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

  • Thread starter freezer
  • Start date
This is because when x gets really big, x^(1/x) gets really close to 1, which is what we want because we're taking the logarithm.In summary, the limit of t^(1/t) as t approaches infinity is 1. This is because as x gets larger, x^(1/x) gets closer to 1, and taking the natural logarithm of both sides shows that the limit of ln y is 0, which means the limit of y is 1.
  • #1
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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  • #2
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?
 
  • #4
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?
 
  • #5
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?
 
  • #6
Yes, ln y = 0, so y = e0 = 1. If you use my corrected definition of y, then you have your answer.
 
  • #8
"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:
  • #9
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:
  • #10

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

1. What is the limit of t^(1/t) as t approaches infinity?

The limit of t^(1/t) as t approaches infinity is equal to 1.

2. How do you solve for the limit of t^(1/t) as t approaches infinity?

To solve for the limit of t^(1/t) as t approaches infinity, we can use the fact that t^(1/t) is equal to the exponential function e^(ln(t)/t). Then, using L'Hopital's rule, we can take the derivative of the numerator and denominator to get the limit as t goes to infinity. The result is 1, as shown in the first question.

3. Is the limit of t^(1/t) as t approaches infinity equal to infinity?

No, the limit of t^(1/t) as t approaches infinity is not equal to infinity. As t gets larger and larger, the value of t^(1/t) approaches 1, not infinity.

4. Can we use the limit of t^(1/t) as t approaches infinity to solve other types of limits?

Yes, we can use the limit of t^(1/t) as t approaches infinity to solve other types of limits, such as limits involving exponentials or logarithms. This is because we can rewrite those functions in terms of t^(1/t) and then use the limit we have already solved for.

5. How does the value of t^(1/t) change as t approaches infinity?

As t approaches infinity, the value of t^(1/t) approaches 1. This means that no matter how large t gets, the value of t^(1/t) will always be close to 1, but never actually reach it.

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