# Prove the limit help

1. Oct 31, 2004

### b0mb0nika

Hi...
I have to prove that the limit as n->oo of n^(1/n) is 1
as n->oo (1/n)-> 0, which means that n^(1/n) ->1
i was wondering though if you could prove this differently using the definition of the limit ( with epsilon ) or maybe take derivatives..?

thanks

2. Oct 31, 2004

### Hurkyl

Staff Emeritus
Actually, your proof is wrong. Notice that your approach would also "prove" that

$$\lim_{n \rightarrow \infty} n = \lim_{n \rightarrow \infty} (n^n)^{(1/n)}$$

would also be 1.

Exponents are messy things -- try applying a logarithm to convert the problem into an easier one.

3. Nov 1, 2004

### pa1o

actualy i think his proof is correct, but your limit is incorrect let me explain:
if you "go" with the limit into the exponent of N (you can do that because N is defined on whole R) the you get->
$$\lim_{n \rightarrow \infty} n(1/n)$$

n goes to infinity and 1/n goes to 0. But you always have to remember that oo is not a number its just a symbol so you cannot say that oo*0 is 0. 0*oo can be any R number even 1. Try to think about it, its a bit difficult but in the in the end its quite understandable

4. Nov 1, 2004

### Hurkyl

Staff Emeritus
You might want to go rechcek the statement of the relevant limit theorems.

5. Nov 1, 2004

### shmoe

You've missed the point of Hurkyl's example and are missing your own advice on treating indeterminate limits with care. b0mb0nika had a limit of the indeterminate form $$\infty^{0}$$. Because it was of this form he (wrongly) concluded that the limit would be 1. Hurkyl gave a limit of the form $$\infty^{0}$$ that is easily manipulated into a limit that is clearly not 1, showing the error in b0mb0nika's approach to limits of this type.

For the original limit try using log's as already suggested, $$\lim_{n\rightarrow \infty}n^{1/n}=\lim_{n\rightarrow\infty}e^{\frac{1}{n}\log{n}}$$

You should be able to find $$\lim_{n\rightarrow\infty}\frac{1}{n}\log{n}$$ without too much trouble.