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.