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

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Homework Help Overview

The discussion revolves around the limit of the expression \( t^{\frac{1}{t}} \) as \( t \) approaches infinity, a topic within calculus and limits.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the limit by taking the natural logarithm of the expression and applying L'Hôpital's rule. There are questions about the interpretation of the result and whether an error has been made in the reasoning.

Discussion Status

Some participants have provided guidance on the interpretation of the limit and the implications of the logarithmic transformation. There is an ongoing exploration of the reasoning behind the limit and the nature of the expression as it approaches infinity.

Contextual Notes

Participants express uncertainty regarding the transition from the logarithmic form back to the original expression and the implications of reaching a limit of zero in the logarithmic context.

freezer
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Homework Statement



[tex]\lim_{t\rightarrow \infty } t^{\frac{1}{t}}[/tex]

Homework Equations


The Attempt at a Solution



[tex]let \ y = x^{\frac{1}{x}}[/tex]

[tex]\ln y = {\frac{1}{x}} ln x[/tex]

[tex]\lim_{x\rightarrow \infty } ln y = \lim_{x\rightarrow \infty } {\frac{ln x}{x}}[/tex]

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

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

Homework Statement



[tex]\lim_{t\rightarrow \infty } t^{\frac{1}{t}}[/tex]

Homework Equations





The Attempt at a Solution



[tex]let \ y = x^{\frac{1}{x}}[/tex]

[tex]\ln y = {\frac{1}{x}} ln x[/tex]

[tex]\lim_{x\rightarrow \infty } ln y = \lim_{x\rightarrow \infty } {\frac{ln x}{x}}[/tex]

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

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

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

freezer said:
[tex]\ln y = \lim_{x\rightarrow \infty } {\frac{\frac{1}{x}}{1}} = \frac{0}{1} = 0[/tex]
(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?
 
eumyang said:
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?
 
Yes, ln y = 0, so y = e0 = 1. If you use my corrected definition of y, then you have your answer.
 
"a^0= 1" has nothing to do with this. If you just replace x with [itex]\infty[/itex], you get [itex]\infty^0[/itex] which is "undetermined".
 
Last edited by a moderator:
HallsofIvy said:
"a^0= 1" has nothing to do with this. If you just replace x with [itex]\infty[/itex], you get [itex]\infty^0[/itex] which is "undetermined".

Then are you saying:
[tex]\lim_{t\rightarrow \infty} t ^ \frac{1}{t} = DNE[/tex]After doing some more searching i found this:
https://www.physicsforums.com/showpost.php?p=3048872&postcount=44
 
Last edited by a moderator:
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