One raised to the infinity power help please

In summary, the question is asking for the evaluation of the limit of f(p) as p approaches 0, where f(p) = ((a^p + b^p)/2)^(1/p). There are different approaches to solving this, such as directly plugging in zero or using an exact approximation in the limit. However, the answer is not simply 1 and further analysis is needed to determine the limit.
  • #1
arpitm08
50
0
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!
 
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  • #2


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?
 
  • #3


00PS said:
why wouldn't it? 1x1x1x1... is still 1 isn't it?

?
[tex]1^{\infty}[/tex] is undefined!
 
  • #4


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

Let then "a" be greater than "b", and rewrite:
[tex](\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}[/tex]
Then,
[tex]\frac{1}{p}=N\frac{\ln(\frac{a}{b})}{N\ln(N)}[/tex]


Therefore, you may rewrite this as:
[tex]((1+\frac{1}{N})^{N})^{\frac{\ln(\frac{a}{b})}{N\ln(N}}}[/tex]
which remains nasty..
 
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  • #5


"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.
 
  • #6


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

1. What does "one raised to the infinity power" mean?

When a number is raised to the power of infinity, it means that the number is multiplied by itself an infinite number of times. In this case, "one raised to the infinity power" is equal to one, since any number multiplied by one remains the same.

2. How do you calculate one raised to the infinity power?

The concept of infinity is not a number, but rather a concept that is used to represent something without an end. Therefore, it is impossible to calculate one raised to the infinity power, as it would require an infinite number of calculations. However, we know that the result is always one.

3. Why is one raised to the infinity power equal to one?

This is because any number raised to the power of zero is equal to one, and infinity can be thought of as a limit approaching zero. In other words, as the exponent gets larger and larger, the result gets closer and closer to one.

4. What is the difference between one raised to the infinity power and infinity raised to the one power?

The difference is in the order of operations. When one is raised to the infinity power, it means that one is multiplied by itself an infinite number of times. However, when infinity is raised to the one power, it means that infinity is multiplied by itself only once. In this case, the result is undefined, as infinity is not a specific number.

5. Can any number be raised to the infinity power?

Technically, any number can be raised to the infinity power, but the result will always be one. This is because, as mentioned before, any number multiplied by one remains the same. However, this concept is often used in limits and calculus to represent a value approaching infinity.

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