What is the Limit of t raised to the Power of 1/t as t Approaches Infinity?

[freezer]

Homework Statement

\lim_{t\rightarrow \infty } t^{\frac{1}{t}}

Homework Equations

The Attempt at a Solution

let \ y = x^{\frac{1}{x}}

\ln y = {\frac{1}{x}} ln x

\lim_{x\rightarrow \infty } ln y = \lim_{x\rightarrow \infty } {\frac{ln x}{x}}

L'H
\lim_{x\rightarrow \infty } ln y = \lim_{x\rightarrow \infty } {\frac{\frac{1}{x}}{1}} = \frac{0}{1} = 0

Wolfram shows the answer to be 1 and intuition says that a^0 = 1 so i am not sure where my error is.

 

[Ray Vickson]

Homework Statement

\lim_{t\rightarrow \infty } t^{\frac{1}{t}}

Homework Equations

The Attempt at a Solution

let \ y = x^{\frac{1}{x}}

\ln y = {\frac{1}{x}} ln x

\lim_{x\rightarrow \infty } ln y = \lim_{x\rightarrow \infty } {\frac{ln x}{x}}

L'H
\lim_{x\rightarrow \infty } ln y = \lim_{x\rightarrow \infty } {\frac{\frac{1}{x}}{1}} = \frac{0}{1} = 0

Wolfram shows the answer to be 1 and intuition says that a^0 = 1 so i am not sure where my error is.

Why do you think you have made an error?

 

[freezer]

Why do you think you have made an error?

http://www.wolframalpha.com/input/?i=lim+t-%3Einf+t^%281%2Ft%29

 

[eumyang]

let \ y = x^{\frac{1}{x}}

It would be better to write it this way:
Let
let \ y = \lim_{x\rightarrow \infty } x^{\frac{1}{x}}

\ln y = {\frac{1}{x}} \ln x

When you take the natural logarithm of both sides, you actually get
\ln y = \ln \left( \lim_{x\rightarrow \infty } x^{\frac{1}{x}} \right)
... but since ln x is continuous, you can rewrite it as
\ln y = \lim_{x\rightarrow \infty } \ln \left(x^{\frac{1}{x}} \right) = \lim_{x\rightarrow \infty } {\frac{1}{x}}\ln x

\ln y = \lim_{x\rightarrow \infty } {\frac{\frac{1}{x}}{1}} = \frac{0}{1} = 0
(EDIT: I removed the "lim" in front of ln y.)

Wolfram shows the answer to be 1 and intuition says that a^0 = 1 so i am not sure where my error is.

That's because you are not finished. You have one more step to go. What does that 0 represent?

 

[freezer]

That's because you are not finished. You have one more step to go. What does that 0 represent?

0 = ln y so e^0 = 1?

 

[eumyang]

Yes, ln y = 0, so y = e⁰ = 1. If you use my corrected definition of y, then you have your answer.

 

[Ray Vickson]

http://www.wolframalpha.com/input/?i=lim+t-%3Einf+t^%281%2Ft%29

That is not what I asked: I know what Wolfram said. I want to know why you think your answer is wrong.

 

[HallsofIvy]

"a^0= 1" has nothing to do with this. If you just replace x with \infty, you get \infty^0 which is "undetermined".

 

[freezer]

"a^0= 1" has nothing to do with this. If you just replace x with \infty, you get \infty^0 which is "undetermined".

Then are you saying:
\lim_{t\rightarrow \infty} t ^ \frac{1}{t} = DNEAfter doing some more searching i found this:
https://www.physicsforums.com/showpost.php?p=3048872&postcount=44

 

[Dick]

Then are you saying:
\lim_{t\rightarrow \infty} t ^ \frac{1}{t} = DNEAfter doing some more searching i found this:
https://www.physicsforums.com/showpost.php?p=3048872&postcount=44

No, you were correct in post 5. If lim ln(y)=0 then lim y=e^0=1. The limit of t^(1/t) is 1.