# U-th root of 1/u as u goes to infinity is 1.

1. Apr 23, 2009

### lukaszh

Is this true? $$n,m\in\mathbb{N}$$
$$\lim_{n\to\infty}\sqrt[n]{\lim_{m\to\infty}\frac{1}{m}}=\lim_{n\to\infty}\lim_{m\to\infty}\sqrt[n]{\frac{1}{m}}=\lim_{u\to\infty}\sqrt{\frac{1}{u}}=1$$
Thanks

2. Apr 24, 2009

### tiny-tim

No, that's completely wrong from start to finish (in particular limu -> ∞√u/u is obviously 0).

What were you trying to do?

3. Apr 24, 2009

### lukaszh

Re: Limit

No, u-th root of 1/u as u goes to infinity is 1.

4. Apr 24, 2009

### HallsofIvy

Re: Limit

Gosh, I wish the LaTex were working!

5. Apr 24, 2009

### tiny-tim

oh … did you mean

6. Apr 25, 2009

### arildno

Re: Limit

Well, lukaszh:
You can't just take two INDEPENDENTLY VARYING parameters, n and m, and convert them into i single parameter!

To see that let m=n^n.

Then, the n-th root of m equals n, and the limit of this as n goes to infinity, is infinity.

Another example. Let m=a^n, where a is some fixed number.

In this case, the limit equals a!

What you need to do if you want to prove the convergence of some expresssion as the two parameters to infinity, is that that limit holds, IRRESPECTIVE of how n and m individually increases towards infinity!