# Lim x->inf t^(1/t)

1. Mar 17, 2013

### freezer

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

2. Mar 17, 2013

### Ray Vickson

Why do you think you have made an error?

3. Mar 17, 2013

### freezer

4. Mar 17, 2013

### eumyang

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

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
$$\ln y = \lim_{x\rightarrow \infty } \ln \left(x^{\frac{1}{x}} \right) = \lim_{x\rightarrow \infty } {\frac{1}{x}}\ln x$$

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

5. Mar 17, 2013

### freezer

0 = ln y so e^0 = 1?

6. Mar 17, 2013

### eumyang

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

7. Mar 17, 2013

### Ray Vickson

8. Mar 18, 2013

### HallsofIvy

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

Last edited by a moderator: Mar 18, 2013
9. Mar 18, 2013

### freezer

Then are you saying:
$$\lim_{t\rightarrow \infty} t ^ \frac{1}{t} = DNE$$

After doing some more searching i found this:
https://www.physicsforums.com/showpost.php?p=3048872&postcount=44

Last edited by a moderator: Mar 18, 2013
10. Mar 18, 2013