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

  • Thread starter lukaszh
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  • #1
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Is this true? [tex]n,m\in\mathbb{N}[/tex]
[tex]\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[/tex]
Thanks
 

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  • #2
tiny-tim
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Is this true? for n,m in N:

limn -> ∞√n limm -> ∞1/m = limn -> ∞limm -> ∞√n/m = limu -> ∞√u/u = 1?

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

What were you trying to do? :confused:
 
  • #3
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(in particular limu -> ∞√u/u is obviously 0).

No, u-th root of 1/u as u goes to infinity is 1.
 
  • #4
HallsofIvy
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Gosh, I wish the LaTex were working!
 
  • #5
tiny-tim
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No, u-th root of 1/u as u goes to infinity is 1.

oh :rolleyes: … did you mean
limn -> ∞(limm -> ∞1/m)1/n = limn -> ∞limm -> ∞m1/n = limu -> ∞u1/u = 1?
 
  • #6
arildno
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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!
 

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