Prove Limit n^(1/n) = 1 as n->oo

[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

 

[Hurkyl]

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

<br /> \lim_{n \rightarrow \infty} n<br /> = \lim_{n \rightarrow \infty} (n^n)^{(1/n)}<br />

would also be 1.



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

 

[pa1o]

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

<br /> \lim_{n \rightarrow \infty} n<br /> = \lim_{n \rightarrow \infty} (n^n)^{(1/n)}<br />

would also be 1.



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

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->
<br /> \lim_{n \rightarrow \infty} n(1/n)<br />

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

 

[Hurkyl]

if you "go" with the limit into the exponent of N (you can do that because N is defined on whole R)

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

 

[shmoe]

actualy i think his proof is correct, but your limit is incorrect let me explain:

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.