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[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]__

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In summary, the conversation discusses the concept of taking the limit of two independent variables, n and m, and converting them into a single variable. However, this approach is incorrect and does not prove the convergence of the expression as the two variables increase towards infinity. The correct method is to ensure that the limit holds regardless of how the variables individually increase towards infinity.

- #1

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[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

Thanks

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- #2

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lukaszh said:Is this true? for n,m in N:

lim_{n -> ∞}√n lim_{m -> ∞}1/m = lim_{n -> ∞}lim_{m -> ∞}√n/m = lim_{u -> ∞}√u/u = 1?

No, that's completely wrong from start to finish (in particular lim

What were you trying to do?

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tiny-tim said:(in particular lim_{u -> ∞}√u/u is obviously 0).

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

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Gosh, I wish the LaTex were working!

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

oh … did you mean

lim_{n -> ∞}(lim_{m -> ∞}1/m)^{1/n}= lim_{n -> ∞}lim_{m -> ∞}m^{1/n}= lim_{u -> ∞}u^{1/u}= 1?

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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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