~exp(x/p)
if p~0 and x~0

by combining these the answer should be clear.

 

[Anthony]

Hint: if your sequence is $$f(p)$$, consider the limit of the related sequence $$g(p)$$ where $$g=\log f$$. Then use some standard properties of continuity.